cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A183034 G.f.: A(x) = exp( Sum_{n>=1} -(-2)^A001511(n)*x^n/n ) where A001511(n) equals the 2-adic valuation of 2n.

Original entry on oeis.org

1, 2, 0, -2, 2, 6, 0, -6, 0, 6, 0, -6, -2, 2, 0, -2, 2, 6, 0, -6, 6, 18, 0, -18, 0, 18, 0, -18, -6, 6, 0, -6, 0, 6, 0, -6, 6, 18, 0, -18, 0, 18, 0, -18, -6, 6, 0, -6, -2, 2, 0, -2, 2, 6, 0, -6, 0, 6, 0, -6, -2, 2, 0, -2, 2, 6, 0, -6, 6, 18, 0, -18, 0, 18, 0, -18, -6, 6, 0, -6, 6, 18, 0, -18, 18, 54, 0
Offset: 0

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Author

Paul D. Hanna, Dec 19 2010

Keywords

Comments

Compare g.f. to B(x), the g.f. of the number of partitions of 2n into powers of 2 (A000123):
B(x) = exp( Sum_{n>=1} 2^A001511(n)*x^n/n ) = (1-x)^(-1)*Product_{n>=0} 1/(1 - x^(2^n)).

Examples

			G.f.: A(x) = 1 + 2*x - 2*x^3 + 2*x^4 + 6*x^5 - 6*x^7 + 6*x^9 -+...
The logarithm of the g.f. begins:
log(A(x)) = 2*x - 4*x^2/2 + 2*x^3/3 + 8*x^4/4 + 2*x^5/5 - 4*x^6/6 + 2*x^7/7 - 16*x^8/8 + 2*x^9/9 - 4*x^10/10 + 2*x^11/11 + 8*x^12/12 + 2*x^13/13 - 4*x^14/14 + 2*x^15/15 + 32*x^16/16 +...
The g.f. may be expressed by the series:
A(x) = 1 + 2*G(x) + 2*G(x^4) + 2*G(x^16) + 2*G(x^64) + 2*G(x^256) +...
where G(x) is the g.f. of A183035:
G(x) = x*(1-x^2)*Product_{n>=1} (1 + x^(4^n))^3
which begins:
G(x) = x - x^3 + 3*x^5 - 3*x^7 + 3*x^9 - 3*x^11 + x^13 - x^15 + 3*x^17 - 3*x^19 + 9*x^21 - 9*x^23 + 9*x^25 - 9*x^27 + 3*x^29 - 3*x^31 +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,-(-2)^valuation(2*m,2)*x^m/m)+x*O(x^n)),n)}
    
  • PARI
    {a(n)=local(L4n=ceil(log(n+1)/log(4)),G=x*(1-x^2)*prod(k=1,L4n,1 + x^(4^k))^3);polcoeff(1+2*sum(k=0,L4n,subst(G,x,x^(4^k)+x*O(x^n))),n)}

Formula

G.f. satisfies: A(x) = A(x^4)*(1+x)^2/(1+x^2).
G.f.: A(x) = 1 + 2*Sum_{n>=0} G(x^(4^n)) where G(x) = x*(1-x^2)*Product_{n>=1} (1 + x^(4^n))^3 is the g.f. of A183035.
a(4n) = a(n); a(4n+2) = 0.