A349466 Expansion of 1/((1-12*x)*(1-16*x)*(1-18*x)*(1-24*x)).

1, 70, 3100, 111160, 3529456, 103663840, 2887307200, 77450369920, 2021488750336, 51703366274560, 1302470537804800, 32436048076257280, 800745898476630016, 19636648385968660480, 479101382689537638400, 11643791435175823114240, 282140675279022464106496

Offset

0,2

Comments

a(n) = p(n+4,4)*(4!)^(n+1) where p(n+4,4) represents the probability that, given n+4 random numbers in [0, 1], there exists a 4-tuple whose sum is smaller than 1. A recurrence formula for p(n,k) is p(n, k) = (1/k)*p(n-1, k-1) + (1-1/k)*p(n-1, k). The generating function for p(n,k) is Sum_{n=k..oo} p(n,k)x^n = (x^k)/(k!*(1-x)*(1-(1/2)*x)*(1-((k-1)/k)*x)). The explicit formula for p(n,k) is p(n,k)= 1+(1/(k-1)!)*Sum_{i=1..(k-1)} ((-1)^(k-i))*binomial(k-1, i) * (i^n)* ((i+1)^(-n+k-1)).

Links

Table of n, a(n) for n=0..16.

Index entries for linear recurrences with constant coefficients, signature (70,-1800,20160,-82944).

Formula

a(n) = 24^(n+1) - (2^n)*(3^(2*n+4)) - (2^(2*n+1))*(3^(n+1)) + 2^(4*n+6).

G.f.: 1/((1-(1/2)*4!*x)*(1-(2/3)*4!*x)*(1-(3/4)*4!*x)*(1-4!*x)).

a(n) = 2^n * A028212(n) = A000079(n) * A028212(n).

E.g.f.: exp(12*x)*(24*exp(12*x) - 81*exp(6*x) + 64*exp(4*x) - 6). - Stefano Spezia, Nov 21 2021

a(n) = 24*24^n + 64*2^(4*n) - 81*18^n - 6*12^n. - Chai Wah Wu, Dec 27 2021

Mathematica

CoefficientList[Series[1/((1 - 12 x) (1 - 16 x) (1 - 18 x) (1 - 24 x)), {x, 0, 20}],  x]
LinearRecurrence[{70, -1800, 20160, -82944}, {1, 70, 3100, 111160}, 20] (* Harvey P. Dale, Apr 30 2023 *)

Prog

(Python)
def A349466(n): return 24*24**n + 64*2**(4*n) - 81*18**n - 6*12**n # Chai Wah Wu, Dec 27 2021

Crossrefs

Other sequences for p(n+k,k)*(k!)^(n+1) include: A000225 (k=2), A016765 (k=3).

Cf. A000079, A028212.

Sequence in context: A213468 A004377 A069296 * A075923 A089274 A266739

Adjacent sequences: A349463 A349464 A349465 * A349467 A349468 A349469

Keyword

nonn,easy

Author

Hsin-Hui Judy Chiang, Yifan Zhang and Wei Wang, Nov 18 2021

Status

approved