A087760 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.

4, 12, 13, 16, 21, 24, 37, 40, 48, 52, 57, 60, 61, 69, 73, 76, 84, 85, 88, 93, 105, 109, 112, 120, 124, 132, 133, 141, 145, 148, 156, 157, 165, 168, 172, 177, 181, 184, 201, 204, 205, 208, 213, 217, 220, 228, 229, 232, 237, 240, 241, 249, 253, 264, 265, 268

Offset

1,1

Comments

Needs to be checked. If p is prime >=m then p^3 divides {binomial(m*p,p)-m}.

Links

Table of n, a(n) for n=1..56.

Example

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.

Prog

(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

Crossrefs

Sequence in context: A074138 A120252 A103053 * A237279 A032498 A144613

Adjacent sequences: A087757 A087758 A087759 * A087761 A087762 A087763

Keyword

nonn

Author

Benoit Cloitre, Oct 02 2003

Extensions

More terms from Michel Marcus, Dec 06 2013

Status

approved