OFFSET
0,5
COMMENTS
Conjectures: a(2^n)=a(2^(n+1)+1)=A033638(n); a(2^n-1)=a(3*2^n-1)=1.
The Gi1 and Gi2 triangle sums, see A180662 for their definitions, of Sierpinski's triangle A047999 equal this sequence. The Gi1 and Gi2 sums can also be interpreted as (i + 4*j = n) and (4*i + j = n) sums, see the Northshield reference. Some A112970(2^n-p) sequences, 0<=p<=32, lead to known sequences, see the crossrefs. - Johannes W. Meijer, Jun 05 2011
LINKS
Sam Northshield, Sums across Pascal’s triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010. [Johannes W. Meijer, Jun 05 2011]
FORMULA
a(n) = Sum_{k=0..n} mod(sum{j=0..n, (-1)^(n-k)*C(j, n-j)*C(k, j-k)}, 2).
From Johannes W. Meijer, Jun 05 2011: (Start)
a(2*n+1) = a(n) and a(2*n) = a(n) + a(n-2) with a(0) = 1, a(1) = 1 and a(n)=0 for n<=-1.
G.f.: Product_{n>=0} (1 + x^(2^n) + x^(4*2^n)). (End)
G.f. A(x) satisfies: A(x) = (1 + x + x^4) * A(x^2). - Ilya Gutkovskiy, Jul 09 2019
MAPLE
MATHEMATICA
a[n_] := a[n] = Which[n<0, 0, n==0 || n==1, 1, Mod[n, 2]==0, a[n/2] + a[n/2-2], True, a[(n-1)/2]];
Table[a[n], {n, 0, 99}] (* Jean-François Alcover, Aug 02 2022 *)
CROSSREFS
Cf. A120562 (Northshield).
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Oct 07 2005
STATUS
approved