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%I #32 Apr 30 2023 11:30:39
%S 1,70,3100,111160,3529456,103663840,2887307200,77450369920,
%T 2021488750336,51703366274560,1302470537804800,32436048076257280,
%U 800745898476630016,19636648385968660480,479101382689537638400,11643791435175823114240,282140675279022464106496
%N Expansion of 1/((1-12*x)*(1-16*x)*(1-18*x)*(1-24*x)).
%C 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)).
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (70,-1800,20160,-82944).
%F a(n) = 24^(n+1) - (2^n)*(3^(2*n+4)) - (2^(2*n+1))*(3^(n+1)) + 2^(4*n+6).
%F G.f.: 1/((1-(1/2)*4!*x)*(1-(2/3)*4!*x)*(1-(3/4)*4!*x)*(1-4!*x)).
%F a(n) = 2^n * A028212(n) = A000079(n) * A028212(n).
%F E.g.f.: exp(12*x)*(24*exp(12*x) - 81*exp(6*x) + 64*exp(4*x) - 6). - _Stefano Spezia_, Nov 21 2021
%F a(n) = 24*24^n + 64*2^(4*n) - 81*18^n - 6*12^n. - _Chai Wah Wu_, Dec 27 2021
%t CoefficientList[Series[1/((1 - 12 x) (1 - 16 x) (1 - 18 x) (1 - 24 x)), {x, 0, 20}], x]
%t LinearRecurrence[{70,-1800,20160,-82944},{1,70,3100,111160},20] (* _Harvey P. Dale_, Apr 30 2023 *)
%o (Python)
%o def A349466(n): return 24*24**n + 64*2**(4*n) - 81*18**n - 6*12**n # _Chai Wah Wu_, Dec 27 2021
%Y Other sequences for p(n+k,k)*(k!)^(n+1) include: A000225 (k=2), A016765 (k=3).
%Y Cf. A000079, A028212.
%K nonn,easy
%O 0,2
%A _Hsin-Hui Judy Chiang_, Yifan Zhang and Wei Wang, Nov 18 2021
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