A226270 Number of primitive permutations with n buds and 3 red or blue elements.

1, 7, 69, 843, 12081, 197127

Offset

0,2

Links

Table of n, a(n) for n=0..5.

Adrian Ocneanu, On the inner structure of a permutation: bicolored partitions and Eulerians, trees and primitives; arXiv preprint arXiv:1304.1263 [math.CO], 2013.

Jianghao Xu and Di Yang, Galilean symmetry of the KdV hierarchy, arXiv:2403.04631 [math-ph], 2024. See p. 6.

Formula

Conjectural recurrence: a(n) = (1/3)*(n+1)*d(n+2) - Sum_{k = 0..n-1} d(n+1-k)*a(k) with a(0)= 1, where d(n) = (2*n-1)!! = A001147(n). If true then the sequence continues [1, 7, 69, 843, 12081, 197127, 3595509, 72393003, 1594224801, 38123341767, 984142175589, ...]. - Peter Bala, Dec 26 2019

From Peter Bala, Jul 07 2022: (Start)

Let A(x) = 1 + 7*x + 69*x^2 + 843*x^3 + 12081*x^4 + 197127*x^5 + ... denote the o.g.f. Then it appears that 1 + 2*x*A(x) = 1/(1 - 2*x/(1 - 5*x/(1 - 4*x/(1 - 7*x/(1 - 6*x/(1 - 9*x/(1 - 8*x/(1 - ...)))))))), a continued fraction of Stieltjes type, and also that 1 + 2*x*A(x) = 1/(1 + 3*x - 5*x/(1 + 5*x - 7*x/ (1 + 7*x - 9*x/ (1 + 9*x - 11*x/ (1 + 11*x - ...))))).

If true, then 1 + x/(1 - 2*x - 6*x^2*A(x)) = 1 + x + 2*x^2 + 10*x^3 + 74*x^4 + 706*x^5 + ... is the g.f. of A000698. (End)

Crossrefs

Cf. A000698, A001147, A167872, A226297, A226298, A226299.

Sequence in context: A025757 A243668 A265033 * A121351 A302353 A059321

Adjacent sequences: A226267 A226268 A226269 * A226271 A226272 A226273

Keyword

nonn,more

Author

N. J. A. Sloane, Jun 05 2013

Status

approved