$a_n=(n-1)a_{n-1}+a_{n-2}$












2














In an answer to a recent question of mine, I was introduced to the recurrence relation $$a_n=(n-1)a_{n-1}+a_{n-2}$$
Where $a_1=0,a_2=1$. I have computational evidence that $a_n=a_{-n}$, as I have computed up to $a_{10}=516901$, and $a_{-10}=516901$. But, so far I don't really see a pattern in the numbers which could help me with my closed form. Of course, there's the patterns one gets by using the general case,
$$a_n=(n-1)bigg[(n-2)a_{n-2}+a_{n-3}bigg]+a_{n-2}$$
$$a_n=(n-1)bigg[(n-2)bigg[(n-3)a_{n-3}+a_{n-4}bigg]+a_{n-3}bigg]+a_{n-2}$$
And so on. These patterns suggest some sort of series of factorials of various types, yet I do not think that going on and on like that would be very productive. As you can see, I am pretty lost.



I heard that one could express this sort of recurrence relation with hypergeometric functions. Does anyone know how to do this? Thanks.



Update:



The OEIS proposes that the relation has the formula
$$a_n=2K_1(2)I_n(-2)+2I_1(2)K_n(2)$$
Where
$$I_alpha(x)=sum_{mgeq0}frac1{m!Gamma(m+alpha+1)}bigg(frac x2bigg)^{2m+alpha}$$
Is the modified Bessel Function of the second kind.



And
$$K_alpha(x)=fracpi2frac{I_{-alpha}(x)-I_alpha(x)}{sinpialpha}$$
Is the modified bessel function of the second kind.



How does one derive this formula?










share|cite|improve this question
























  • Your sequence appears as A001053 in the OEIS. It has some results for the closed form, specifically in terms of Bessel functions and in terms of hypergeometric functions.
    – Semiclassical
    Dec 1 at 0:01










  • @Semiclassical Thanks. The currently posted answer provides the same link. I'll edit the questions to include the closed forms, but I still want to learn how to derive them.
    – clathratus
    Dec 1 at 0:06










  • You can prove $a_n = a_{-n}$ by induction on $n$ straightforwardly (since the recursion yields $a_{-n+2} = left(-n+1right) a_{-n+1} + a_{-n}$, which rewrites as $a_{-n} = left(n-1right) a_{-n+1} + a_{-n+2}$). The same holds for any recursion of the form $a_n = Fleft(n-1right) a_{n-1} + a_{n-2}$, where $F$ is any odd function from $mathbb{Z}$ to an additive abelian group, provided that it holds for $n = 1$.
    – darij grinberg
    Dec 1 at 1:46


















2














In an answer to a recent question of mine, I was introduced to the recurrence relation $$a_n=(n-1)a_{n-1}+a_{n-2}$$
Where $a_1=0,a_2=1$. I have computational evidence that $a_n=a_{-n}$, as I have computed up to $a_{10}=516901$, and $a_{-10}=516901$. But, so far I don't really see a pattern in the numbers which could help me with my closed form. Of course, there's the patterns one gets by using the general case,
$$a_n=(n-1)bigg[(n-2)a_{n-2}+a_{n-3}bigg]+a_{n-2}$$
$$a_n=(n-1)bigg[(n-2)bigg[(n-3)a_{n-3}+a_{n-4}bigg]+a_{n-3}bigg]+a_{n-2}$$
And so on. These patterns suggest some sort of series of factorials of various types, yet I do not think that going on and on like that would be very productive. As you can see, I am pretty lost.



I heard that one could express this sort of recurrence relation with hypergeometric functions. Does anyone know how to do this? Thanks.



Update:



The OEIS proposes that the relation has the formula
$$a_n=2K_1(2)I_n(-2)+2I_1(2)K_n(2)$$
Where
$$I_alpha(x)=sum_{mgeq0}frac1{m!Gamma(m+alpha+1)}bigg(frac x2bigg)^{2m+alpha}$$
Is the modified Bessel Function of the second kind.



