cp's OEIS Frontend

A178811 The smallest integer that begins the longest run of consecutive integers with the prime signature of A025487(n).

%I A178811 #48 Dec 02 2023 09:50:27
%S A178811 1,2,4,33,8,10093613546512321,16,28375,1309,32,36,7939375,
%T A178811 932537185321,64
%N A178811 The smallest integer that begins the longest run of consecutive integers with the prime signature of A025487(n).
%C A178811 From _Bernard Schott_, Feb 17 2021: (Start)
%C A178811 The corresponding lengths of these longest runs of consecutive integers are in A178810.
%C A178811 If a(n) = 2^k for some k <> 1, then a(n) = A025487(n) and A178810(n) = 1; for k = 1, a(2) = A025487(2) = A178810(2) = 2 because there exists a run of two consecutive primes (2,3).
%C A178811 a(18) = 128, a(22) = 203433. [corrected by _Jon E. Schoenfield_, Nov 30 2023] (End)
%C A178811 From _Jon E. Schoenfield_, Dec 02 2023: (Start)
%C A178811 a(16) = 3302209375 = 5^5 * 1056707. (3302209376 = 2^5 * 103194043, 3302209377 = 3^5 * 13589339.) No run of four consecutive integers of the form p^5 * q with p,q distinct primes can exist: one of the two even numbers would be 32*q and the other would be 2*p^5, and they would differ by 2, yielding either (1) 2*p^5 + 2 = 32*q -> p^5 + 1 = 16*q -> (p^4 - p^3 + p^2 - p + 1)*(p+1) = 16*q, so p+1 = 16, but then p = 15 would not be a prime, or (2) 2*p^5 - 2 = 32*q -> p^5 - 1 = 16*q -> (p^4 + p^3 + p^2 + p + 1)*(p-1) = 16*q, so p-1 = 16, so p = 17, but then the odd number between 2*p^5 and 32*q would be 2*17^5 - 1 = 2839713 = 3 * 37 * 25583 (which would not have the required prime signature).
%C A178811 a(17) <= 921198089181020748838245 (which starts a run of seven consecutive integers of the form p^3*q*r; no run of eight or more can exist, as any set of eight consecutive integers includes an odd multiple of 4).
%C A178811 a(n) = A025487(n) if A025487(n) is a proper power (i.e., a number of the form b^e where b,e > 1). (Thus a(3) = 4, a(5) = 8, a(7) = 16, a(10) = 32, a(11) = 36, a(14) = 64, a(18) = 128, a(19) = 144, a(23) = 216, a(25) = 256, a(32) = 512, a(33) = 576, a(38) = 900, a(40) = 1024, a(44) = 1296, a(48) = 1728, a(51) = 2048, a(53) = 2304.)
%C A178811 Conjecture: a(n) = A025487(n) if A025487(n) is a powerful number (A001694); i.e., if A025487(n) is a powerful number, then there exists no run of two or more consecutive integers with the same prime signature as that of A025487(n). (E.g., if this conjecture holds, a(15) = 72 (cf. A367781), a(26) = 288, a(30) = 432, a(37) = 864, a(42) = 1152, a(49) = 1800.) (End)
%H A178811 Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a1-pot-pourri/2911-a1845-les-squelettes">A1845, Les squelettes</a> (in French).
%e A178811 For n = 3, A025487(3) = 4, corresponding to a prime signature of {2}. Since the maximum number of consecutive integers with that prime signature is 1, a(3) is 4, the smallest integer that starts a "run" of 1.
%e A178811 A025487(4) = 6 whose prime signature is {1,1}; a(4) = 33 because 33 is the smallest integer where starts a run of A178810(4) = 3 consecutive integers with prime signature {1,1}: (33=3*11, 34=2*17, 35=5*7). - _Bernard Schott_, Feb 16 2021
%Y A178811 Cf. A001694, A025487, A060355, A178810 (maximum size of such runs), A141621.
%K A178811 more,nonn
%O A178811 1,2
%A A178811 _Will Nicholes_, Jun 16 2010
%E A178811 Minor edits by _Ray Chandler_, Jul 29 2010
%E A178811 a(6) corrected by _Bobby Jacobs_, Sep 25 2016
%E A178811 a(12) from _Hugo van der Sanden_, May 20 2019
%E A178811 a(13)-a(14) from _Bernard Schott_, Feb 16 2021