A186927 - OEIS

%I #23 Nov 11 2016 04:38:06

%S 1,2,3,8,27,243,2048,524288,129140163,68630377364883,

%T 36472996377170786403,19342813113834066795298816,

%U 706965049015104706497203195837614914543357369,13703277223523221219433362313025801636536040755174924956117940937101787

%N Lesser of two consecutive 3-smooth numbers having no common divisors.

%C a(n) = A003586(A186771(n)); A186928(n) = A003586(A186771(n) + 1).

%C Subsequence of A006899: all terms are either powers of 2 or of 3.

%C Najman improves an algorithm of Bauer & Bennett for computing the function that measures the minimal gap size f(k) in the sequence of integers at least one of whose prime factors exceeds k. This allows us to compute values of f(k) for larger k and obtain new values of f(k). - _Jonathan Vos Post_, Aug 18 2011

%H Charles R Greathouse IV, <a href="/A186927/b186927.txt">Table of n, a(n) for n = 1..21</a>

%H M. Bauer and M. A. Bennett, <a href="http://www.math.ubc.ca/~bennett/erdos-graham3.pdf">Prime factors of consecutive integers</a>, Mathematics of Computation 77 (2008), pp. 2455-2459.

%H Charles R Greathouse IV, <a href="/A186927/a186927.txt">Illustration of n, a(n) for n = 1..33</a>

%H Filip Najman, <a href="http://arxiv.org/abs/1108.3710">Large strings of consecutive smooth integers</a>, Aug 18, 2011

%t smoothNumbers[p_, max_] := Module[{a, aa, k, pp, iter}, k = PrimePi[p]; aa = Array[a, k]; pp = Prime[Range[k]]; iter = Table[{a[j], 0, PowerExpand @ Log[pp[[j]], max/Times @@ (Take[pp, j - 1]^Take[aa, j - 1])]}, {j, 1, k}]; Table[Times @@ (pp^aa), Sequence @@ iter // Evaluate] // Flatten // Sort]; sn = smoothNumbers[3, 10^100]; Reap[For[i = 1, i <= Length[sn] - 1, i++, If[CoprimeQ[sn[[i]], sn[[i + 1]]], Sow[sn[[i]]]]]][[2, 1]] (* _Jean-François Alcover_, Nov 11 2016 *)

%Y Cf. A186711.

%K nonn

%O 1,2

%A _Charles R Greathouse IV_ and _Reinhard Zumkeller_, Mar 01 2011