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%I #40 Apr 06 2017 02:23:15
%S 1,6,6,12,6,16,12,24,6,36,16,32,12,36,24,48,6,36,36,72,16,56,32,64,12,
%T 72,36,72,24,68,48,96,6,36,36,72,36,96,72,144,16,96,56,112,32,100,64,
%U 128,12,72,72,144,36,120,72,144,24,144,68,136
%N Number of odd terms in expansion of (1 + x + x^2 + x^3 + x^4 + x^5)^n.
%C All the following are of the same type: A001316, A071053, A134660, A134661, A134662, A255485, A247649, A255486. It would be nice to have some unifying formula or recurrence. (Restating the definition, these are the Hamming weights of the n-th powers of the corresponding polynomials over GF(2). - _Joerg Arndt_, Mar 02 2015)
%H Alois P. Heinz, <a href="/A255488/b255488.txt">Table of n, a(n) for n = 0..8191</a>
%e From _Omar E. Pol_, Mar 01 2015: (Start)
%e Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
%e 1;
%e 6;
%e 6,12;
%e 6,16,12,24;
%e 6,36,16,32,12,36,24,48;
%e 6,36,36,72,16,56,32,64,12,72,36,72,24,68,48,96;
%e 6,36,36,72,36,96,72,144,16,96,56,112,32,100,64,128,12,72,72,144,36,120,72,144,24,144,68,136...
%e ...
%e In each row the first quarter of the terms (and no more) are equal to 6 times the beginning of the sequence itself (corrected after Sloane's comment in A247649, Mar 03 2015).
%e (End)
%p r1:=proc(f) local g,n; g:=n->nops(expand(f^n) mod 2); [seq(g(n),n=0..90)]; end;
%p r1(1+x+x^2+x^3);
%t a[n_] := Count[CoefficientList[(1 + x + x^2 + x^3 + x^4 + x^5)^n, x], _?OddQ];
%t Table[a[n], {n, 0, 90}] (* _Jean-François Alcover_, Apr 06 2017 *)
%o (PARI) a(n) = {my(pol=(1+x+x^2+x^3+x^4+x^5)*Mod(1,2)); subst(lift(pol^n), x, 1);} \\ _Michel Marcus_, Mar 01 2015
%Y Cf. A001316, A071053, A134660, A134661, A134662, A255485, A247649, A255486.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Mar 01 2015
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