A178792 Dot product of the rows of triangle A046899 with vector (1,2,4,8,...) (= A000079).
1, 5, 31, 209, 1471, 10625, 78079, 580865, 4361215, 32978945, 250806271, 1916280833, 14698053631, 113104519169, 872801042431, 6751535300609, 52337071357951, 406468580343809, 3162019821780991, 24634626678980609, 192179216026959871
Offset: 0
Examples
a(3) = (1,4,10,20)dot(1,2,4,8) = 209.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- J. Abate and W. Whitt, Brownian motion and generalized Catalan numbers, Journal of Integer Sequences, Vol. 14 (2011).
- Karl Dilcher and Maciej Ulas, Arithmetic properties of polynomial solutions of the Diophantine equation P(x)x^(n+1)+Q(x)(x+1)^(n+1) = 1, arXiv:1909.11222 [math.NT], 2019.
- Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO], 2023.
Programs
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Maple
a := n -> binomial(2*n+2,n+1)*hypergeom([-n, n + 1], [n + 2], -1)/2: seq(simplify(a(n)), n=0..20); # Peter Luschny, Feb 21 2017
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Mathematica
CoefficientList[Series[(3-Sqrt[1-8*x])/(2*(1+x)*Sqrt[1-8*x]), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *) Table[Sum[2^k*Binomial[n + k, k], {k, 0, n}], {n, 0, 20}] (* Michael De Vlieger, Oct 28 2016 *) a[n_] := (-1)^(n + 1) - 2^(n + 1) (2n + 1) Binomial[2n, n] Hypergeometric2F1[1, 2n + 2, n + 2, 2]/(n + 1); Array[a, 22, 0] (* Robert G. Wilson v, Jul 21 2018 *)
Formula
a(n) = Sum_{k = 0..n} A046899(n,k)*2^k = Sum_{k = 0..n} 2^k * binomial(n+k,k).
G.f.: (1/3)*(4/sqrt(1 - 8*x) - 1/(1 - x*c(2*x))) with c(x) the g.f. of the Catalan numbers A000108.
a(n) + a(n+1) = 6*2^n*A001700(n).
O.g.f.: (3 - sqrt(1 - 8*x))/(2*(1 + x)*sqrt(1 - 8*x)). - Peter Bala, Apr 10 2012
a(n) = 2^n *binomial(2+2*n,1+n)*2F1(1, 2+2*n; 2+n;-1). - Olivier Gérard, Aug 19 2012
D-finite with recurrence n*a(n) = (7*n - 4)*a(n-1) + 4*(2*n - 1)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(3n+2)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 20 2012
a(n) = Sum_{k = 0..n} binomial(k+n,n) * binomial(2*n+1,n-k). - Vladimir Kruchinin, Oct 28 2016
a(n) = 1/2*(n + 1)*binomial(2*n+2,n+1)*Sum_{k = 0..n} binomial(n,k)/(n + k + 1). - Peter Bala, Feb 21 2017
a(n) = binomial(2*n+2,n+1)*hypergeom([-n, n+1], [n+2], -1)/2. - Peter Luschny, Feb 21 2017
a(n) = (-1)^(n+1) - 2^(n+1)*(2*n+1)*binomial(2*n,n)*hypergeom([1, 2*n+2], [n+2], 2)/(n+1). - John M. Campbell, Jul 14 2018
From Akiva Weinberger, Dec 06 2024: (Start)
a(n) = (2*n + 1)!/(n!^2) * Integral_{t=0..1} (t + t^2)^n dt.
a(n) = (Integral_{t=0..1} (t + t^2)^n dt) / (Integral_{t=0..1} (t - t^2)^n dt). (End)
a(n) = (-1)^n * Sum_{k=0..n} binomial(2*n+1,k) * (-2)^k. - André M. Timpanaro, Dec 15 2024
From Seiichi Manyama, Aug 04 2025: (Start)
a(n) = [x^n] (1+x)^(2*n+1)/(1-x)^(n+1).
a(n) = [x^n] 1/((1-x) * (1-2*x)^(n+1)). (End)
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