100 and more conjectures from the OEIS

Abstract

' This book is a sequel to Ralf Stephan's Prove or Disprove. 100 Conjectures from the OEIS. It contains a collection of conjectures of varying difficulty from easy to difficult from The On-Line Encyclopedia of Integer Sequences (OEIS), a database of integer sequences by Neil Sloane. If the reader wishes, he might help solving one or the other of these conjectures and thus improve the database.

Introduction

The OEIS is a database containing initial terms, along with description, formulae and references of over 100,000 sequences. It was started in teh 1960's and is still maintained by Neil Sloane. Many of the sequences concern unsolved problems and conjectures. The purpose of this book is to find solutions of some of the riddles.

Here we go.

The Problems

Problem 1

A002421	Expansion of $(1-4x{)}^{(3/2)}$.
$(1-4x{)}^{3/2}=1-6x+\sum _{n>1}\frac{12(2n-4)!}{n!(n-2)!}{x}^n.$
Proof by Ralf Stephan, 2004-Sep-27

Problem 2

A071721	Expansion of $(1+x^2*C^2)*C^2$, where $C=(1-(1-4*x{)}^{(1/2)})/(2*x)$ is g.f. for Catalan numbers,A000108.
$(1+x^{2}C(x{)}^2)C(x{)}^2=\sum _{n\ge 0}\frac{6n(2n)!}{n!(n+1)!(n+2)}{x}^n,C(x)=\frac{1-\sqrt{1-4x}}{2x}.$
Solution through $[x^n](-1+x)\sqrt{1-4x}=-C(2n,n)/(1-2n)+C(2n,n)/(3-2n)$. -- Ralf Stephan, 2004-Sep-27

Problem 3

A004130	Numerators in expansion of $(1-x{)}^{-1/4}$
Please show how to get the reduced numerator/denominator in: $(1-x{)}^{1/4}=$$\sum _{n\ge 0}\frac{\prod _{k=1}^n(5-4k)2^{v_2(k)}/k}{2^{3n-e_1(n)}}{x}^n.$
Proved by Mitch Harris, 2004-Oct-11

Problem 4

A001800	Coefficients of Legendre polynomials.
$[x^n]P_{n+2}(x)=\frac{1}{2^{n+2}}(n+1)(\genfrac{}{}{0ex}{}{2n+2}{n+1}),$($P_n$ the Legendre polynomials)
Equation 28 of Mathworld's Legendre Polynomial. --researched by Mitch Harris, 2004-Oct-12

Problem 5

A077071	Row sums of triangle T(n,k) in A077070.
$\sum _{k=0}^n-v_2([x^{2k}]P_{2n}(x))=2n^2+2n-2\sum _{i=0}^n{e}_1(i).$
Proved by Nikolaus Meyberg, 2004-Oct-15

Problem 6

A001896	Numerators of cosecant numbers $-2*(2^{2*n-1}-1)*Bernoulli(2*n)$; also of Bernoulli(2n,1/2) and Bernoulli(2n,1/4).
$B(2n,\frac{1}{2})=\frac{a}{b},B(2n,\frac{1}{4})=\frac{c}{d}\Rightarrow a=c,B(n,x)$\ Bernoulli polynomials}.
Consequence of the stronger statement that $B(2n,1/2)=2^{2n}B(2n,1/4)$, proved via g.f., together with the fact that the numerator of $B(2n,1/2)$ is odd. --Ira Gessel, 2004-Sep-29

