A271929 G.f. A(x) satisfies: A(x)^3 = A(x^3) / (1 - 3*x).
1, 1, 2, 5, 12, 31, 83, 224, 615, 1708, 4777, 13455, 38110, 108428, 309714, 887666, 2551575, 7353423, 21240460, 61478489, 178269670, 517784717, 1506162369, 4387201004, 12795170784, 37359689295, 109199349181, 319493390481, 935616592227, 2742209152877, 8043500169958, 23610710680582, 69354125493930, 203852682699869, 599549063015417, 1764338532368820
Offset: 1
Keywords
Examples
G.f.: A(x) = x + x^2 + 2*x^3 + 5*x^4 + 12*x^5 + 31*x^6 + 83*x^7 + 224*x^8 + 615*x^9 + 1708*x^10 + 4777*x^11 + 13455*x^12 +... where A(x)^3 = A(x^3) / (1 - 3*x). Also, when expressed as the EULER transform of A046211, A(x) = x/( (1-x) * (1-x^2) * (1-x^3)^3 * (1-x^4)^6 * (1-x^5)^16 * (1-x^6)^39 * (1-x^7)^104 * (1-x^8)^270 * (1-x^9)^729 *...* (1-x^n)^A046211(n) *...). RELATED SERIES. A(x)^3 = x^3 + 3*x^4 + 9*x^5 + 28*x^6 + 84*x^7 + 252*x^8 + 758*x^9 + 2274*x^10 + 6822*x^11 + 20471*x^12 + 61413*x^13 + 184239*x^14 +...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..400
Crossrefs
Cf. A123916.
Programs
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PARI
{a(n) = my(A=x); for(i=1, n, A = ( subst(A, x, x^3)/(1 - 3*x +x*O(x^n)))^(1/3)); polcoeff(G=A, n)} for(n=1, 50, print1(a(n), ", "))
Formula
The EULER transform of A046211, where A046211(n) is the number of ternary Lyndon words whose digits sum to 1 (or 2) mod 3.
a(n) ~ c * 3^n / n^(2/3), where c = 0.1260671867244258410294918... . - Vaclav Kotesovec, Apr 18 2016
Comments