Table $ 2n-1 $ from $ n=2 $ to $ 50 $ except when $ n=3m-1 $ [duplicate]











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How to Table $ 2n-1 $ for $ n $ from $ 2 $ to $ 50 $, except when $ n=3m-1 $ for another integer $ m $?










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    How to Table $ 2n-1 $ for $ n $ from $ 2 $ to $ 50 $, except when $ n=3m-1 $ for another integer $ m $?










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      How to Table $ 2n-1 $ for $ n $ from $ 2 $ to $ 50 $, except when $ n=3m-1 $ for another integer $ m $?










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      How to Table $ 2n-1 $ for $ n $ from $ 2 $ to $ 50 $, except when $ n=3m-1 $ for another integer $ m $?





      This question already has an answer here:




      • How to generate a table with i != j

        6 answers








      list-manipulation table






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      edited Nov 22 at 12:12









      Αλέξανδρος Ζεγγ

      3,5281927




      3,5281927






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      asked Nov 22 at 6:47









      Amr

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          4 Answers
          4






          active

          oldest

          votes

















          up vote
          7
          down vote













          Using Drop, Riffle and Complement:



          Drop[Range[5, 2 50 - 1, 2], {3, -1, 3} ]



          {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
          55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




          Riffle[#, # + 2] &[2 Range[3, 50, 3] - 1]



          same result




          2 Complement[#, #[[;; ;; 3]]] & @ Range[2, 50] - 1;



          same result




          All three are faster than Pick+ Unitize combination from Henrik's answer:



          nmax = 1000000;

          e0 = Drop[Range[5, 2 nmax - 1, 2], {3, -1, 3} ]; // AbsoluteTiming



          0.0102131




          e1 = Riffle[#, # + 2] &[2 Range[3, nmax, 3] - 1]; // AbsoluteTiming // First



          0.0126898




          e2 = 2 Complement[#, #[[;; ;; 3]]] & @ Range[2, nmax] - 1; //  AbsoluteTiming // First



          0.0354908




          versus Henrik's



          d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First



          0.0526116




          and chyanog's method:



          (k = Floor[(2 nmax + 1)/3] - 1; 
          res = Range[4, 3 k + 1, 3] + BitAnd[Range[k], 1];) // AbsoluteTiming // First



          0.0125891




          e0 == e1 == e2 == d == res



          True







          share|improve this answer























          • Very clever! +1
            – ciao
            Nov 22 at 9:30


















          up vote
          5
          down vote













          The condition to avoid can be rewritten using Mod, hence:



          Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, 50}]


          Or you can use Map /@ instead of Table:



          If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, 50]


          (Thanks to Lukas Lang for the easier use of Nothing).






          share|improve this answer



















          • 1




            You could also return Nothing in the else case to remove the unwanted elements directly
            – Lukas Lang
            Nov 22 at 7:03


















          up vote
          4
          down vote













          An idiomatic way with Sow and Reap:



          Reap[
          Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, 50}]
          ][[2, 1]]



          {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
          55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




          A harder to read, vectorized way: First create the list and pick the valid elements with Pick.



          Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, 50]



          {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
          55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




          The vectorized version is about 40 times faster:



          nmax = 1000000;
          a = Reap[Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, nmax}]][[2, 1]]; // AbsoluteTiming // First
          b = Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, nmax}]; // AbsoluteTiming // First
          c = If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, nmax]; // AbsoluteTiming // First
          d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First
          a == b == c == d



          0.943828



          0.907419



          1.34615



          0.018833



          True







          share|improve this answer






























            up vote
            2
            down vote













            More efficient way



            Clear["f*"];
            f1[m_] := Module[{r = Range[4, 2 m - Mod[m, 3], 3]}, r[[1 ;; ;; 2]] += 1; r];
            f2[m_] := Drop[Range[5, 2 m - 1, 2], {3, -1, 3}];

            r1 = f1[10^7]; // RepeatedTiming
            r2 = f2[10^7]; // RepeatedTiming

            r1 == r2



            {0.11, Null}



            {0.14, Null}



            True







            share|improve this answer






























              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              7
              down vote













              Using Drop, Riffle and Complement:



              Drop[Range[5, 2 50 - 1, 2], {3, -1, 3} ]



              {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
              55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




