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A307877 Number of ways of partitioning the set of the first n positive squares into two subsets whose sums differ at most by 1.

%I A307877 #52 Dec 07 2020 13:59:55
%S A307877 1,1,0,0,0,0,2,1,1,5,2,1,5,13,43,43,57,193,274,239,430,1552,3245,2904,
%T A307877 5419,18628,31048,27813,50213,188536,372710,348082,649300,2376996,
%U A307877 4197425,3913496,7287183,27465147,53072709,50030553,93696497,351329160,650125358
%N A307877 Number of ways of partitioning the set of the first n positive squares into two subsets whose sums differ at most by 1.
%H A307877 Alois P. Heinz, <a href="/A307877/b307877.txt">Table of n, a(n) for n = 0..200</a>
%H A307877 Wikipedia, <a href="https://en.wikipedia.org/wiki/Partition_problem">Partition problem</a>
%F A307877 a(n) = A083527(n) if n == 0 or 3 (mod 4).
%e A307877 a(6) = 2: 1,9,36/4,16,25; 1,4,16,25/9,36.
%e A307877 a(7) = 1: 1,4,16,49/9,25,36.
%p A307877 s:= proc(n) s(n):= `if`(n=0, 1, n^2+s(n-1)) end:
%p A307877 b:= proc(n, i) option remember; `if`(i=0, `if`(n<=1, 1, 0),
%p A307877       `if`(n>s(i), 0, (p-> b(n+p, i-1)+b(abs(n-p), i-1))(i^2)))
%p A307877     end:
%p A307877 a:= n-> ceil(b(0, n)/2):
%p A307877 seq(a(n), n=0..45);
%t A307877 s[n_] := s[n] = If[n == 0, 1, n^2 + s[n - 1]];
%t A307877 b[n_, i_] := b[n, i] = If[i == 0, If[n <= 1, 1, 0], If[n > s[i], 0, Function[p, b[n + p, i - 1] + b[Abs[n - p], i - 1]][i^2]]];
%t A307877 a[n_] := Ceiling[b[0, n]/2];
%t A307877 a /@ Range[0, 45] (* _Jean-François Alcover_, Dec 07 2020, after _Alois P. Heinz_ *)
%Y A307877 Cf. A000290, A069918, A083527, A158092, A306443, A308682.
%K A307877 nonn
%O A307877 0,7
%A A307877 _Alois P. Heinz_, Jun 04 2019