User talk:Thomas Scheuerle/Fun and Games/Make It Square!

This is a Game

Rules for this sequence :

1. All values are greater than zero.

2. The sum of all values of this sequence and all the related finite differences sum to a square number.

Example:

$\sum _{}{\Bigl \{}{\begin{matrix}{\quad 1,\quad 18}\\{\quad 17}\end{matrix}}{\Bigr \}}=6^{2}$ and $\sum _{}{\Biggl \{}{\begin{matrix}{\quad 1,\quad 18,\quad 7}\\{\quad 17,-11}\\{-28}\end{matrix}}{\Biggr \}}=2^{2}$

$s_{m}^{2}=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}a(j+n-k)$

This means the first square is a(1). The second: a(1)+a(2)+(a(2)-a(1)) = 2*a(2). The third: a(1)+a(2)+a(3)+(a(2)-a(1))+(a(3)-a(2))+((a(3)-a(2))-(a(2)-a(1))) = a(1)-a(2)+3*a(3) ...

3. Everybody may extend this sequence or fix errors and mistakes.

4. This may require to change some numbers already present in this sequence, because it may reach a dead end. If while extending this sequence, numbers which where previously already present are changed such that the former last value becomes smaller, then will this be counted as a double success in this game.

Here is the sequence so far:

1,18,7,4,2,2,5,3,17,106,248,215,33,986,3937,2207,1478,115812,447098,269,9971,30703632,138185910,821,512479,15005336029

Hall of Fame

The first number counts 1 Point the second number 2 Point's and so forth. If an edit extends the sequence and reduces the former last number, all Points are doubled. If an error in the sequence becomes fixed, this counts 10 Points.

User	Score	Any remark
--Thomas Scheuerle (talk) 15:25, 28 January 2022 (EST)	1. to 26. number -> 351 Points	:-)

A related problem

$m^{2}=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}a(j+n-k)$ This can be solved with the rational number sequence:

a(x) = {0;(1/2);(3/2);3;(49/10);7;(127/14);11;(129/10);15;(379/22);19;(18419/910);23;(57/2);27;(1313/170);31;(157935/798);35;(-170541/110);39;...} with $2(n-2)+1-a(n)=-3B_{n}=-\sum _{k=0}^{n}(-1)^{k}{n \choose k}a(n-k)$for all n > 1.

$(m+1)^{2}=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}a(j+n-k)$ This can be solved with the rational number sequence:

a(x) = {1;2;(10/3);5;(104/15);9;(232/21);13;(224/15);17;(632/33);21;(30704/1365);25;(88/3);29;...} with $2n-1-a(n)=-2B_{n}=-\sum _{k=0}^{n}(-1)^{k}{n \choose k}a(n-k)$for all n > 0.

Interestingly for all n > 0, a(1+2*n) is an integer and a(1+2*(n+1)) - a(1+2*n) = 4 while a(2*n) describes a complicated curve with changing sign.

$m=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}a(j+n-k)$ This can be solved with the rational number sequence:

a(x) = {0;(1/2);(5/6);1;(31/30);1;(41/42);1;(61/66);1;(3421/2730);1;...} $=1-B_{x}$ the binomial transform of this sequence generates again Bernoulli numbers: $B_{n}=-\sum _{k=0}^{n}(-1)^{k}{n \choose k}a(n-k)$

$1=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}B_{j+n-k}$

We also get this identity: A032346$(m+1)=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}$A032346$(j+n-k)$

If we modify this: $a(m)=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}a(j+n-k)$ we find this as a possible solution:

a(n) = {-1;0;(1/2);(1/2);(1/8);(-7/20);(-9/16);(-21/80);(257/640);(373/480);(1283/6400);...} a(n) = ${\frac {-a(n-1)+\sum _{m=0}^{n-1}\sum _{k=0}^{m}(-1)^{k}{m+1 \choose k}a(m+k)}{n}}={\frac {-a(n-1)+\sum _{k=0}^{n-1}A130595(n+1,k)*a(k)}{n}},a(0)=-1$

This nested sum is an operator which maps a polynomial of order w to a polynomial of order w+1:

${\frac {1}{w+1}}m^{w+1}+\sum _{v=1}^{w}{\frac {(v+B_{v})(w)_{v}}{v!}}m^{w+1-v}-$A052875$(w)=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}(j+n-k)^{w},m>w-1$ and $(m)_{v}$ is the falling factorial.

This can be generalized to m <= w-1, if we build the sequence $O_{m}(w$) by the exponential generating function ${\frac {(e^{x}-1)^{2+m}}{2-e^{x}}}$ and develop the sequence from this e.g.f. for each m individually.

If we add $O_{m}(w)$ to the formula, it will hold for m <= w-1 too.

$2*(Fibonacci(2*\lfloor {\frac {m+1}{2}}\rfloor -1)-1)=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}Fibonacci(j+n-k)$

$Lucas(2*\lfloor {\frac {m}{2}}\rfloor +2)-2=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}Lucas(j+n-k)$

A145766$(\lfloor {\frac {m+1}{2}}\rfloor )=\sum _{j=0}^{m}\sum _{n=0}^{m-j}\sum _{k=0}^{n}(-1)^{k}{n \choose k}Jacobsthal(j+n-k)$

The coefficients for the nested summations result into this triangle:

1

0, 2

1,-1, 3

0, 3, -3, 4

1,-2, 7, -6, 5

0, 4, -8, 14,-10, 6

1,-3, 13,-21, 25,-15, 7

0, 5,-15, 35,-45, 41,-21, 8

1,-4, 21,-49, 81,-85, 63,-28, 9

...

The falling diagonals are:

A000027 Positive integers

A000217 Triangular numbers

A004006

A051744