login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A260878
Number of set partitions of {1, 2, ..., 2*n} with sizes in {[n, n], [2n]}.
10
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
0,1
COMMENTS
Third column in A260876.
LINKS
Hans Salié, Über die Koeffizienten der Blasiusschen Reihe, Math. Nachr. 1955, (14, 4-6), 241--248.
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)
EXAMPLE
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}}.
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
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 *)
PROG
(Sage)
print([A260876(n, 2) for n in (0..30)])
(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
CROSSREFS
a(n) = A112849(n) for n >= 2. - Alois P. Heinz, Aug 06 2015
a(n) = A052473(n+2) - 1.
a(n) = A088218(n) + 1.
a(n) = (-1)^n*A110556(n) + 1.
a(n+1) - a(n) = A097613(n+1) for n > 0.
Cf. A323230 (d=0), this sequence (d=1), A323229 (d=2).
Sequence in context: A153950 A153947 A298778 * A064880 A115011 A296688
KEYWORD
nonn,easy
AUTHOR
Peter Luschny, Aug 02 2015
STATUS
approved