A112970 A generalized Stern sequence.

1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 2, 2, 1, 5, 3, 3, 2, 5, 2, 3, 1, 6, 4, 3, 2, 6, 2, 3, 1, 7, 5, 4, 3, 8, 3, 5, 2, 8, 5, 4, 2, 8, 3, 3, 1, 9, 6, 5, 4, 9, 3, 6, 2, 9, 6, 4, 2, 9, 3, 3, 1, 10, 7, 6, 5, 11, 4, 8, 3, 12, 8, 6, 3, 13, 5, 5, 2, 13, 8, 7, 5, 12, 4, 7, 2, 12, 8, 5, 3, 11, 3, 4, 1, 12, 9, 7, 6

Offset: 0

Author

Paul Barry, Oct 07 2005

Keywords

• easy
• nonn

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

Crossrefs

Cf. A002487, A112971.

Cf. A120562 (Northshield).

Cf. A033638 (p=0), A000012 (p=1), A004526 (p=2, p=3, p=5, p=9, p=17), A002620 (p=4, p=7, p=13, p=25), A000027 (p=6, p=11, p=21), A004116 (p=8, p=15, p=29), A035106 (p=10, p=19), A024206 (p=14, p=27), A007494 (p=18), A014616 (p=22), A179207 (p=26). - Johannes W. Meijer, Jun 05 2011

Programs

Maple

A112970:=proc(n) option remember; if n <0 then A112970(n):=0 fi: if (n=0 or n=1) then 1 elif n mod 2 = 0 then A112970(n/2) + A112970((n/2)-2) else A112970((n-1)/2); fi; end: seq(A112970(n),n=0..99); # Johannes W. Meijer, Jun 05 2011

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 *)

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