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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: A353094 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 June 14 14:51 EDT 2024. Contains 373400 sequences. (Running on oeis4.)