A114058 Start of record gap in even semiprimes (A100484).

4, 6, 14, 46, 178, 226, 1046, 1774, 2258, 2654, 19102, 31366, 39218, 62794, 311842, 721306, 740522, 984226, 2699066, 2714402, 4021466, 9304706, 34103414, 41662646, 94653386, 244329494, 379391318, 383825566, 774192266

Offset: 1

Author

Jonathan Vos Post, Feb 02 2006

Keywords

• easy
• nonn

Comments

5 of the first 6 values of record gaps in even semiprimes are also record merits = (A100484(k+1)-A100484(k))/log_10(A100484(k)), namely: (6 - 4) / log_10(4) = 3.32192809; (10 - 6) / log_10(6) = 5.14038884; (22 - 14) / log_10(14) = 6.98002296; (58 - 46) / log_10(46) = 7.21692586; (254 - 226) / log_10(226) = 11.8940995. It is easy to prove that there are gaps of arbitrary length in even semiprimes (A100484), as 2*(n!+2), 2*(n!+3), 2*(n!+4), ..., 2*(n!+n) gives (n-1) consecutive even nonsemiprimes. Can we prove that there are gaps of arbitrary length in odd semiprimes (A046315) and in semiprimes (A001358)?

For every n, a(n) = 2*A002386(n). - John W. Nicholson, Jul 26 2012

Examples

			gap[a(1)] = A100484(2)-A100484(1) = 6 - 4 = 2.
gap[a(2)] = A100484(3)-A100484(2) = 10 - 6 = 4.
gap[a(3)] = A100484(5)-A100484(4) = 22 - 14 = 8.
gap[a(4)] = A100484(10)-A100484(9) = 58 - 46 = 12.
gap[a(5)] = A100484(25)-A100484(24) = 194 - 178 = 16.
gap[a(6)] = A100484(31)-A100484(30) = 254 - 226 = 28.
		

Links

• John W. Nicholson, Table of n, a(n) for n = 1..75

Crossrefs

Cf. A001358, A046315, A065516, A085809, A100484, A114412, A114021. Maximal gap small prime A002386.

Programs

Mathematica

f[n_] := Block[{k = n + 2}, While[ Plus @@ Last /@ FactorInteger@k != 2, k += 2]; k]; lst = {}; d = 0; a = b = 4; Do[{a, b} = {b, f[a]}; If[b - a > d, d = b - a; AppendTo[lst, a]], {n, 10^8}]; lst (* Robert G. Wilson v *)

Formula

a(n) = A100484(k) such that A100484(k+1)-A100484(k) is a record.

Extensions

a(7)-a(25) from Robert G. Wilson v, Feb 03 2006

a(26)-a(31) from Donovan Johnson, Mar 14 2010