1,1

Consider the prime sequence starting from 2: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, ... The first difference sequence is 1, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, ... There exists no prime sequence > 2 which has a first difference of 1. Therefore begin with the n+1 prime (in this case 3 and its first forward difference of 2) and look for the first prime above 3 with a forward difference of 2. That number is 5, so a(1) = 5.

Next start with now the third prime - 5, with two forward differences of 2,4,... The next prime above 5 which starts out with differences of 2,4 is 11. So a(2) = 11.

Next start with the fourth prime - 7, with three forward differences of 4,2,4,... The next prime above 7 which starts out with those difference is 13. So a(3) = 13.

a(9) > 1734338611, a(10) = 51448351, a(11) > 131045797.

a(4) = 101: the four primes above 101 are 103, 107, 109 & 113 with forward differences of 2, 4, 2 & 4 that matches with the differences corresponding to 11, 13, 17, 19 & 23.

Do[s = Table[Prime[i + 1] - Prime[i], {i, n + 1, 2n}]; p = 0; q = 0; a = s; k = n + 2; While[p = q; q = Prime[k]; a = Drop[a, 1]; a = Append[a, q - p]; s != a, k++ ]; Print[Prime[PrimePi[q] - n]], {n, 1, 8}]

Sequence in context: A034924 A018607 A032481 this_sequence A084864 A102801 A079030

Adjacent sequences: A073612 A073613 A073614 this_sequence A073616 A073617 A073618

hard,nonn

Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Aug 06 2002

Edited and extended by Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 09 2002