A103505 Denominator in expansion of (1-x)*log(1-x).

1, 1, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450

Offset: 0

Author

Paul Barry, Feb 09 2005

Keywords

• easy
• nonn
• frac

Comments

Apart from initial terms, same as A002378.

See A002378 for many more comments and references.

Denominators for the sequence with o.g.f. (1-x)*log(1-x). Numerators are given by 1 - 0^n - 2(C(1,n) - C(0,n)). Also denominators for the sequence with o.g.f. (1+x)*log(1+x). This sequence has numerators (-1)^n - 0^n + 2(C(1,n) - C(0,n)).

Also the denominator of the least distance between two adjacent Farey fractions of order n. The numerator is 1. - Robert G. Wilson v, Apr 13 2014

For n>0, a(n) are the Engel expansion of A096789. - Benedict W. J. Irwin, Dec 15 2016

Number of permutations of length n>=0 avoiding the partially ordered pattern (POP) {1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second one. - Sergey Kitaev, Dec 08 2020

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
• Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.
• Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Crossrefs

Cf. A000384.

Programs

Magma

[0^n+Binomial(1,n)-Binomial(0,n)+2*Binomial(n,2): n in [0..60]]; // Vincenzo Librandi, Dec 18 2016

Mathematica

CoefficientList[Series[(1-2x+2x^2+2x^3-x^4)/(1-x)^3,{x,0,50}],x] (* or *) Denominator/@CoefficientList[Normal[Series[(1-x)Log[1-x], {x,0,50}]], x]  (* Harvey P. Dale, Apr 20 2011 *)

Formula

G.f.: (1-2*x+2*x^2+2*x^3-x^4) / (1-x)^3;

a(n) = 0^n + C(1, n) - C(0, n) + 2*C(n, 2).