cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A083527 a(n) is the number of times that sums 1+-4+-9+-16+-...+-n^2 of the first n squares is zero. There are 2^(n-1) choices for the sign patterns.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 5, 0, 0, 43, 57, 0, 0, 239, 430, 0, 0, 2904, 5419, 0, 0, 27813, 50213, 0, 0, 348082, 649300, 0, 0, 3913496, 7287183, 0, 0, 50030553, 93696497, 0, 0, 611793542, 1161079907, 0, 0, 8009933135, 15176652567, 0, 0
Offset: 1

Views

Author

T. D. Noe, Apr 29 2003

Keywords

Comments

The frequency of each possible sum is computed by the Mathematica program without explicitly computing the individual sums.
a(n) is the maximal number of subsets of the first n squares that share the same sum. Cf. A025591, A083309.
a(n)=0 when n==1 or 2 (mod 4).

Examples

			a(7) = 1 because there is only one sign pattern of the first seven squares that yields zero: 1+4-9+16-25-36+49.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local m;
          m:= (1+(3+2*i)*i)*i/6;
          `if`(n>m, 0, `if`(n=m, 1, b(abs(n-i^2), i-1) +b(n+i^2, i-1)))
        end:
    a:= n-> `if`(irem(n-1, 4)<2, 0, b(n^2, n-1)):
    seq(a(n), n=1..40);  # Alois P. Heinz, Oct 31 2011
  • Mathematica
    d={1, 1}; nMax=60; zeroLst={0}; Do[p=n^2; d=PadLeft[d, Length[d]+p]+PadRight[d, Length[d]+p]; If[1==Mod[Length[d], 2], AppendTo[zeroLst, d[[(Length[d]+1)/2]]], AppendTo[zeroLst, 0]], {n, 2, nMax}]; zeroLst/2
    p = 1; t = {}; Do[p = Expand[p(x^(n^2) + x^(-n^2))]; AppendTo[t, Select[p, NumberQ[ # ] &]/2], {n, 51}]; t (* Robert G. Wilson v, Oct 31 2005 *)
  • PARI
    a(n)=sum(i=0,2^(n-1)-1,sum(j=1,n-1,(-1)^bittest(i,j-1)*j^2)==n^2) \\ Charles R Greathouse IV, Nov 05 2012

Formula

a(n) is half the coefficient of x^0 in the product_{k=1..n} x^(k^2)+x^(k^-2).
a(n) = A158092(n)/2.
a(n) = [x^(n^2)] Product_{k=1..n-1} (x^(k^2) + 1/x^(k^2)). - Ilya Gutkovskiy, Feb 01 2024