              Riffle[#, # + 2] &[2 Range[3, 50, 3] - 1]



              same result




              2 Complement[#, #[[;; ;; 3]]] & @ Range[2, 50] - 1;



              same result




              All three are faster than Pick+ Unitize combination from Henrik's answer:



              nmax = 1000000;

              e0 = Drop[Range[5, 2 nmax - 1, 2], {3, -1, 3} ]; // AbsoluteTiming



              0.0102131




              e1 = Riffle[#, # + 2] &[2 Range[3, nmax, 3] - 1]; // AbsoluteTiming // First



              0.0126898




              e2 = 2 Complement[#, #[[;; ;; 3]]] & @ Range[2, nmax] - 1; //  AbsoluteTiming // First



              0.0354908




              versus Henrik's



              d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First



              0.0526116




              and chyanog's method:



              (k = Floor[(2 nmax + 1)/3] - 1; 
              res = Range[4, 3 k + 1, 3] + BitAnd[Range[k], 1];) // AbsoluteTiming // First



              0.0125891




              e0 == e1 == e2 == d == res



              True







              share|improve this answer























              • Very clever! +1
                – ciao
                Nov 22 at 9:30















              up vote
              7
              down vote













              Using Drop, Riffle and Complement:



              Drop[Range[5, 2 50 - 1, 2], {3, -1, 3} ]



              {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
              55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




              Riffle[#, # + 2] &[2 Range[3, 50, 3] - 1]



              same result




              2 Complement[#, #[[;; ;; 3]]] & @ Range[2, 50] - 1;



              same result




              All three are faster than Pick+ Unitize combination from Henrik's answer:



              nmax = 1000000;

              e0 = Drop[Range[5, 2 nmax - 1, 2], {3, -1, 3} ]; // AbsoluteTiming



              0.0102131




              e1 = Riffle[#, # + 2] &[2 Range[3, nmax, 3] - 1]; // AbsoluteTiming // First



              0.0126898




              e2 = 2 Complement[#, #[[;; ;; 3]]] & @ Range[2, nmax] - 1; //  AbsoluteTiming // First



              0.0354908




              versus Henrik's



              d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First



              0.0526116




              and chyanog's method:



              (k = Floor[(2 nmax + 1)/3] - 1; 
              res = Range[4, 3 k + 1, 3] + BitAnd[Range[k], 1];) // AbsoluteTiming // First



              0.0125891




              e0 == e1 == e2 == d == res



              True







              share|improve this answer























              • Very clever! +1
                – ciao
                Nov 22 at 9:30













              up vote
              7
              down vote










              up vote
              7
              down vote









              Using Drop, Riffle and Complement:



              Drop[Range[5, 2 50 - 1, 2], {3, -1, 3} ]



              {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
              55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




              Riffle[#, # + 2] &[2 Range[3, 50, 3] - 1]



              same result




              2 Complement[#, #[[;; ;; 3]]] & @ Range[2, 50] - 1;



              same result




              All three are faster than Pick+ Unitize combination from Henrik's answer:



              nmax = 1000000;

              e0 = Drop[Range[5, 2 nmax - 1, 2], {3, -1, 3} ]; // AbsoluteTiming



              0.0102131




              e1 = Riffle[#, # + 2] &[2 Range[3, nmax, 3] - 1]; // AbsoluteTiming // First



              0.0126898




              e2 = 2 Complement[#, #[[;; ;; 3]]] & @ Range[2, nmax] - 1; //  AbsoluteTiming // First



              0.0354908




              versus Henrik's



              d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First



              0.0526116




              and chyanog's method:



              (k = Floor[(2 nmax + 1)/3] - 1; 
              res = Range[4, 3 k + 1, 3] + BitAnd[Range[k], 1];) // AbsoluteTiming // First



              0.0125891




              e0 == e1 == e2 == d == res



              True







              share|improve this answer














              Using Drop, Riffle and Complement:



              Drop[Range[5, 2 50 - 1, 2], {3, -1, 3} ]



              {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
              55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




              Riffle[#, # + 2] &[2 Range[3, 50, 3] - 1]



              same result




              2 Complement[#, #[[;; ;; 3]]] & @ Range[2, 50] - 1;



              same result




              All three are faster than Pick+ Unitize combination from Henrik's answer:



              nmax = 1000000;

              e0 = Drop[Range[5, 2 nmax - 1, 2], {3, -1, 3} ]; // AbsoluteTiming



              0.0102131




              e1 = Riffle[#, # + 2] &[2 Range[3, nmax, 3] - 1]; // AbsoluteTiming // First



              0.0126898




              e2 = 2 Complement[#, #[[;; ;; 3]]] & @ Range[2, nmax] - 1; //  AbsoluteTiming // First



