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a(n) = binomial(2n+1, n+1) - 1.
18

%I #67 Feb 19 2022 14:24:14

%S 0,2,9,34,125,461,1715,6434,24309,92377,352715,1352077,5200299,

%T 20058299,77558759,300540194,1166803109,4537567649,17672631899,

%U 68923264409,269128937219,1052049481859,4116715363799,16123801841549,63205303218875,247959266474051

%N a(n) = binomial(2n+1, n+1) - 1.

%C (With a different offset:) p divides a(p) for prime p. p^2 divides a(p) for prime p > 2. p^3 divides a(p) for prime p > 3 (implied by Wolstenholme's theorem). Wolstenholme's quotients are listed in A034602(n) = a(prime(n))/prime(n)^3 = {1, 5, 265, 2367, 237493, 2576561, 338350897, ...} = a(p)/p^3 for prime p > 3. p^3 divides a(p^k) for prime p > 3 and integer k > 0. Primes in a(n) are listed in A112862(n) = {2, 461, 92377, 269128937219, ...} Primes of the form (2*n)!/(2*(n!)^2) - 1. Numbers n such that a(n) is prime are listed in A112861(n) = {2, 6, 10, 21, 45, 63, 306, 404, 437, 471, 646, ...}. - _Alexander Adamchuk_, Jan 05 2007

%C a(n-1) is the number of weak compositions of n into n parts in which at least one part is zero. a(3)=34 since 4 can be written as 4+0+0+0 (4 such compositions); 3+1+0+0 (12 such compositions); 2+2+0+0 (6 such compositions); 2+1+1+0 (12 such compositions). All these weak compositions contain at least one zero. - _Enrique Navarrete_, Jan 09 2022

%H Vincenzo Librandi, <a href="/A010763/b010763.txt">Table of n, a(n) for n = 0..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/WolstenholmesTheorem.html">Wolstenholme's Theorem</a>

%H Jianqiang Zhao, <a href="http://arxiv.org/abs/1412.8044">Uniform Approach to Double Shuffle and Duality Relations of Various q-Analogs of Multiple Zeta Values via Rota-Baxter Algebras</a>, arXiv preprint arXiv:1412.8044 [math.NT], 2014.

%F a(n) = (n/(2n+2))*Sum_{k = 1..n+1} C(2n+2, k)/C(n+1, k). - _Benoit Cloitre_, Aug 20 2002

%F a(n) = Sum_{i = 1..n} C(n + i, n). - _Benoit Cloitre_, Oct 15 2002

%F a(n + 1) = C(2n - 1, n - 1) - 1. - _Alonso del Arte_, Dec 15 2012

%F From _Ilya Gutkovskiy_, Feb 07 2017: (Start)

%F O.g.f.: (1 - sqrt(1 - 4*x))/(2*x*sqrt(1 - 4*x)) - 1/(1 - x).

%F E.g.f.: exp(2*x)*(BesselI(0,2*x) + BesselI(1,2*x)) - exp(x). (End)

%p A010763:=n->binomial(2*n+1, n+1) - 1: seq(A010763(n), n=0..30); # _Wesley Ivan Hurt_, Sep 05 2015

%t Table[Binomial[2n - 1, n - 1] - 1, {n, 20}] (* _Alonso del Arte_, Dec 13 2012 *)

%t CoefficientList[Series[Exp[2*x]*(BesselI[0,2*x] + BesselI[1,2*x]) - Exp[x], {x, 0, 20}], x]*Table[n!, {n, 0, 20}] (* _Stefano Spezia_, Dec 02 2018 *)

%o (Magma) [Binomial(2*n-1,n-1)-1: n in [1..30]]; // _Vincenzo Librandi_, Mar 21 2013

%o (PARI) a(n) = binomial(2*n+1, n+1) - 1;

%o vector(30, n, a(n-1)) \\ _Michel Marcus_, Sep 05 2015

%o (PARI) first(n) = x='x+O('x^n); Vec((1 - sqrt(1 - 4*x))/(2*x*sqrt(1 - 4*x)) - 1/(1 - x), -n) \\ _Iain Fox_, Dec 19 2017 (corrected by _Iain Fox_, Oct 24 2018)

%Y Cf. A001700, A001701.

%Y Cf. A001008, A007406, A112861, A112862, A034602.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_