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A106271 Row sums of number triangle A106270. 2
1, 0, -2, -7, -21, -63, -195, -624, -2054, -6916, -23712, -82498, -290510, -1033410, -3707850, -13402695, -48760365, -178405155, -656043855, -2423307045, -8987427465, -33453694485, -124936258125, -467995871775, -1757900019099, -6619846420551, -24987199492703 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

From Petros Hadjicostas, Jul 15 2019: (Start)

To prove R. J. Mathar's conjecture, let A(x) be the g.f. of the current sequence. We note first that

Sum_{n >= 2} (n+1)*a(n)*x^n = (x*A(x))' - 1,

Sum_{n >= 2} (1-5*n)*a(n-1)*x^n = x*A(x) - 5*x*(x*A(x))' + 4*x, and

Sum_{n >= 2} 2*(2*n-1)*a(n-2)*x^n = 4*x*(x^2*A(x))' - 2*x^2*A(x).

Adding these equations (side by side), we get

Sum_{n >= 2} ((n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2))*x^n = 0,

which proves the conjecture. (End)

LINKS

Michael De Vlieger, Table of n, a(n) for n = 0..1669

FORMULA

G.f.: c(x)*sqrt(1 - 4x)/(1 - x), where c(x) is the g.f. of A000108.

a(n) = Sum_{k = 0..n} 2*0^(n-k) - C(n-k), where C(m) = A000108(m) (Catalan numbers).

a(n) = 2 - A014137(n) for n >= 0 and a(n) = 1 - A014138(n) for n >= 0. - Alexander Adamchuk, Feb 23 2007, corrected by Vaclav Kotesovec, Jul 22 2019

Conjecture: (n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 09 2012

a(n) ~ -2^(2*n + 2) / (3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jul 22 2019

MATHEMATICA

Table[1 - Sum[(2n)!/n!/(n+1)!, {n, 1, k}], {k, 0, 30}] (* Alexander Adamchuk, Feb 23 2007 *)

CROSSREFS

Cf. A014138, A014137 (partial sums of Catalan numbers (A000108)).

Cf. A106270.

Sequence in context: A005666 A291411 A159972 * A027990 A037520 A218836

Adjacent sequences:  A106268 A106269 A106270 * A106272 A106273 A106274

KEYWORD

easy,sign

AUTHOR

Paul Barry, Apr 28 2005

EXTENSIONS

More terms from Alexander Adamchuk, Feb 23 2007

STATUS

approved

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Last modified September 23 04:52 EDT 2021. Contains 347609 sequences. (Running on oeis4.)