A110496 Least k such that prime(n)^3 divides binomial(2k,k).

7, 14, 63, 172, 666, 1099, 2457, 3430, 6084, 12195, 14896, 25327, 34461, 39754, 51912, 74439, 102690, 113491, 150382, 178956, 194509, 246520, 285894, 352485, 456337, 515151, 546364, 612522, 647515, 721449, 1024192, 1124046, 1285677

Offset: 1

Author

T. D. Noe, Jul 22 2005

Keywords

• nonn

Comments

For prime p > (2n)^(1/3), p^3 does not divide binomial(2n,n).

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A000984, A030078.

Cf. A110495 (binomial(2k, k) is cubefree).

Programs

Mathematica

t3=Table[f=FactorInteger[Binomial[2n, n]]; s=Select[f, #[[2]]>2&]; If[s=={}, 0, s[[ -1, 1]]], {n, 15000}]; Table[p=Prime[i]; First[Flatten[Position[t3, p]]], {i, PrimePi[Max[t3]]}]
lst={7};Do[AppendTo[lst,(DivisorSigma[3,Prime[n]])/2],{n,2,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 11 2009 *)

Formula

a(n) = (prime(n)^3 + 1)/2 = (1+A030078(n))/2 for n>1.

Product_{n>=1} (1 - 1/a(n)) = (54/49)*zeta(6)/zeta(3)^2. - Amiram Eldar, Jun 08 2022