A066796 a(n) = Sum_{i=1..n} binomial(2*i,i).

2, 8, 28, 98, 350, 1274, 4706, 17576, 66196, 250952, 956384, 3660540, 14061140, 54177740, 209295260, 810375650, 3143981870, 12219117170, 47564380970, 185410909790, 723668784230, 2827767747950, 11061198475550, 43308802158650

Offset: 1

Author

Benoit Cloitre, Jan 18 2002

Keywords

• nonn

Comments

Comments from Alexander Adamchuk, Jul 02 2006: (Start)

Every a(n) is divisible by prime 2, a(n)/2 = A079309(n).

a(n) is divisible by prime 3 only for n=12,30,36,84,90,108,120,... A083096.

a(p) is divisible by p^2 for primes p=5,11,17,23,29,41,47,... Primes of form 6n-1. A007528.

a(p-1) is divisible by p^2 for primes p=7,13,19,31,37,43,... Primes of form 6n+1. A002476.

Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p=7,13,19,31,37,43,... Primes of form 6n+1. A002476.

Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p>3.

a(p^2-1), a(p^2-2) and a(p^2-3) are divisible by p^2 for prime p>3.

a(p^2-4) is divisible by p^2 for prime p>5.

a(p^2-5) is divisible by p^2 for prime p>7.

a(p^2-6) is divisible by p^2 for prime p>7.

a(p^2-7) is divisible by p^2 for prime p>11.

a(p^2-8) is divisible by p^2 for prime p>13.

a(p^3) is divisible by p^2 for prime 2 and prime p=5,11,... Primes of form 6n-1. A007528.

a(p^3-1) is divisible by p^2 for prime p=7,13,... Primes of form 6n+1. A002476.

a(p^4-1) is divisible by p^2 for prime p>3. (End)

Mod[ a(3^k), 9 ] = 1 for integer k>0. Smallest number k such that 2^n divides a(k) is k(n) = {1,2,2,11,11,46,46,707,707,707,...}. Smallest number k such that 3^n divides a(k) is k(n) = {12,822,2466,...}. a(2(p-1)/3) is divisible by p^2 for prime p = {7,13,19,31,37,43,61,...} = A002476 Primes of form 6n+1. Every a(n) from a(p^2-(p+1)/2) to a(p^2-1) is divisible by p^2 for prime p>3. Every a(n) from a((4p+3)(p-1)/6) to a((2p+3)(p-1)/3) is divisible by p^2 for prime p = {7,13,19,31,37,43,61,...} = A002476 Primes of form 6n+1. - Alexander Adamchuk, Jan 04 2007

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000 (Terms 1 to 200 computed by Harry J. Smith; terms 201 to 1000 computed by G. C. Greubel, Jan 15 2017)
• Guo-Shuai Mao, Proof of a conjecture of Adamchuk, arXiv:2003.09810 [math.NT], 2020.
• Guo-Shuai Mao, On a supercongruence conjecture of Z.-W. Sun, arXiv:2003.14221 [math.NT], 2020.
• Guo-Shuai Mao, On some supercongruence conjectures of Z.-W. Sun, Nanjing Univ. Info. Sci. Tech. (China, 2023).
• Guo-Shuai Mao, Proof of some congruences via the hypergeometric identities, Nanjing Univ. Info. Sci. Tech. (China, 2023).
• Guo-Shuai Mao, On two pairs of congruence conjectures of Z.-W. Sun, Nanjing Univ. Sci. Tech. (China), ResearchGate, 2024.
• Guo-Shuai Mao and Roberto Tauraso, Three pairs of congruences concerning sums of central binomial coefficients, arXiv:2004.09155 [math.NT], 2020.
• Guo-Shuai Mao and Dong-Hui Zhang, On some congruences involving harmonic numbers and Bernoulli polynomials, ResearchGate, 2023. See pp. 1-2, 12.
• Z.-W. Sun, Fibonacci numbers modulo cubes of primes, arXiv:0911.3060 [math.NT], 2009-2013; Taiwanese J. Math., to appear 2013. - From _N. J. A. Sloane_, Mar 01 2013
• Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
• Eric Weisstein's World of Mathematics, Binomial Sums.

Crossrefs

Essentially the same as A079309 and A054114.

Equals A006134 - 1.

Cf. A002476, A006134, A007528, A079309, A083096.

Programs

Mathematica

Table[Sum[(2k)!/(k!)^2,{k,1,n}],{n,1,50}] (* Alexander Adamchuk, Jul 02 2006 *)
Table[Sum[Binomial[2k,k],{k,1,n}],{n,1,30}] (* Alexander Adamchuk, Jan 04 2007 *)

PARI

{ a=0; for (n=1, 200, write("b066796.txt", n, " ", a+=binomial(2*n, n)) ) } \\ Harry J. Smith, Mar 27 2010

PARI

a(n) = sum(i=1, n, binomial(2*i,i)); \\ Michel Marcus, Jan 04 2016

Formula

a(n) = A006134(n) - 1; generating function: (sqrt(1-4*x)-1)/(sqrt(1-4*x)*(x-1)) - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 11 2003, corrected by Vaclav Kotesovec, Nov 06 2012

a(n) = Sum_{k=1..n}(2k)!/(k!)^2. - Alexander Adamchuk, Jul 02 2006

a(n) = Sum_{k=1..n}binomial(2k,k). - Alexander Adamchuk, Jan 04 2007

a(n) ~ 2^(2*n+2)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Nov 06 2012