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%I #16 Aug 14 2020 17:32:02
%S 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,
%T 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,
%U 2,3,6,9,2,4,3,9
%N Sum of binomial coefficients C(i+j,i) modulo 2 over all pairs (i,j) of positive integers satisfying 5i+j=n.
%C 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.
%C Some A191373(2^n-p) sequences, 0<=p<=32, lead to known sequences, see the crossrefs.
%H Sam Northshield, <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.530.857">Sums across Pascal’s triangle modulo 2</a>, Congressus Numerantium, 200, pp. 35-52, 2010.
%F 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.
%F a(n) = Le1{1,5}(n) = add(T(n-4*k,k),k=0..floor(n/5))
%F a(n) = Le1{1,5}(n) = sum(binomial(i + j, i) mod 2 | (i + 5*j) = n)
%F a(n) = Le2{5,1}(n) = add(T(n-4*k,n-5*k),k=0..floor(n/5))
%F a(n) = Le2{5,1}(n) = sum(binomial(i + j, i) mod 2 | (5*i + j) = n)
%F G.f.: prod((1+x^(2^n)+x^(5*2^n), n=0..infinity)
%F G.f. A(x) satisfies: A(x) = (1 + x + x^5) * A(x^2). - _Ilya Gutkovskiy_, Jul 09 2019
%p 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);
%Y Cf. A001316 (1,1), A002487 (2,1), A120562 (3,1), A112970 (4,1), A191373 (5,1)
%Y 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).
%K nonn
%O 0,6
%A _Johannes W. Meijer_, Jun 05 2011
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