A124133 a(n) = (-1/2)*Sum_{i1 + i2 + i3 = 2*n} ((2*n)!/(i1! i2! i3!))*B(i1), where B are the Bernoulli numbers (with i1, i2, i3 >= 1).

0, 5, 28, 123, 506, 2041, 8184, 32759, 131062, 524277, 2097140, 8388595, 33554418, 134217713, 536870896, 2147483631, 8589934574, 34359738349, 137438953452, 549755813867, 2199023255530, 8796093022185, 35184372088808

Offset

1,2

Links

Table of n, a(n) for n=1..23.

Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]

Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.

Eric Weisstein's World of Mathematics, Bernoulli number.

Wikipedia, Bernoulli number.

Index entries for linear recurrences with constant coefficients, signature (6,-9,4).

Formula

a(n) = 2^(2*n - 1) - n - 1.

From Colin Barker, Dec 01 2012: (Start)

a(n) = 6*a(n-1) - 9*a(n-2) + 4*a(n-3).

G.f.: x^2*(2*x - 5)/((x - 1)^2*(4*x - 1)). (End)

Example

From Petros Hadjicostas, Jul 25 2020: (Start)
For n = 1, a(1) = 0 because there are no integers i1, i2, i3 >= 1 s.t. i1 + i2 + i3 = 2.
For n = 2, a(2) = 5 because we have 1 + 1 + 2 = 1 + 2 + 1 = 2 + 1 + 1 = 4, multinomial(4, [1, 1, 2]) = multinomial(4, [1, 2, 1]) = multinomial(4, [2, 1, 1]) = 12, B(1) = -1/2, and B(2) = 1/6. Thus, a(2) = (-1/2)*(12*(-1/2) + 12*(-1/2) + 12*(1/6)) = 5. (End)

Mathematica

LinearRecurrence[{6, -9, 4}, {0, 5, 28}, 30] (* Harvey P. Dale, Dec 02 2014 *)

Prog

(PARI) a(n)=2^(2*n-1)-n-1
/* Second PARI program using the definition */
lista(nn) = {my(a=vector(nn)); for(n=1, nn, a[n] = (-1/2)*sum(i1=1, 2*n-2, sum(i2=1, 2*n-i1-1, binomial(2*n, i1)*binomial(2*n-i1, i2)*bernfrac(i1)))); a; } \\ Petros Hadjicostas, Jul 25 2020

Crossrefs

Cf. A001469.

Sequence in context: A088727 A027016 A034582 * A261670 A258628 A054148

Adjacent sequences: A124130 A124131 A124132 * A124134 A124135 A124136

Keyword

nonn,easy

Author

Benoit Cloitre, Nov 30 2006

Status

approved