proposed

approved

editing

proposed

a(2*n) = A038664(n). - Michel Marcus, Apr 29 2023

a(3) = 0 since no two consecutive primes differ by 3.

a(6) = 9 since k = 9 is the smallest k making pprime(k+1)-pprime(k) = 6. a(3) = 0 since no two primes differ by 3.

a(3) = 0 since no two primes differ by 3.

Cf. A001223 (prime gaps), A038664.

a(n) = least k such that f(k) = n, where f is the prime gaps function given by f(m) = pprime(m+1)-pprime(m) and pprime(m) denotes the m-th prime, if k exists; 0 otherwise.

Cf. A001223 (prime gaps).

approved

editing

_Joseph L. Pe (joseph_l_pe(AT)hotmail.com), _, Jan 03 2002

a(n) = least k such that f(k) = n, where f is the prime gaps function given by f(m) = p(m+1)-p(m) and p(m) denotes the m-th prime, if k exists; 0 otherwise.

1, 2, 0, 4, 0, 9, 0, 24, 0, 34, 0, 46, 0, 30, 0, 282, 0, 99, 0, 154, 0, 189, 0, 263, 0, 367, 0, 429, 0, 590, 0, 738, 0, 217, 0, 1183, 0, 3302, 0, 2191, 0, 1879, 0, 1831, 0, 7970, 0, 3077, 0, 3427

1,2

Obviously, a(n) = 0 for every odd n except 1. From the list, it appears that a(n) is nonzero for every even n; is this true in general? That is, for each even n, are there primes which differ by n?

a(6) = 9 since k = 9 is the smallest k making p(k+1)-p(k) = 6. a(3) = 0 since no two primes differ by 3.

f[n_] := Prime[n + 1] - Prime[n]; g[n_] := Min[Select[Range[1, 10^4], f[ # ] == n &]]; Table[g[i], {i, 1, 50}]

nonn

Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Jan 03 2002

approved