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%I #12 Jun 13 2022 11:05:49
%S 4,12,13,16,21,24,37,40,48,52,57,60,61,69,73,76,84,85,88,93,105,109,
%T 112,120,124,132,133,141,145,148,156,157,165,168,172,177,181,184,201,
%U 204,205,208,213,217,220,228,229,232,237,240,241,249,253,264,265,268
%N Numbers n such that the sequence of k>=2 such that k^3 divides {binomial(n*k,k)-n} coincides exactly with the sequence of prime numbers.
%C Needs to be checked. If p is prime >=m then p^3 divides {binomial(m*p,p)-m}.
%e The sequence of k>=2 such that k^3 divides {binomial(12*k,k)-12} is : 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53... which is exactly the sequence of prime numbers. Hence 12 is in the sequence.
%o (PARI) isok(n) = {my(nb = 0); for (k=2, n+10, if (!((binomial(n*k, k) - n) % k^3), nb++; if (k != prime(nb), return (0); )); ); return (1); } \\ _Michel Marcus_, Dec 06 2013; corrected Jun 13 2006
%K nonn
%O 1,1
%A _Benoit Cloitre_, Oct 02 2003
%E More terms from _Michel Marcus_, Dec 06 2013