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

Author

T. D. Noe, Jul 22 2005

Keywords

• nonn
• easy

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)

Crossrefs

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

Programs

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 *)

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

Formula

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

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