And
$$K_alpha(x)=fracpi2frac{I_{-alpha}(x)-I_alpha(x)}{sinpialpha}$$
Is the modified bessel function of the second kind.



How does one derive this formula?










share|cite|improve this question
























  • Your sequence appears as A001053 in the OEIS. It has some results for the closed form, specifically in terms of Bessel functions and in terms of hypergeometric functions.
    – Semiclassical
    Dec 1 at 0:01










  • @Semiclassical Thanks. The currently posted answer provides the same link. I'll edit the questions to include the closed forms, but I still want to learn how to derive them.
    – clathratus
    Dec 1 at 0:06










  • You can prove $a_n = a_{-n}$ by induction on $n$ straightforwardly (since the recursion yields $a_{-n+2} = left(-n+1right) a_{-n+1} + a_{-n}$, which rewrites as $a_{-n} = left(n-1right) a_{-n+1} + a_{-n+2}$). The same holds for any recursion of the form $a_n = Fleft(n-1right) a_{n-1} + a_{n-2}$, where $F$ is any odd function from $mathbb{Z}$ to an additive abelian group, provided that it holds for $n = 1$.
    – darij grinberg
    Dec 1 at 1:46
















2












2








2


1





In an answer to a recent question of mine, I was introduced to the recurrence relation $$a_n=(n-1)a_{n-1}+a_{n-2}$$
Where $a_1=0,a_2=1$. I have computational evidence that $a_n=a_{-n}$, as I have computed up to $a_{10}=516901$, and $a_{-10}=516901$. But, so far I don't really see a pattern in the numbers which could help me with my closed form. Of course, there's the patterns one gets by using the general case,
$$a_n=(n-1)bigg[(n-2)a_{n-2}+a_{n-3}bigg]+a_{n-2}$$
$$a_n=(n-1)bigg[(n-2)bigg[(n-3)a_{n-3}+a_{n-4}bigg]+a_{n-3}bigg]+a_{n-2}$$
And so on. These patterns suggest some sort of series of factorials of various types, yet I do not think that going on and on like that would be very productive. As you can see, I am pretty lost.



I heard that one could express this sort of recurrence relation with hypergeometric functions. Does anyone know how to do this? Thanks.



Update:



The OEIS proposes that the relation has the formula
$$a_n=2K_1(2)I_n(-2)+2I_1(2)K_n(2)$$
Where
$$I_alpha(x)=sum_{mgeq0}frac1{m!Gamma(m+alpha+1)}bigg(frac x2bigg)^{2m+alpha}$$
Is the modified Bessel Function of the second kind.



And
$$K_alpha(x)=fracpi2frac{I_{-alpha}(x)-I_alpha(x)}{sinpialpha}$$
Is the modified bessel function of the second kind.



How does one derive this formula?










share|cite|improve this question















In an answer to a recent question of mine, I was introduced to the recurrence relation $$a_n=(n-1)a_{n-1}+a_{n-2}$$
Where $a_1=0,a_2=1$. I have computational evidence that $a_n=a_{-n}$, as I have computed up to $a_{10}=516901$, and $a_{-10}=516901$. But, so far I don't really see a pattern in the numbers which could help me with my closed form. Of course, there's the patterns one gets by using the general case,
$$a_n=(n-1)bigg[(n-2)a_{n-2}+a_{n-3}bigg]+a_{n-2}$$
$$a_n=(n-1)bigg[(n-2)bigg[(n-3)a_{n-3}+a_{n-4}bigg]+a_{n-3}bigg]+a_{n-2}$$
And so on. These patterns suggest some sort of series of factorials of various types, yet I do not think that going on and on like that would be very productive. As you can see, I am pretty lost.



I heard that one could express this sort of recurrence relation with hypergeometric functions. Does anyone know how to do this? Thanks.



Update:



The OEIS proposes that the relation has the formula
$$a_n=2K_1(2)I_n(-2)+2I_1(2)K_n(2)$$
Where
$$I_alpha(x)=sum_{mgeq0}frac1{m!Gamma(m+alpha+1)}bigg(frac x2bigg)^{2m+alpha}$$
Is the modified Bessel Function of the second kind.



