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Let $$f_n=\sum_{k=0}^{n}\binom{n}{k}^3,$$ Franel showed that $$(n+1)^2f_{n+1}=(7n^2+7n+2)f_n+8n^2f_{n-1}$$ Now, my question is if i am able to show that $$Af_{n+1}=Bf_n+Cf_{n-1}+Df_{n-2}$$ for some integers $A, B, C$ and $D$, would this be a publishable result?

Update: based on some answers given below, i would like to reveal the expressions for $A,B,C$ and $D$ $$A=n(3n-1)(n+1)^2$$ $$B=n(9n^3-6n^2-9n-2)$$ $$C=108n^4+48n^3-60n^2-16n+16$$ $$D=32(3n^2+5n+2)(n-1)^2$$

Is this too complex to be taken seriously?

Shuaib Lateef
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  • One might have supposed that if such a relation were true it would long since have been found. Have you tested it for small values of $n$? – Oscar Lanzi Aug 20 '22 at 09:47
  • @OscarLanzi I have actually proven it to be true. I just want to know if i won't be wasting my time preparing a paper for it. – Shuaib Lateef Aug 20 '22 at 10:01
  • @RobertZ No, they are not They are expressed in terms of $n$ just like Franel formula. – Shuaib Lateef Aug 20 '22 at 10:02
  • Have you decided where to publish? You could write the paper and send it to a journal. They will review it and if the formula is known might point you to where to find it. If it's not and the result is considered worthwhile, it will be published. – Raskolnikov Aug 20 '22 at 12:50

1 Answers1

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Generally one would like to have a linear recurrence of minimal length. The known one has length $2$ and yours has length $3$.

Moreover, in order to get it published, your recurrence should not be implied by the known one. Have you checked that?

For instance, by shifting the index we have $$n^2f_{n}=(7(n-1)^2+7(n-1)+2)f_{n-1}+8(n-1)^2f_{n-2}$$ and therefore $$(n+1)^2f_{n+1}=(7n^2+7n+2)f_n+8n^2f_{n-1}\\= (7n+2)f_n+8n^2f_{n-1}+7((7(n-1)^2+7(n-1)+2)f_{n-1}+8(n-1)^2f_{n-2})$$ which is of the form $Af_{n+1}=Bf_n+Cf_{n-1}+Df_{n-2}$ where $A,B,C,D$ are integers (polynomials in $n$ of degree at most two).

Robert Z
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