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A110494 Least k such that prime(n)^2 divides binomial(2k,k). 8
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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

LINKS

T. D. Noe and 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

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Last modified September 15 04:09 EDT 2019. Contains 327062 sequences. (Running on oeis4.)