1,1

Comments from Alexander Adamchuk (alex(AT)kolmogorov.com), 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 (alex(AT)kolmogorov.com), Jan 04 2007

Eric Weisstein's World of Mathematics, Central Binomial Coefficient.

Eric Weisstein's World of Mathematics, Binomial Sums.

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

a(n) = Sum[ (2k)!/(k!)^2, {k,1,n} ]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 02 2006

a(n) = Sum[ Binomial[2k,k], {k,1,n} ]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jan 04 2007

Table[Sum[(2k)!/(k!)^2, {k, 1, n}], {n, 1, 50}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 02 2006

Table[Sum[Binomial[2k, k], {k, 1, n}], {n, 1, 30}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jan 04 2007

Essentially the same as A079309 and A054114.

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

Sequence in context: A090426 A060995 A106731 this_sequence A104934 A056711 A114590

Adjacent sequences: A066793 A066794 A066795 this_sequence A066797 A066798 A066799

nonn

Benoit Cloitre (benoit7848c(AT)orange.fr), Jan 18 2002