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.

A182185 G.f.: exp( Sum_{n>=1} 3^b(n) * x^n/n ) where b(n) = highest exponent of 3 in 2^n+1.

Original entry on oeis.org

1, 3, 5, 9, 15, 21, 29, 39, 49, 63, 81, 99, 123, 153, 183, 219, 261, 303, 353, 411, 469, 537, 615, 693, 781, 879, 977, 1089, 1215, 1341, 1485, 1647, 1809, 1989, 2187, 2385, 2607, 2853, 3099, 3375, 3681, 3987, 4323, 4689, 5055, 5457, 5895, 6333, 6813, 7335, 7857, 8421
Offset: 0

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Author

Paul D. Hanna, Apr 18 2012

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 5*x^2 + 9*x^3 + 15*x^4 + 21*x^5 + 29*x^6 + 39*x^7 +...
The g.f. satisfies:
A(x)/A(x^3) = 1 + 3*x + 5*x^2 + 6*x^3 + 6*x^4 + 6*x^5 +...+ 6*x^n +...
The logarithm of the g.f. begins:
log(A(x)) = 3*x + x^2/2 + 9*x^3/3 + x^4/4 + 3*x^5/5 + x^6/6 + 3*x^7/7 + x^8/8 + 27*x^9/9 + x^10/10 + 3*x^11/11 + x^12/12 +...+ 3^b(n)*x^n/n +...
where b(n) = highest exponent of 3 in 2^n+1, for n>=1, and begins:
b = [1,0,2,0,1,0,1,0,3,0,1,0,1,0,2,0,1,0,1,0,2,0,1,0,1,0,4,...].
The g.f.s of the TRISECTIONS begin:
T_0(x) = 1 + 9*x + 29*x^2 + 63*x^3 + 123*x^4 + 219*x^5 + 353*x^6 +...
T_1(x) = 3 + 15*x + 39*x^2 + 81*x^3 + 153*x^4 + 261*x^5 + 411*x^6 +...
T_2(x) = 5 + 21*x + 49*x^2 + 99*x^3 + 183*x^4 + 303*x^5 + 469*x^6 +...
where T_1(x)/T_0(x) = 3*(1+x)/(1+5*x), T_2(x)/T_0(x) = (5+x)/(1+5*x).
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(m=1, n+1, 3^valuation(2^m+1, 3)*x^m/m)+x*O(x^n)), n)}
    for(n=0, 65, print1(a(n), ", "))
    
  • PARI
    {a(n)=local(A=1+x);for(i=1,ceil(log(n+1)/log(3)),A=(1-x^2)*(1-x^3)/(1-x)^3*subst(A,x,x^3+x*O(x^n)));polcoeff(A,n)}

Formula

G.f. satisfies: A(x) = (1-x^2)*(1-x^3)/(1-x)^3 * A(x^3).
Define TRISECTIONS: A(x) = T_0(x^3) + x*T_1(x^3) + x^2*T_2(x^3), then:
(1) T_1(x)/T_0(x) = 3*(1+x)/(1+5*x),
(2) T_2(x)/T_0(x) = (5+x)/(1+5*x),
(3) T_0(x)/T_0(x^3) = (1+x)*(1+5*x)*(1-x^3)^2 / ((1-x)^3*(1+5*x^3)),
(4) T_1(x)/T_1(x^3) = (1+x)^2*(1-x^3)^2 / ((1-x)^3*(1+x^3)),
(5) T_2(x)/T_2(x^3) = (1+x)*(5+x)*(1-x^3)^2 / ((1-x)^3*(5+x^3)),
(6) A(x) = (1-x)/(1+5*x)*T_0(x) = (1-x)/(1+x)*T_1(x)/3 = (1-x)/(5+x)*T_2(x).