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A107716
Inverse INVERT transform of triple factorial numbers (3*n-2)!!! (A007559).
7
1, 3, 21, 219, 2973, 49323, 964173, 21680571, 551173053, 15633866379, 489583062381, 16780438408539, 624935780160285, 25131869565110571, 1085528359404039117, 50124679063548821499, 2464153823558024331645, 128500643820213560377803, 7085182933810282490250285
OFFSET
0,2
COMMENTS
Column 0 of triangle A107717.
LINKS
FORMULA
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).
a(n) = Sum_{k, 0<=k<=n} A089949(n, k)*3^k . - Philippe Deléham, Aug 15 2005
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
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
From Peter Bala, May 23 2017: (Start)
G.f. A(x) = 1/(1 + x - 4*x/(1 + 4*x - 7*x/(1 + 7*x - 10*x/(1 + 10*x - ...)))).
A(x) = 1/(1 + x - 4*x/(1 - 3*x/(1 - 7*x/(1 - 6*x/(1 - 10*x/(1 - 9*x - ...)))))). (End)
EXAMPLE
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:
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 -...).
MAPLE
b:= proc(n) b(n):= `if`(n=0, 1, b(n-1)*(3*n+1)) end:
a:= proc(n) a(n):= -`if`(n<0, 1, add(a(n-i-1)*b(i), i=0..n)) end:
seq(a(n), n=0..20); # Alois P. Heinz, May 23 2017
MATHEMATICA
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 *)
PROG
(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)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, May 23 2005
STATUS
approved