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

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, 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

Offset: 0

Author

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

Keywords

• easy
• nonn

Comments

a(n) is the 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.

A120562 can be generated from triangle A177444. Partial sums of A120562 = A177445. - Gary W. Adamson, May 08 2010

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 crossrefs. - Johannes W. Meijer, Jun 05 2011

Examples

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

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• K. Anders, Counting Non-Standard Binary Representations, JIS vol 19 (2016) #16.3.3 example 6.
• Cristina Ballantine and George Beck, Partitions enumerated by self-similar sequences, arXiv:2303.11493 [math.CO], 2023.
• S. 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. A177444, A177445. - Gary W. Adamson, May 08 2010

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). - Johannes W. Meijer, Jun 05 2011

Same recurrence for odd n as A000930.

Programs

Maple

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:
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); # Johannes W. Meijer, Jun 05 2011

Mathematica

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 *)
Nest[Append[#1, If[EvenQ@ #2, #1[[#2/2 + 1]], Total@ #1[[#2 ;; #2 + 1]] & @@ {#1, (#2 - 1)/2}]] & @@ {#, Length@ #} &, {1, 1}, 10^4 - 1] (* Michael De Vlieger, Feb 19 2019 *)

Formula

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

G.f.: A(x) = Product_{i>=0} (1 + x^(2^i) + x^(3*2^i)) = (1 + x + x^3)*A(x^2).

a(n-1) << n^x with x = log_2(phi) = 0.69424... - Charles R Greathouse IV, Dec 27 2011

Extensions

Reference edited and link added by Jason G. Wurtzel, Aug 22 2010