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A113619 Heptagon-free numbers. 1

%I #14 Jun 24 2017 02:34:04

%S 1,2,3,4,5,6,8,9,10,11,12,13,15,16,17,19,20,22,23,24,25,26,27,29,30,

%T 31,32,33,37,38,39,40,41,43,44,45,46,47,48,50,51,52,53,57,58,59,60,61,

%U 62,64,65,66,67,69,71,73,74,75,76,78,79,80,82,83,85,86,87,88,89,92,93,94,95,96,97,99,100

%N Heptagon-free numbers.

%C Heptagonal number analogy of A112886 (the triangle-free positive integers).

%H J. V. Post, <a href="http://www.magicdragon.com/poly.html">Table of Polytope Numbers, Sorted, Through 1,000,000.</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HeptagonalNumber.html">Heptagonal Number.</a>

%F {a(n)} = {integers k>1: no divisor of k is a heptagonal number n*(5*n-3)/2}>1}.

%F {a(n)} = {integers k>1: no divisor of k is in A000566 and >1}.

%e 7 = Hep(2) is the first nontrivial heptagonal number, so neither 7 nor any multiple of 7 is in this sequence. 18 = Hep(3) is the first nontrivial heptagonal number, so neither 18 nor any multiple of 18 is in this sequence. 34 = Hep(4) is the first nontrivial heptagonal number, so neither 34 nor any multiple of 34 is in this sequence. Similarly for Hep(5) = 55, Hep(6) = 81, Hep(7) = 112. Hence three consecutive integers are excluded with 54 = 3*18, 55 = Hep(5), 56 = 7*8.

%t upto=100;Module[{maxhep=Floor[(3+Sqrt[9+40upto])/10],heps}, heps= Rest[ Table[(n(5n-3))/2,{n,maxhep}]];Complement[Range[upto],Union[ Flatten[ Table[n*heps,{n,Ceiling[upto/7]}]]]]] (* _Harvey P. Dale_, May 19 2012 *)

%Y Cf. A000566, A112886, A113502.

%K easy,nonn

%O 1,2

%A _Jonathan Vos Post_, Jan 14 2006

%E Corrected by _Harvey P. Dale_, May 19 2012

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