A151641 Number of permutations of 4 indistinguishable copies of 1..n with exactly 3 adjacent element pairs in decreasing order.

0, 16, 8464, 724320, 37229920, 1558185200, 59416090096, 2167506244544, 77394535148480, 2734912695301840, 96159966699204560, 3372863224609356576, 118169571125488257824, 4137881135327148408240, 144857367811462402307760, 5070515828676757812456320

Offset: 1

Author

R. H. Hardin, May 29 2009

Keywords

• nonn

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..200
• Index entries for linear recurrences with constant coefficients, signature (84,-2651,41784,-364146,1838880,-5445950,9357000,-9128125,4687500,-984375).

Crossrefs

Column k=3 of A236463.

Programs

Magma

[(&+[(-1)^j*Binomial(4*n+1,j)*Binomial(7-j,4)^n: j in [0..3]]): n in [1..30]]; // G. C. Greubel, Sep 08 2022

Mathematica

With[{B=Binomial}, Table[Sum[(-1)^j*B[4*n+1,j]*B[7-j,4]^n, {j,0,3}], {n, 30}]] (* G. C. Greubel, Sep 08 2022 *)

PARI

a(n) = {35^n - (4*n + 1)*15^n + binomial(4*n+1, 2)*5^n - binomial(4*n+1, 3)} \\ Andrew Howroyd, May 07 2020

SageMath

def A151641(n): return sum((-1)^j*binomial(4*n+1,j)*binomial(7-j,4)^n for j in (0..3))
[A151641(n) for n in (1..30)] # G. C. Greubel, Sep 08 2022

Formula

a(n) = 35^n - (4*n + 1)*15^n + binomial(4*n+1, 2)*5^n - binomial(4*n+1, 3). - Andrew Howroyd, May 07 2020

From G. C. Greubel, Sep 08 2022: (Start)

a(n) = Sum_{j=0..3} (-1)^j*binomial(4*n+1, j)*binomial(7-j, 4)^n.

G.f.: 16*x^2*(1 +445*x +3485*x^2 -115215*x^3 +200675*x^4 +798375*x^5 -1890625*x^6 -703125*x^7)/( Product_{j=0..3} (1 - binomial(j+4,4)*x)^(4-j) ).

E.g.f.: exp(35*x) -(1+60*x)*exp(15*x) +50*x*(1+4*x)*exp(5*x) -(2/3)*x*(15 +48*x +16*x^2)*exp(x). (End)

Extensions

Terms a(8) and beyond from Andrew Howroyd, May 07 2020