Problem #

Anum	Aname.
Formula
State

1. $\arctan (\tanh x\tan x)=\sum _{n\ge 0}(-1{)}^n{2}^{6n+2}(2^{4n+2}-1)\frac{B_{4n+2}}{2n+1}\cdot \frac{x^{4n+2}}{(4n+2)!}.$
2. $\tau (2^n)=[x^{2^n}]x\prod _{k\ge 1}(1-x^k{)}^{24}=[x^n]\frac{1}{2048x^2+24x+1}.$
3. $7|[\frac{x^{6k+4}}{(6k+4)!}]\frac{1}{2-\cosh (x)}.$
4. $4|[\frac{x^{6k+4}}{(6k+4)!}]\exp (\cos x-1),11|\left[\frac{x^{10k}}{(10k)!}\right]\exp (\cos x-1)\dots .$
5. Res, $(x^n-1,4x^2-1)=4^n-2^n-(-2{)}^n+(-1{)}^n.$
6. $\left\{n\right|n^2-1|(\genfrac{}{}{0ex}{}{2n}{n})\}\setminus \{2\}\subset \left\{n\right|n!(n-1)!|2(2n-3)!\}.$
7. $\sum _{k=0}^n(k+1)\sum _{l=0}^k{2}^l(\genfrac{}{}{0ex}{}{k}{l})(\genfrac{}{}{0ex}{}{n-k}{l})=[x^n]\frac{1-x}{(1-2x-x^2{)}^2}.$
8. $\sum _{k=0}^{n+1}(k+1)[(\genfrac{}{}{0ex}{}{2n+1}{k})-(\genfrac{}{}{0ex}{}{2n+1}{k-1})]=\frac{n+2}{2}(\genfrac{}{}{0ex}{}{2n+2}{n+1})-4^n.$
9. $a(n)=\sum _{k=0}^n\left[(\genfrac{}{}{0ex}{}{n}{k})mod2\right]\cdot 2^k\Rightarrow a_{2n+1}=3a_{2n}.$
10. $\sum _{k=0}^{\lfloor n/2\rfloor }{D}^k(\genfrac{}{}{0ex}{}{n}{2k+1})=[x^n]\frac{x}{1-2x+(1-D)x^2}.$
11. $((\genfrac{}{}{0ex}{}{2n}{n}),(\genfrac{}{}{0ex}{}{3n}{n}),\cdot ,(\genfrac{}{}{0ex}{}{(n-1)n}{n}))=1\iff n+1=\sum _i{p}_i^{e_i}\wedge m=\underset{i}{\max }{p}_i^{e_i}\wedge \frac{n+1}{m}>m.$
12. $\{a_n|$Least term in period of cont.~frac. of $\sqrt{a_n}=20\}=100n^2+n.$
13. Define $PCF(n)$ the period of the continued fraction expansion for $\sqrt{n}$. Then

$PCF(n)=PCF(n+1)\equiv 1mod2\Rightarrow n\equiv 1mod24.$

1. The largest term in the periodic part of the cont.~frac. of $\sqrt{3^n+1}$ is $2\cdot \lfloor (\sqrt{3}{)}^n\rfloor .$
2. The numerators of the continued fraction convergents to $\sqrt{27}$ are

$[x^n]\frac{5+26x+5x^2-x^3}{1-52x^2+x^4}.$

20. $n=5^{i}11^j\Rightarrow n|\sum _{k=1}^{10}{k}^n.$

21. $n+1|d(n{!}^n).$

22. Group the natural numbers such that the product of the terms of the n-th group is divisible by n!. Let $a_n$ the first term of the n-th group. Then $a_n=\lfloor \frac{(n-1{)}^2}{2}+1\rfloor .$

23. #{cubic residues mod $8^n$}=$\frac{4\cdot 8^n+3}{7}.$

24. lcm $(3n+1,3n+2,3n+3)=\frac{3}{4}(9n^3+18n^2+11n+2)(3+(-1{)}^n).$ Disproved at n=61

(26a) Let $a_n=\prod _{k=1}^n$ lcm $(k,n-k+1)$. Then $a_n=n^2(n-1){!}^2$ for n even, n+1 prime. Also, if n is odd and >3 $2(n+1)a_n$ is a perfect square, the root of which has the factor $\frac{1}{2}n(n-1)((n-1)/2)!$.

Solutions

1. This problem was solved at 2004-Sep-27 by Ralf Stephan himself:$[x^n](1-4x)\sqrt{1-4x}=\frac{C(2n,n)}{1-2n}-\frac{4*C(2n-2,n-1)}{3-2n}.$ Details
2. Also solved by Ralf Stephan 2004-Sep-27: Solution through $[x^n](-1+x)\sqrt{1-4x}=-C(2n,n)/(1-2n)+C(2n,n)/(3-2n).$
3. Equation 28 of Mathworld's Legendre Polynomial, as researched by Mitch Harris, 2004-Oct-12.
4. Proved by Nikolaus Meyberg, 2004-Oct-15.
   Notation:
   $v(n)=$dyadic valuation of n,
   $e(n)=$ number of ones in binary representation of n,
   $[x^n]p(x)=n$'th coefficient of polynomial $p(x)$,
   $P_n(x)=1/(2^n*n!)*d^n/d{x}^n[(x^2-1{)}^n]$: the $n$'th Legendre polynomial. $2n,2k$ means $2*n,2*k$, resp.(!) means "has to be shown".
   
   Proposition:
   $\sum _{k=3}^n-v([x^{(}2k)]P_{(}2n)(x))=2*n^2+2n-2*\sum _{i=0}^{n}i=e(i)$ for $n>=3.$
   (to be continued)