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Start of record gap in even semiprimes (A100484).
1

%I #18 Oct 31 2013 12:17:36

%S 4,6,14,46,178,226,1046,1774,2258,2654,19102,31366,39218,62794,311842,

%T 721306,740522,984226,2699066,2714402,4021466,9304706,34103414,

%U 41662646,94653386,244329494,379391318,383825566,774192266

%N Start of record gap in even semiprimes (A100484).

%C 5 of the first 6 values of record gaps in even semiprimes are also record merits = [A100484(k+1)-A100484(k)]/log(A100484(k)], namely: (6 - 4) / log(4) = 3.32192809; (10 - 6) / log(6) = 5.14038884; (22 - 14) / log(14) = 6.98002296; (58 - 46) / log(46) = 7.21692586; (254 - 226) / log(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)?

%C For every n, a(n) = 2*A002386(n). [John W. Nicholson, Jul 26 2012]

%H John W. Nicholson, <a href="/A114058/b114058.txt">Table of n, a(n) for n = 1..75</a>

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

%e gap[a(1)] = A100484(2)-A100484(1) = 6 - 4 = 2.

%e gap[a(2)] = A100484(3)-A100484(2) = 10 - 6 = 4.

%e gap[a(3)] = A100484(5)-A100484(4) = 22 - 14 = 8.

%e gap[a(4)] = A100484(10)-A100484(9) = 58 - 46 = 12.

%e gap[a(5)] = A100484(25)-A100484(24) = 194 - 178 = 16.

%e gap[a(6)] = A100484(31)-A100484(30) = 254 - 226 = 28.

%t 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_ *)

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

%K easy,nonn

%O 1,1

%A _Jonathan Vos Post_, Feb 02 2006

%E a(7)-a(25) from _Robert G. Wilson v_, Feb 03 2006

%E a(26)-a(31) from _Donovan Johnson_, Mar 14 2010