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Numbers such that n(n+1)(n+2)...(n+12) / (n+(n+1)+(n+2)+...+(n+12)) is a multiple of n.
10

%I #10 Dec 17 2016 17:30:01

%S 1,2,3,4,5,6,8,9,10,11,12,14,15,16,17,18,19,21,22,23,24,25,27,28,29,

%T 30,31,32,34,35,36,37,38,40,41,42,43,44,45,47,48,49,50,51,53,54,55,56,

%U 57,58,60,61,62,63,64,66,67,68,69,70,71,73,74,75,76,77,79,80,81,82,83,84

%N Numbers such that n(n+1)(n+2)...(n+12) / (n+(n+1)+(n+2)+...+(n+12)) is a multiple of n.

%C Equals natural numbers minus '7,13,20,26,33,...' (= previous term +6,+7,+6,+7,...).

%F From _Chai Wah Wu_, Dec 17 2016: (Start)

%F a(n) = a(n-1) + a(n-11) - a(n-12) for n > 12.

%F G.f.: x*(x^11 + x^10 + x^9 + x^8 + x^7 + 2*x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)/(x^12 - x^11 - x + 1). (End)

%Y Cf. A032765-A032798.

%K nonn

%O 1,2

%A _Patrick De Geest_, May 15 1998