And
$$K_alpha(x)=fracpi2frac{I_{-alpha}(x)-I_alpha(x)}{sinpialpha}$$
Is the modified bessel function of the second kind.



How does one derive this formula?







sequences-and-series combinatorics recurrence-relations closed-form hypergeometric-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 at 0:41

























asked Nov 30 at 23:49









clathratus

2,947328




2,947328












  • Your sequence appears as A001053 in the OEIS. It has some results for the closed form, specifically in terms of Bessel functions and in terms of hypergeometric functions.
    – Semiclassical
    Dec 1 at 0:01










  • @Semiclassical Thanks. The currently posted answer provides the same link. I'll edit the questions to include the closed forms, but I still want to learn how to derive them.
    – clathratus
    Dec 1 at 0:06










  • You can prove $a_n = a_{-n}$ by induction on $n$ straightforwardly (since the recursion yields $a_{-n+2} = left(-n+1right) a_{-n+1} + a_{-n}$, which rewrites as $a_{-n} = left(n-1right) a_{-n+1} + a_{-n+2}$). The same holds for any recursion of the form $a_n = Fleft(n-1right) a_{n-1} + a_{n-2}$, where $F$ is any odd function from $mathbb{Z}$ to an additive abelian group, provided that it holds for $n = 1$.
    – darij grinberg
    Dec 1 at 1:46




















  • Your sequence appears as A001053 in the OEIS. It has some results for the closed form, specifically in terms of Bessel functions and in terms of hypergeometric functions.
    – Semiclassical
    Dec 1 at 0:01










  • @Semiclassical Thanks. The currently posted answer provides the same link. I'll edit the questions to include the closed forms, but I still want to learn how to derive them.
    – clathratus
    Dec 1 at 0:06










  • You can prove $a_n = a_{-n}$ by induction on $n$ straightforwardly (since the recursion yields $a_{-n+2} = left(-n+1right) a_{-n+1} + a_{-n}$, which rewrites as $a_{-n} = left(n-1right) a_{-n+1} + a_{-n+2}$). The same holds for any recursion of the form $a_n = Fleft(n-1right) a_{n-1} + a_{n-2}$, where $F$ is any odd function from $mathbb{Z}$ to an additive abelian group, provided that it holds for $n = 1$.
    – darij grinberg
    Dec 1 at 1:46


















Your sequence appears as A001053 in the OEIS. It has some results for the closed form, specifically in terms of Bessel functions and in terms of hypergeometric functions.
– Semiclassical
Dec 1 at 0:01




Your sequence appears as A001053 in the OEIS. It has some results for the closed form, specifically in terms of Bessel functions and in terms of hypergeometric functions.
– Semiclassical
Dec 1 at 0:01












@Semiclassical Thanks. The currently posted answer provides the same link. I'll edit the questions to include the closed forms, but I still want to learn how to derive them.
– clathratus
Dec 1 at 0:06




@Semiclassical Thanks. The currently posted answer provides the same link. I'll edit the questions to include the closed forms, but I still want to learn how to derive them.
– clathratus
Dec 1 at 0:06












You can prove $a_n = a_{-n}$ by induction on $n$ straightforwardly (since the recursion yields $a_{-n+2} = left(-n+1right) a_{-n+1} + a_{-n}$, which rewrites as $a_{-n} = left(n-1right) a_{-n+1} + a_{-n+2}$). The same holds for any recursion of the form $a_n = Fleft(n-1right) a_{n-1} + a_{n-2}$, where $F$ is any odd function from $mathbb{Z}$ to an additive abelian group, provided that it holds for $n = 1$.
– darij grinberg
Dec 1 at 1:46






You can prove $a_n = a_{-n}$ by induction on $n$ straightforwardly (since the recursion yields $a_{-n+2} = left(-n+1right) a_{-n+1} + a_{-n}$, which rewrites as $a_{-n} = left(n-1right) a_{-n+1} + a_{-n+2}$). The same holds for any recursion of the form $a_n = Fleft(n-1right) a_{n-1} + a_{n-2}$, where $F$ is any odd function from $mathbb{Z}$ to an additive abelian group, provided that it holds for $n = 1$.
– darij grinberg
Dec 1 at 1:46












