A260878 Number of set partitions of {1, 2, ..., 2*n} with sizes in {[n, n], [2n]}.
2, 2, 4, 11, 36, 127, 463, 1717, 6436, 24311, 92379, 352717, 1352079, 5200301, 20058301, 77558761, 300540196, 1166803111, 4537567651, 17672631901, 68923264411, 269128937221, 1052049481861, 4116715363801, 16123801841551, 63205303218877, 247959266474053
Offset: 0
Examples
The set partitions counted by a(3) = 11 are: {{1, 2, 3, 4, 5, 6}},
{{1, 2, 4}, {3, 5, 6}}, {{1, 2, 3}, {4, 5, 6}}, {{1, 3, 4}, {2, 5, 6}},
{{1, 3, 5}, {2, 4, 6}}, {{1, 4, 5}, {2, 3, 6}}, {{1, 5, 6}, {2, 3, 4}},
{{1, 4, 6}, {2, 3, 5}}, {{1, 3, 6}, {2, 4, 5}}, {{1, 2, 6}, {3, 4, 5}},
{{1, 2, 5}, {3, 4, 6}}.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Hans Salié, Über die Koeffizienten der Blasiusschen Reihe, Math. Nachr. 1955, (14, 4-6), 241--248.
Crossrefs
Programs
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Maple
a := proc(n) option remember; if n < 2 then [2, 2][n+1] else ((4*n - 2)*a(n-1) - 3*n + 2)/n fi end: seq(a(n), n=0..26); # Or: egf := n -> exp(exp(x)*(1 - (GAMMA(n,x)/GAMMA(n)))): a := n -> `if`(n<2, 2, (2*n)!*coeff(series(egf(n), x, 2*n+1), x, 2*n)): seq(a(n), n=0..26); # Peter Luschny, Aug 02 2019
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Mathematica
Table[Binomial[2 n - 1, n] + 1, {n, 0, 26}] (* or *) CoefficientList[Series[(4 x^2 - 13 x + 3 + Sqrt[(1 - 4 x) (x - 1)^2])/(2 (4 x - 1) (x - 1)), {x, 0, 26}], x] (* Michael De Vlieger, Feb 26 2017 *) -
Sage
print([A260876(n,2) for n in (0..30)])
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Sage
# Alternative: def A260878(): a, f, s, n = 2, 2, 1, 1 yield a while True: yield a f += 4; s += 3; n += 1 a = (f*a - s)/n a = A260878() print([next(a) for n in range(27)]) # Peter Luschny, Aug 02 2019
Formula
G.f.: (4*x^2 - 13*x + 3 + sqrt((1 - 4*x)*(x - 1)^2))/(2*(4*x - 1)*(x - 1)). - Alois P. Heinz, Aug 06 2015
a(n) = Binomial(2*n-1, n) + 1. - Vladimir Kruchinin, Feb 26 2017
The generating function G(x) satisfies the differential equation x^3 + 2*x = (4*x^4 - 9*x^3 + 6*x^2 - x)*diff(G(x), x) + (2*x^3 - 4*x^2 + 2*x)*G(x). - Peter Luschny, Feb 12 2019
From Peter Luschny, Aug 02 2019: (Start)
a(n) = ((4*n - 2)*a(n-1) - 3*n + 2)/n for n >= 2.
a(n) = (2*n)! * [x^(2*n)] exp(exp(x)*(1 - (Gamma(n,x)/Gamma(n)))) for n >= 2.
a(n) ~ 4^n/sqrt(4*Pi*n). More precise asymptotic estimates are:
1 + (4^n/sqrt(n*Pi)) * (1/2 - 1/(16*n) * (1 - 1/(16*n))), and
1 + 4^n*(2 - 2/N^2 + 21/N^4 - 671/N^6) / sqrt(2*N*Pi) with N = 8*n + 2.
Let b(n) = binomial(2*(n-1), n-1) + 1 = A323230(n) for n >= 0. Then by Salié:
p divides a(p+k) - b(k+1) if p is a prime > k and 0 <= k <= 4.
Conjecture: p divides a(p+5) - b(6) if p is a prime > b(6).
If p is a prime divisor of n then a(n) == a(n/p) (mod p) (by Salié, theorem 2).
(End)
From Peter Bala, Apr 20 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^k * 3*n/(2*n + k) * binomial(2*n+k, n-k) for n >= 1.
a(n) = Sum_{k = 0..n} (-1)^k * 3*n/(n + 2*k) * binomial(2*n+k-1, n-k) for n >= 1.
(-1)^n * a(n) equals the n-th order Taylor polynomial (centered at 0) of 1/c(x)^(3*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108. (End)
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