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 A120562 Sum of binomial coefficients C(i+j,i) modulo 2 over all pairs (i,j) of positive integers satisfying 3i+j=n. 9

%I

%S 1,1,1,2,1,2,2,3,1,3,2,3,2,4,3,5,1,4,3,4,2,5,3,5,2,5,4,6,3,7,5,8,1,6,

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

%N Sum of binomial coefficients C(i+j,i) modulo 2 over all pairs (i,j) of positive integers satisfying 3i+j=n.

%C a(n)=number of 'vectors' (...,e_k, e_{k-1},...,e_0) with e_k in {0,1,3} such that sum_k e_k 2^k=n. a(2^n-1)=F(n+1) a(2^{k+1}+j)+a(j)=a(2^k+j)+a(2^{k-1}+j) if 2^k>4j. This sequence corresponds to the pair (3,1) as Stern's diatomic sequence [A002487] corresponds to (2,1) and Gould's sequence [A001316] corresponds to (1,1). There are many interesting similarities to [A000119], the number of representations of n as a sum of distinct Fibonacci numbers.

%C A120562 can be generated from triangle A177444. Partial sums of A120562 = A177445. [From Gary W. Adamson, May 08 2010]

%C The Ca1 and Ca2 triangle sums, see A180662 for their definitions, of Sierpinski's triangle A047999 equal this sequence. Some A120562(2^n-p) sequences, 0<=p<=32, lead to known sequences, see the cross-refs. [From Johannes W. Meijer, Jun 05 2011]

%H S. Northshield, <a href="http://digitalcommons.plattsburgh.edu/mathematics_facpubs/9/">Sums across Pascal's triangle modulo 2</a>, Congressus Numerantium, 200, pp. 35-52, 2010.

%F Recurrence; a(0)=a(1)=1, a(2*n)=a(n) and a(2*n+1)=a(n)+a(n-1).

%F G.f.: A(x) = prod(i>=0, 1+x^(2^i)+x^(3*2^i) ) = (1+x+x^3)*A(x^2).

%F a(n-1) << n^x with x = lg(phi) = 0.69424... - _Charles R Greathouse IV_, Dec 27 2011

%e a(2^n)=1 since a(2n)=a(n).

%p p := product((1+x^(2^i)+x^(3*2^i)), i=0..25): s := series(p, x, 1000): for k from 0 to 250 do printf(`%d, `, coeff(s, x, k)) od:

%p A120562:=proc(n) option remember; if n <0 then A120562(n):=0 fi: if (n=0 or n=1) then 1 elif n mod 2 = 0 then A120562(n/2) else A120562((n-1)/2) + A120562((n-3)/2); fi; end: seq(A120562(n),n=0..64); [From Johannes W. Meijer, Jun 05 2011]

%t a[0] = a[1] = 1; a[n_?EvenQ] := a[n] = a[n/2]; a[n_?OddQ] := a[n] = a[(n-1)/2] + a[(n-1)/2 - 1]; Table[a[n], {n, 0, 64}] (* _Jean-François Alcover_, Sep 29 2011 *)

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

%Y Cf. A177444, A177445 [From _Gary W. Adamson_, May 08 2010]

%Y Cf. A000012 (p=0), A000045 (p=1, p=2, p=4, p=8, p=16, p=32), A000071 (p=3, p=6, p=12, p=13, p=24, p=26), A001610 (p=5, p=10, p=20), A001595 (p=7, p=14, p=28), A014739 (p=11, p=22, p=29), A111314 (p=15, p=30), A027961 (p=19), A154691 (p=21), A001911 (p=23) [From Johannes W. Meijer, Jun 05 2011]

%K easy,nonn

%O 0,4

%A Sam Northshield (samuel.northshield(AT)plattsburgh.edu), Aug 07 2006

%E Reference edited and link added by _Jason G. Wurtzel_, Aug 22 2010

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