2 Answers
2






active

oldest

votes


















2














The modified Bessel functions of the first kind have the recurrence relation



$$
frac{2alpha}{x}C_{alpha}(x) = C_{alpha - 1}(x) - C_{alpha + 1}(x) tag{1}
$$



where $C_alpha$ denotes $I_alpha$, $e^{alpha i pi} K_alpha$. Take $alpha = n-1$ and $x = 2$



$$
C_{n} (2) = -(n - 1)C_{n-1}(2) + C_{n-2}(2) tag{2}
$$



And you can see from here that this looks already looks a lot like your problem. For example, if you take $I_n$, then you have



$$
I_n(2) = -(n - 1)I_{n-1}(2) + I_{n-2}(2) tag{3}
$$



The problem is the sign in the first term, but that one is easy to figure out, because the $I_n$ has the parity of $n$. So if $n$ is even, $n$ is odd and you can move the minus to the argument. If $n$ is odd, $n-1$ is even, and you can multiply by $-1$ everything and move the sign to the argument again. You can repeat the same analysis for $K$ and reach the conclusion that



$$
a_n = c_1 I_n(-2) + c_2 K_n(2) tag{4}
$$



The constants $c_1$ and $c_2$ you can find with the seeds $a_1 = 0$, $a_2 = 1$






share|cite|improve this answer





















  • Perfect! thank you very much.
    – clathratus
    Dec 1 at 0:42



















1














Hint to show $a_n=a_{-n}$: Let $b_n=a_{-n}$, and see that $b_0=1$, $b_1=0$, $b_2=2$. Now



$$a_n=(n-1)a_{n-1}+a_{n-2}$$



Substituting $n=-k+2$,



$$a_{-k+2}=(-k+1)a_{-k+1}+a_{-k}$$



$$b_{k-2}=(-k+1)b_{k-1}+b_k$$



$$b_k=(k-1)b_{k-1}+b_{k-2}.$$



So, $b_n$ has the same first recurrence relation as $a_n$ and it also has the same first two terms, so we can show (by induction) that $b_n=a_n$ for all $n$.



Finding a closed form is likely to be difficult: for example, the OEIS entry has no such form.






share|cite|improve this answer





















  • Thanks for the proof. It makes sense. (+1)
    – clathratus
    Dec 1 at 0:01













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020826%2fa-n-n-1a-n-1a-n-2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














The modified Bessel functions of the first kind have the recurrence relation



$$
frac{2alpha}{x}C_{alpha}(x) = C_{alpha - 1}(x) - C_{alpha + 1}(x) tag{1}
$$



where $C_alpha$ denotes $I_alpha$, $e^{alpha i pi} K_alpha$. Take $alpha = n-1$ and $x = 2$



$$
C_{n} (2) = -(n - 1)C_{n-1}(2) + C_{n-2}(2) tag{2}
$$



And you can see from here that this looks already looks a lot like your problem. For example, if you take $I_n$, then you have



$$
I_n(2) = -(n - 1)I_{n-1}(2) + I_{n-2}(2) tag{3}
$$



The problem is the sign in the first term, but that one is easy to figure out, because the $I_n$ has the parity of $n$. So if $n$ is even, $n$ is odd and you can move the minus to the argument. If $n$ is odd, $n-1$ is even, and you can multiply by $-1$ everything and move the sign to the argument again. You can repeat the same analysis for $K$ and reach the conclusion that



$$
a_n = c_1 I_n(-2) + c_2 K_n(2) tag{4}
$$



The constants $c_1$ and $c_2$ you can find with the seeds $a_1 = 0$, $a_2 = 1$






share|cite|improve this answer





















  • Perfect! thank you very much.
    – clathratus
    Dec 1 at 0:42
















2














The modified Bessel functions of the first kind have the recurrence relation



$$
frac{2alpha}{x}C_{alpha}(x) = C_{alpha - 1}(x) - C_{alpha + 1}(x) tag{1}
$$



where $C_alpha$ denotes $I_alpha$, $e^{alpha i pi} K_alpha$. Take $alpha = n-1$ and $x = 2$



