A110494 Least k such that prime(n)^2 divides binomial(2k,k).

3, 5, 13, 25, 61, 85, 145, 181, 265, 421, 481, 685, 841, 925, 1105, 1405, 1741, 1861, 2245, 2521, 2665, 3121, 3445, 3961, 4705, 5101, 5305, 5725, 5941, 6385, 8065, 8581, 9385, 9661, 11101, 11401, 12325, 13285, 13945, 14965, 16021, 16381, 18241, 18625, 19405

Offset

1,1

Comments

For prime p > sqrt(2n), p^2 does not divide binomial(2n,n).

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 60 terms from T. D. Noe)

Formula

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

a(n) = A066885(n), n > 1. - R. J. Mathar, Aug 18 2008

Mathematica

t=Table[f=FactorInteger[Binomial[2n, n]]; s=Select[f, #[[2]]>1&]; If[s=={}, 0, s[[ -1, 1]]], {n, 100}]; Table[p=Prime[i]; First[Flatten[Position[t, p]]], {i, PrimePi[Max[t]]}]
lk[n_]:=Module[{k=1, c=Prime[n]^2}, While[!Divisible[Binomial[2k, k], c], k=k+2]; k]; Array[lk, 40] (* Harvey P. Dale, Oct 10 2012 *)

Prog

(PARI) fv(n, p)=my(s); while(n\=p, s+=n); s
a(n)=my(p=prime(n), k=p^2\2+1); while(fv(2*k, p)-2*fv(k, p)<2, k++); k \\ Charles R Greathouse IV, Mar 27 2014
(PARI) a(n)=prime(n)^2\2+1 \\ Charles R Greathouse IV, Mar 27 2014

Crossrefs

Cf. A110493 (largest prime p such that p^2 divides binomial(2n, n)).

Sequence in context: A219699 A320330 A159290 * A098615 A026720 A026003

Adjacent sequences: A110491 A110492 A110493 * A110495 A110496 A110497

Keyword

nonn,easy

Author

T. D. Noe, Jul 22 2005

Status

approved