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A110494 Least k such that prime(n)^2 divides binomial(2k,k). 8

%I

%S 3,5,13,25,61,85,145,181,265,421,481,685,841,925,1105,1405,1741,1861,

%T 2245,2521,2665,3121,3445,3961,4705,5101,5305,5725,5941,6385,8065,

%U 8581,9385,9661,11101,11401,12325,13285,13945,14965,16021,16381,18241,18625,19405

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

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

%H T. D. Noe and Charles R Greathouse IV, <a href="/A110494/b110494.txt">Table of n, a(n) for n = 1..10000</a> (first 60 terms from T. D. Noe)

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

%F a(n) = A066885(n), n > 1. - _R. J. Mathar_, Aug 18 2008

%t 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]]}]

%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 *)

%o (PARI) fv(n,p)=my(s); while(n\=p, s+=n); s

%o 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

%o (PARI) a(n)=prime(n)^2\2+1 \\ _Charles R Greathouse IV_, Mar 27 2014

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

%K nonn,easy

%O 1,1

%A _T. D. Noe_, Jul 22 2005

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Last modified January 23 17:33 EST 2022. Contains 350514 sequences. (Running on oeis4.)