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%I #22 Jan 12 2020 11:34:43
%S 4,1728,2903040,12541132800,115880067072000,69528040243200000,
%T 1807729046323200000,43295255277764345856000000,
%U 20188846756043686829592191472500736000000000,989253491046140654650017382152536064000000000
%N Integral quotients of products of first k consecutive composites divided by their sums: products (dividends).
%C Based on A141092.
%C Compare with A140761 A159578 A140763 A116536.
%C Take the first k composite numbers. If their product divided by their sum results in an integer, their product is a term of the sequence. - _Harvey P. Dale_, Apr 29 2018
%H Amiram Eldar, <a href="/A141090/b141090.txt">Table of n, a(n) for n = 1..93</a>
%F Find the products and sums of first k consecutive composites. When the product divided by the sum produces an integral quotient, add product to sequence.
%e a(3) = 2903040 because 4*6*8*9*10*12*14 = 2903040 and 4+6+8+9+10+12+14 = 63; 2903040/63 = 46080, integral -- 2903040 is added to the sequence.
%t With[{c=Select[Range[100],CompositeQ]},Table[If[IntegerQ[ Times@@Take[ c,n]/Total[ Take[ c,n]]], Times@@ Take[ c,n],0],{n,Length[c]}]]/.(0-> Nothing) (* _Harvey P. Dale_, Apr 29 2018 *)
%Y Cf. A196415, A141089, A141091, A141092.
%K easy,nonn
%O 1,1
%A _Enoch Haga_, Jun 01 2008
%E Checked by _N. J. A. Sloane_, Oct 02 2011
%E Edited by _N. J. A. Sloane_, Apr 29 2018
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