A191373 Sum of binomial coefficients C(i+j,i) modulo 2 over all pairs (i,j) of positive integers satisfying 5i+j=n.

1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 4, 1, 2, 2, 4, 2, 3, 3, 5, 1, 3, 2, 5, 2, 3, 4, 6, 1, 3, 2, 6, 2, 3, 4, 6, 2, 4, 3, 7, 3, 5, 5, 8, 1, 4, 3, 8, 2, 3, 5, 8, 2, 4, 3, 8, 4, 6, 6, 9, 1, 5, 3, 9, 2, 3, 6, 9, 2, 4, 3, 9

Offset: 0

Author

Johannes W. Meijer, Jun 05 2011

Keywords

• nonn

Comments

The Le1{1,5} and Le2{5,1} triangle sums of Sierpinski’s triangle A047999 equal this sequence; see the formulas for their definitions. The Le1{1,5} and Le2{5,1} triangle sums are similar to the Kn11 and Kn21 sums, the Ca1 and Ca2 sums, and the Gi1 and Gi2 sums, see A180662.

Some A191373(2^n-p) sequences, 0<=p<=32, lead to known sequences, see the crossrefs.

Links

• Sam Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.

Crossrefs

Cf. A001316 (1,1), A002487 (2,1), A120562 (3,1), A112970 (4,1), A191373 (5,1)

Cf. A000012 (p=0), A006498 (p=1, p=2, p=4, p=8, p=16, p=32), A070550 (p=3, p=6, p=12, p=24), A000071 (p=15, p=30), A115008 (p=23).

Programs

Maple

A191373:=proc(n) option remember; if n <0 then A191373(n):=0 fi: if (n=0 or n=1) then 1 elif n mod 2 = 0 then A191373(n/2) else A191373((n-1)/2) + A191373(((n-1)/2)-2); fi; end: seq(A191373(n),n=0..75);

Formula

a(2*n) = a(n) and a(2*n+1) = a(n) + a(n-2) with a(0) = 1, a(1) = 1 and a(n)=0 for n<=-1.

a(n) = Le1{1,5}(n) = add(T(n-4*k,k),k=0..floor(n/5))

a(n) = Le1{1,5}(n) = sum(binomial(i + j, i) mod 2 | (i + 5*j) = n)

a(n) = Le2{5,1}(n) = add(T(n-4*k,n-5*k),k=0..floor(n/5))

a(n) = Le2{5,1}(n) = sum(binomial(i + j, i) mod 2 | (5*i + j) = n)

G.f.: Product_{n>=0} (1+x^(2^n)+x^(5*2^n)).

G.f. A(x) satisfies: A(x) = (1 + x + x^5) * A(x^2). - Ilya Gutkovskiy, Jul 09 2019