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A129515
Numbers n such that binomial(2n,n) has the same prime factors as binomial(2k,k) for some k>n.
1
87, 199, 237, 467, 607, 967, 1127, 1319, 1483, 1903, 1943, 2012, 2047, 2287, 2348, 2359, 2464, 2479, 2495, 2507, 2623, 2645, 2719, 3349, 3467, 3514, 3568, 3629, 3633, 3712, 3847, 3919, 4088, 4224, 4287, 4360, 4479, 4927, 4987, 5087, 5167, 5224, 5669
OFFSET
1,1
COMMENTS
The Erdos paper mentions 87 and 607. The paper conjectures that the sequence is infinite. For the n listed here, k=n+1. Note that we need only examine k such that pi(2n) = pi(2k), where pi is the prime counting function.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (corrected and extended from original by T. D. Noe)
P. Erdős, R. L. Graham, I. Z. Russa and E. G. Straus, On the prime factors of C(2n,n), Math. Comp. 29 (1975), 83-92.
MATHEMATICA
s={}; nLst={}; t={}; Do[p=Transpose[FactorInteger[Binomial[2n, n]]][[1]]; If[s!={} && p[[ -1]]!=s[[ -1, -1]], s={}; nLst={}]; pos=Position[s, p, 1, 1]; If[pos!={}, m=pos[[1, 1]]; AppendTo[t, nLst[[m]]], AppendTo[s, p]; AppendTo[nLst, n]], {n, 10000}]; t
PROG
(PARI) valp(n, p)=my(s); while(n\=p, s+=n); s
f(n, p)=valp(2*n, p)==2*valp(n, p)
is(n)=for(k=n+1, nextprime(2*n)\2, forprime(p=2, 2*n, if(f(n, p)!=f(k, p), next(2))); return(k)); 0 \\ Charles R Greathouse IV, Oct 18 2017
CROSSREFS
Cf. A067434 (number of distinct prime factors in binomial(2n, n)).
Sequence in context: A063349 A101259 A063336 * A133524 A304509 A020314
KEYWORD
nonn
AUTHOR
T. D. Noe, Apr 18 2007
STATUS
approved