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%I #55 Sep 08 2022 08:44:41
%S 1,12,97,660,4081,23772,133057,724260,3863761,20308332,105558817,
%T 544039860,2785713841,14192221692,72020501377,364354427460,
%U 1838822866321,9262446387852,46585947584737
%N Expansion of 1/((1-3*x)*(1-4*x)*(1-5*x)).
%C As (0,0,1,12,97,...) this is the fourth binomial transform of cosh(x)-1. It is the binomial transform of A016269, when this has two leading zeros. Its e.g.f. is then exp(4x)cosh(x) - exp(4x). - _Paul Barry_, May 13 2003
%C This gives the third column of the Sheffer triangle A143495 (3-restricted Stirling2 numbers). See the e.g.f. below, and A193685 for comments on the general case. - _Wolfdieter Lang_, Oct 08 2011
%C From _Kevin Long_, Mar 25 2017: (Start)
%C In the power set poset 2^(n+2), a(n) gives the number of size 3 subposets {A,B,C} such that A subset of C, B subset of C, and A||B. By symmetry, it also counts the size 3 subposets {A,B,C} such that C subset of A, C subset of B, and A||B.
%C By the power set poset, I mean the subsets of [n+2] ordered by inclusion. A||B means A and B are incomparable.
%C The result can be proved by showing that the formula holds. 5^n counts triples (A,B,C) of subsets of [n] where A subset of C and B subset of C, since for each x in [n], it is either in C only, in A and C, in B and C, in all three, or in none. However, this also counts the cases where A subset of B and where B subset of A, and we want A||B.
%C Each case can be counted by 4^n, since if A subset of B⊆C, then each element x of [n] is either in all three, in B and C, in only C, or in none. Hence we subtract 2*4^n from 5^n. These two cases intersect, however, when A = B subset of C, which can be counted by 3^n, since each element x of [n] can be either in all three sets, in only C, or in none.
%C For the purposes of inclusion-exclusion, we add these sets back in to get 5^n-2*4^n+3^n to count all triples (A,B,C) where A subset of C, B subset of C, and A||B. We want sets, not triples, so this double-counts the sets since interchanging A and B give the same set, so we divide this by 2. Hence the formula for a(n) counts these subposets for 2^(n+2). (End)
%H G. C. Greubel, <a href="/A016753/b016753.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (12,-47,60).
%F a(n) = 5^(n+2)/2 - 4^(n+2) + 3^(n+2)/2. - _Paul Barry_, May 13 2003
%F If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-2) = f(n,2,3), (n >= 2). - _Milan Janjic_, Apr 26 2009
%F a(n) = 9*a(n-1) - 20*a(n-2) + 3^n, n >= 2. - _Vincenzo Librandi_, Mar 20 2011
%F O.g.f.: 1/((1-3*x)*(1-4*x)*(1-5*x)).
%F E.g.f.: (d^2/dx^2) (exp(3*x)*((exp(x)-1)^2)/2!). - _Wolfdieter Lang_, Oct 08 2011
%F a(n) = A245019(n+2)/2. - _Kevin Long_, Mar 24 2017
%t CoefficientList[ Series[ 1/((1 - 3x)(1 - 4x)(1 - 5x)), {x, 0, 25} ], x ]
%t LinearRecurrence[{12,-47,60}, {1, 12, 97}, 30] (* _G. C. Greubel_, Sep 15 2018 *)
%o (PARI) Vec(1/((1-3*x)*(1-4*x)*(1-5*x))+O(x^99)) \\ _Charles R Greathouse IV_, Sep 26 2012
%o (Magma) [(5^(n+2) - 2*4^(n+2) + 3^(n+2))/2: n in [0..30]]; // _G. C. Greubel_, Sep 15 2018
%Y Cf. A000244, A005061, A016753, A132495, A193685, A245019.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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