$$
C_{n} (2) = -(n - 1)C_{n-1}(2) + C_{n-2}(2) tag{2}
$$



And you can see from here that this looks already looks a lot like your problem. For example, if you take $I_n$, then you have



$$
I_n(2) = -(n - 1)I_{n-1}(2) + I_{n-2}(2) tag{3}
$$



The problem is the sign in the first term, but that one is easy to figure out, because the $I_n$ has the parity of $n$. So if $n$ is even, $n$ is odd and you can move the minus to the argument. If $n$ is odd, $n-1$ is even, and you can multiply by $-1$ everything and move the sign to the argument again. You can repeat the same analysis for $K$ and reach the conclusion that



$$
a_n = c_1 I_n(-2) + c_2 K_n(2) tag{4}
$$



The constants $c_1$ and $c_2$ you can find with the seeds $a_1 = 0$, $a_2 = 1$






share|cite|improve this answer





















  • Perfect! thank you very much.
    – clathratus
    Dec 1 at 0:42














2












2








2






The modified Bessel functions of the first kind have the recurrence relation



$$
frac{2alpha}{x}C_{alpha}(x) = C_{alpha - 1}(x) - C_{alpha + 1}(x) tag{1}
$$



where $C_alpha$ denotes $I_alpha$, $e^{alpha i pi} K_alpha$. Take $alpha = n-1$ and $x = 2$



$$
C_{n} (2) = -(n - 1)C_{n-1}(2) + C_{n-2}(2) tag{2}
$$



And you can see from here that this looks already looks a lot like your problem. For example, if you take $I_n$, then you have



$$
I_n(2) = -(n - 1)I_{n-1}(2) + I_{n-2}(2) tag{3}
$$



The problem is the sign in the first term, but that one is easy to figure out, because the $I_n$ has the parity of $n$. So if $n$ is even, $n$ is odd and you can move the minus to the argument. If $n$ is odd, $n-1$ is even, and you can multiply by $-1$ everything and move the sign to the argument again. You can repeat the same analysis for $K$ and reach the conclusion that



$$
a_n = c_1 I_n(-2) + c_2 K_n(2) tag{4}
$$



The constants $c_1$ and $c_2$ you can find with the seeds $a_1 = 0$, $a_2 = 1$






share|cite|improve this answer












The modified Bessel functions of the first kind have the recurrence relation



$$
frac{2alpha}{x}C_{alpha}(x) = C_{alpha - 1}(x) - C_{alpha + 1}(x) tag{1}
$$



where $C_alpha$ denotes $I_alpha$, $e^{alpha i pi} K_alpha$. Take $alpha = n-1$ and $x = 2$



$$
C_{n} (2) = -(n - 1)C_{n-1}(2) + C_{n-2}(2) tag{2}
$$



And you can see from here that this looks already looks a lot like your problem. For example, if you take $I_n$, then you have



$$
I_n(2) = -(n - 1)I_{n-1}(2) + I_{n-2}(2) tag{3}
$$



The problem is the sign in the first term, but that one is easy to figure out, because the $I_n$ has the parity of $n$. So if $n$ is even, $n$ is odd and you can move the minus to the argument. If $n$ is odd, $n-1$ is even, and you can multiply by $-1$ everything and move the sign to the argument again. You can repeat the same analysis for $K$ and reach the conclusion that



$$
a_n = c_1 I_n(-2) + c_2 K_n(2) tag{4}
$$



The constants $c_1$ and $c_2$ you can find with the seeds $a_1 = 0$, $a_2 = 1$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 at 0:39









