A129515 Numbers m such that binomial(2*m, m) has the same prime factors as binomial(2*k, k) for some k > m.

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

Author

T. D. Noe, Apr 18 2007

Keywords

• nonn
• changed

Comments

The Erdős paper mentions 87 and 607. The paper conjectures that the sequence is infinite. For the m listed here, k=m+1. Note that we need only examine k such that pi(2*m) = pi(2*k), 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)
• Thomas Bloom, Problem 730, Erdős Problems.
• 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.

Crossrefs

Cf. A067434 (number of distinct prime factors in binomial(2n, n)).

Programs

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

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