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%I #29 May 01 2019 07:52:28
%S 1,3,21,219,2973,49323,964173,21680571,551173053,15633866379,
%T 489583062381,16780438408539,624935780160285,25131869565110571,
%U 1085528359404039117,50124679063548821499,2464153823558024331645,128500643820213560377803,7085182933810282490250285
%N Inverse INVERT transform of triple factorial numbers (3*n-2)!!! (A007559).
%C Column 0 of triangle A107717.
%H Alois P. Heinz, <a href="/A107716/b107716.txt">Table of n, a(n) for n = 0..380</a>
%F G.f.: A(x) = 1 - 1/[1 + Sum_{n>=1} (3*n-2)!!! * x^n ] where (3*n-2)!!! = Product_{k=0..n-1} (3*k+1).
%F a(n) = Sum_{k, 0<=k<=n} A089949(n, k)*3^k . - _Philippe Deléham_, Aug 15 2005
%F G.f.: (1 - Q(0))/x where Q(k) = 1 - x*(3*k+1)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Mar 20 2013
%F G.f.: 1/x - 2 - 2/x/G(0), where G(k)= 1 + 1/(1 - x*(3*k+3)/(x*(3*k+4) + 1/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 25 2013
%F From _Peter Bala_, May 23 2017: (Start)
%F G.f. A(x) = 1/(1 + x - 4*x/(1 + 4*x - 7*x/(1 + 7*x - 10*x/(1 + 10*x - ...)))).
%F A(x) = 1/(1 + x - 4*x/(1 - 3*x/(1 - 7*x/(1 - 6*x/(1 - 10*x/(1 - 9*x - ...)))))). (End)
%e The triple factorials begin: {1,4,28,280,3640,58240,...}; thus the inverse INVERT transform of the triple factorials can be calculated by the g.f.s:
%e 1/(1 + x + 4*x^2 + 28*x^3 + 280*x^4 + 3640*x^5 + 58240*x^6 +...) = (1 - x - 3*x^2 - 21*x^3 - 219*x^4 - 2973*x^5 - 49323*x^6 -...).
%p b:= proc(n) b(n):= `if`(n=0, 1, b(n-1)*(3*n+1)) end:
%p a:= proc(n) a(n):= -`if`(n<0, 1, add(a(n-i-1)*b(i), i=0..n)) end:
%p seq(a(n), n=0..20); # _Alois P. Heinz_, May 23 2017
%t m = 20; f3[n_] := Product[3k+1, {k, 0, n-1}]; A[x_] = 1-1/(1+Sum[f3[n] x^n, {n, 1, m}]); CoefficientList[A[x] + O[x]^m, x] // Rest (* _Jean-François Alcover_, May 01 2019 *)
%o (PARI) a(n)=polcoeff(1-(1+sum(k=1,n+1,prod(j=0,k-1,3*j+1)*x^k)+x^2*O(x^n))^-1,n+1)
%Y Cf. A007559, A000698, A107717.
%K nonn,easy
%O 0,2
%A _Paul D. Hanna_, May 23 2005
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