              0.0354908




              versus Henrik's



              d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First



              0.0526116




              and chyanog's method:



              (k = Floor[(2 nmax + 1)/3] - 1; 
              res = Range[4, 3 k + 1, 3] + BitAnd[Range[k], 1];) // AbsoluteTiming // First



              0.0125891




              e0 == e1 == e2 == d == res



              True








              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Nov 22 at 13:45

























              answered Nov 22 at 8:58









              kglr

              173k8194400




              173k8194400












              • Very clever! +1
                – ciao
                Nov 22 at 9:30


















              • Very clever! +1
                – ciao
                Nov 22 at 9:30
















              Very clever! +1
              – ciao
              Nov 22 at 9:30




              Very clever! +1
              – ciao
              Nov 22 at 9:30










              up vote
              5
              down vote













              The condition to avoid can be rewritten using Mod, hence:



              Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, 50}]


              Or you can use Map /@ instead of Table:



              If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, 50]


              (Thanks to Lukas Lang for the easier use of Nothing).






              share|improve this answer



















              • 1




                You could also return Nothing in the else case to remove the unwanted elements directly
                – Lukas Lang
                Nov 22 at 7:03















              up vote
              5
              down vote













              The condition to avoid can be rewritten using Mod, hence:



              Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, 50}]


              Or you can use Map /@ instead of Table:



              If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, 50]


              (Thanks to Lukas Lang for the easier use of Nothing).






              share|improve this answer



















              • 1




                You could also return Nothing in the else case to remove the unwanted elements directly
                – Lukas Lang
                Nov 22 at 7:03













              up vote
              5
              down vote










              up vote
              5
              down vote









              The condition to avoid can be rewritten using Mod, hence:



              Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, 50}]


              Or you can use Map /@ instead of Table:



              If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, 50]


              (Thanks to Lukas Lang for the easier use of Nothing).






              share|improve this answer














              The condition to avoid can be rewritten using Mod, hence:



              Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, 50}]


              Or you can use Map /@ instead of Table:



              If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, 50]


              (Thanks to Lukas Lang for the easier use of Nothing).







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Nov 22 at 14:52

























              answered Nov 22 at 6:56









              bill s

              52.4k375149




              52.4k375149








              • 1




                You could also return Nothing in the else case to remove the unwanted elements directly
                – Lukas Lang
                Nov 22 at 7:03














              • 1




                You could also return Nothing in the else case to remove the unwanted elements directly
                – Lukas Lang
                Nov 22 at 7:03








              1




              1




              You could also return Nothing in the else case to remove the unwanted elements directly
              – Lukas Lang
              Nov 22 at 7:03




              You could also return Nothing in the else case to remove the unwanted elements directly
              – Lukas Lang
              Nov 22 at 7:03










              up vote
              4
              down vote













              An idiomatic way with Sow and Reap:



              Reap[
              Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, 50}]
              ][[2, 1]]



              {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
              55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




              A harder to read, vectorized way: First create the list and pick the valid elements with Pick.



              Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, 50]



              {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
              55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




              The vectorized version is about 40 times faster:



              nmax = 1000000;
              a = Reap[Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, nmax}]][[2, 1]]; // AbsoluteTiming // First
              b = Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, nmax}]; // AbsoluteTiming // First
              c = If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, nmax]; // AbsoluteTiming // First
              d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First
              a == b == c == d



              0.943828



              0.907419



              1.34615



              0.018833



              True







              share|improve this answer



























                up vote
                4
                down vote













                An idiomatic way with Sow and Reap:



                Reap[
                Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, 50}]
                ][[2, 1]]



                {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
                55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




                A harder to read, vectorized way: First create the list and pick the valid elements with Pick.



                Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, 50]



                {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
                55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




                The vectorized version is about 40 times faster:



                nmax = 1000000;
                a = Reap[Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, nmax}]][[2, 1]]; // AbsoluteTiming // First
                b = Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, nmax}]; // AbsoluteTiming // First
                c = If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, nmax]; // AbsoluteTiming // First
                d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First
                a == b == c == d



                0.943828



                0.907419



                1.34615



                0.018833



                True







                share|improve this answer

























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  An idiomatic way with Sow and Reap:



                  Reap[
                  Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, 50}]
                  ][[2, 1]]



                  {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
                  55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




                  A harder to read, vectorized way: First create the list and pick the valid elements with Pick.