caverac

13.1k21029




13.1k21029












  • Perfect! thank you very much.
    – clathratus
    Dec 1 at 0:42


















  • Perfect! thank you very much.
    – clathratus
    Dec 1 at 0:42
















Perfect! thank you very much.
– clathratus
Dec 1 at 0:42




Perfect! thank you very much.
– clathratus
Dec 1 at 0:42











1














Hint to show $a_n=a_{-n}$: Let $b_n=a_{-n}$, and see that $b_0=1$, $b_1=0$, $b_2=2$. Now



$$a_n=(n-1)a_{n-1}+a_{n-2}$$



Substituting $n=-k+2$,



$$a_{-k+2}=(-k+1)a_{-k+1}+a_{-k}$$



$$b_{k-2}=(-k+1)b_{k-1}+b_k$$



$$b_k=(k-1)b_{k-1}+b_{k-2}.$$



So, $b_n$ has the same first recurrence relation as $a_n$ and it also has the same first two terms, so we can show (by induction) that $b_n=a_n$ for all $n$.



Finding a closed form is likely to be difficult: for example, the OEIS entry has no such form.






share|cite|improve this answer





















  • Thanks for the proof. It makes sense. (+1)
    – clathratus
    Dec 1 at 0:01


















1














Hint to show $a_n=a_{-n}$: Let $b_n=a_{-n}$, and see that $b_0=1$, $b_1=0$, $b_2=2$. Now



$$a_n=(n-1)a_{n-1}+a_{n-2}$$



Substituting $n=-k+2$,



$$a_{-k+2}=(-k+1)a_{-k+1}+a_{-k}$$



$$b_{k-2}=(-k+1)b_{k-1}+b_k$$



$$b_k=(k-1)b_{k-1}+b_{k-2}.$$



So, $b_n$ has the same first recurrence relation as $a_n$ and it also has the same first two terms, so we can show (by induction) that $b_n=a_n$ for all $n$.



Finding a closed form is likely to be difficult: for example, the OEIS entry has no such form.






share|cite|improve this answer





















  • Thanks for the proof. It makes sense. (+1)
    – clathratus
    Dec 1 at 0:01
















1












1








1






Hint to show $a_n=a_{-n}$: Let $b_n=a_{-n}$, and see that $b_0=1$, $b_1=0$, $b_2=2$. Now



$$a_n=(n-1)a_{n-1}+a_{n-2}$$



Substituting $n=-k+2$,



$$a_{-k+2}=(-k+1)a_{-k+1}+a_{-k}$$



$$b_{k-2}=(-k+1)b_{k-1}+b_k$$



$$b_k=(k-1)b_{k-1}+b_{k-2}.$$



So, $b_n$ has the same first recurrence relation as $a_n$ and it also has the same first two terms, so we can show (by induction) that $b_n=a_n$ for all $n$.



Finding a closed form is likely to be difficult: for example, the OEIS entry has no such form.






share|cite|improve this answer












Hint to show $a_n=a_{-n}$: Let $b_n=a_{-n}$, and see that $b_0=1$, $b_1=0$, $b_2=2$. Now



$$a_n=(n-1)a_{n-1}+a_{n-2}$$



Substituting $n=-k+2$,



$$a_{-k+2}=(-k+1)a_{-k+1}+a_{-k}$$



$$b_{k-2}=(-k+1)b_{k-1}+b_k$$



$$b_k=(k-1)b_{k-1}+b_{k-2}.$$



So, $b_n$ has the same first recurrence relation as $a_n$ and it also has the same first two terms, so we can show (by induction) that $b_n=a_n$ for all $n$.



Finding a closed form is likely to be difficult: for example, the OEIS entry has no such form.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 at 23:58









Carl Schildkraut

11.1k11441




11.1k11441












  • Thanks for the proof. It makes sense. (+1)
    – clathratus
    Dec 1 at 0:01




















  • Thanks for the proof. It makes sense. (+1)
    – clathratus
    Dec 1 at 0:01


















Thanks for the proof. It makes sense. (+1)
– clathratus
Dec 1 at 0:01






Thanks for the proof. It makes sense. (+1)
– clathratus
Dec 1 at 0:01




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020826%2fa-n-n-1a-n-1a-n-2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Berounka

Sphinx de Gizeh

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...