A094664 Row sums of triangle A094344.
1, 1, 2, 7, 38, 286, 2756, 32299, 444998, 7038898, 125620652, 2495811814, 54618201884, 1305184303996, 33812846036552, 943878836768947, 28242424937855558, 901709392642750186, 30597227032818965276, 1099566630423067201234, 41718229482624755005748
Offset: 0
Examples
a(3) = 7, a(4) = 38, since top row of M^3 = (7, 7, 9, 15) with 38 = (7 + 7 + 9 + 15).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..200
Programs
-
Mathematica
nmax = 20; CoefficientList[Series[1/Fold[(1 - #2/#1) &, 1, Reverse[(2*Range[nmax + 1] - 2*Floor[Range[nmax + 1]/2] - 1)*x]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 05 2017 *) -
PARI
{a(n)=local(CF=1+x*O(x^n)); for(k=0, n, CF=1/(1-(2*n-2*k+1)*x/(1-(2*n-2*k+1)*x*CF))); polcoeff(CF, n, x)} /* Paul D. Hanna, Sep 17 2011 */ -
PARI
{a(n)=local(A=1+x);for(i=1,n,A=1+x*(2*A-A^2)+2*x^2*A'+x*O(x^n));polcoeff(A,n)} \\ Paul D. Hanna, Mar 09 2013
Formula
a(n) = Sum_{k = 0..n} A094344(n, k).
From Gary W. Adamson, Jul 26 2011: (Start)
a(n) = upper left term in M^n, a(n+1) = sum of top row terms in M^n; M = the following infinite square production matrix:
1, 1, 0, 0, 0, ...
1, 1, 3, 0, 0, ...
1, 1, 1, 5, 0, ...
1, 1, 1, 1, 7, ...
... (End)
G.f.: 1/(1 - x/(1 - x/(1 - 3*x/(1 - 3*x/(1 - 5*x/(1 - 5*x/(1 - 7*x/(1 - 7*x/(1-...))))))))) (continued fraction). - Paul D. Hanna, Sep 17 2011
G.f. A(x) satisfies A(x) = 1 + x*(2*A(x)-A(x)^2) + 2*x^2*A'(x). - Paul D. Hanna, Mar 09 2013
From Sergei N. Gladkovskii, Oct 15 2012 - Aug 14 2013: (Start)
Continued fractions:
G.f.: 1/U(0) where U(k) = 1 - x*(2*k+1)/(1 - x*(2*k+1)/U(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+3)/Q(k+1) ).
G.f.: Q(0)/x - 1/x, where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+1)/Q(k+1)).
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(4*k+2)/(x*(4*k+2)-1+ x*(4*k+2)/G(k+1))).
G.f.: G(0)/2/x - 1/x + 2, where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) - 1 + 2*x*(2*k-1)/G(k+1))).
G.f.: G(0), where G(k) = 1-x*(2*k+1)/(x*(2*k+1)-1/(1-x*(2*k+1)/(x*(2*k+1)- 1/G(k+1)))).
G.f.: 2 - 1/x - G(0)/x, where G(k) = 2*x - 2*x*k - 1 - x*(2*k-1)/G(k+1).
(End)
a(n) ~ 2^n * (n-1)! / Pi. - Vaclav Kotesovec, Sep 05 2017
Conjecture: a(n) = R(n-1, 0) for n > 0 with a(0) = 1 where R(n, q) = (2*q + 1)*R(n-1, q+1) + Sum_{j=0..q} R(n-1, j) for n > 0, q >= 0 with R(0, q) = 1 for q >= 0. - Mikhail Kurkov, Jun 19 2023