                  Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, 50]



                  {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
                  55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




                  The vectorized version is about 40 times faster:



                  nmax = 1000000;
                  a = Reap[Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, nmax}]][[2, 1]]; // AbsoluteTiming // First
                  b = Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, nmax}]; // AbsoluteTiming // First
                  c = If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, nmax]; // AbsoluteTiming // First
                  d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First
                  a == b == c == d



                  0.943828



                  0.907419



                  1.34615



                  0.018833



                  True







                  share|improve this answer














                  An idiomatic way with Sow and Reap:



                  Reap[
                  Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, 50}]
                  ][[2, 1]]



                  {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
                  55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




                  A harder to read, vectorized way: First create the list and pick the valid elements with Pick.



                  Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, 50]



                  {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53,
                  55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97}




                  The vectorized version is about 40 times faster:



                  nmax = 1000000;
                  a = Reap[Do[If[Mod[n, 3] != 2, Sow[2 n - 1]], {n, 2, nmax}]][[2, 1]]; // AbsoluteTiming // First
                  b = Table[If[Mod[n, 3] != 2, 2 n - 1, Nothing], {n, 2, nmax}]; // AbsoluteTiming // First
                  c = If[Mod[#, 3] != 2, 2 # - 1, Nothing] & /@ Range[2, nmax]; // AbsoluteTiming // First
                  d = Pick[2 # - 1, Unitize[Mod[#, 3] - 2], 1] &@Range[2, nmax]; // AbsoluteTiming // First
                  a == b == c == d



                  0.943828



                  0.907419



                  1.34615



                  0.018833



                  True








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 22 at 15:58

























                  answered Nov 22 at 7:28









                  Henrik Schumacher

                  45.6k466132




                  45.6k466132






















                      up vote
                      2
                      down vote













                      More efficient way



                      Clear["f*"];
                      f1[m_] := Module[{r = Range[4, 2 m - Mod[m, 3], 3]}, r[[1 ;; ;; 2]] += 1; r];
                      f2[m_] := Drop[Range[5, 2 m - 1, 2], {3, -1, 3}];

                      r1 = f1[10^7]; // RepeatedTiming
                      r2 = f2[10^7]; // RepeatedTiming

                      r1 == r2



                      {0.11, Null}



                      {0.14, Null}



                      True







                      share|improve this answer



























                        up vote
                        2
                        down vote













                        More efficient way



                        Clear["f*"];
                        f1[m_] := Module[{r = Range[4, 2 m - Mod[m, 3], 3]}, r[[1 ;; ;; 2]] += 1; r];
                        f2[m_] := Drop[Range[5, 2 m - 1, 2], {3, -1, 3}];

                        r1 = f1[10^7]; // RepeatedTiming
                        r2 = f2[10^7]; // RepeatedTiming

                        r1 == r2



                        {0.11, Null}



                        {0.14, Null}



                        True







                        share|improve this answer

























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          More efficient way



                          Clear["f*"];
                          f1[m_] := Module[{r = Range[4, 2 m - Mod[m, 3], 3]}, r[[1 ;; ;; 2]] += 1; r];
                          f2[m_] := Drop[Range[5, 2 m - 1, 2], {3, -1, 3}];

                          r1 = f1[10^7]; // RepeatedTiming
                          r2 = f2[10^7]; // RepeatedTiming

                          r1 == r2



                          {0.11, Null}



                          {0.14, Null}



                          True







                          share|improve this answer














                          More efficient way



                          Clear["f*"];
                          f1[m_] := Module[{r = Range[4, 2 m - Mod[m, 3], 3]}, r[[1 ;; ;; 2]] += 1; r];
                          f2[m_] := Drop[Range[5, 2 m - 1, 2], {3, -1, 3}];

                          r1 = f1[10^7]; // RepeatedTiming
                          r2 = f2[10^7]; // RepeatedTiming

                          r1 == r2



                          {0.11, Null}



                          {0.14, Null}



                          True








                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Nov 23 at 3:23

























                          answered Nov 22 at 11:25









                          chyanog

                          6,80921546




                          6,80921546















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