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%I #20 Nov 04 2021 15:54:19
%S 1,4,7,8,10,12,13,14,16,18,19,20,22,23,24,25,26,28,30,31,32,33,34,36,
%T 37,38,40,42,43,44,46,47,48,49,50,52,53,54,55,56,58,60,61,62,63,64,66,
%U 67,68,70,72,73,74,75,76,78,79,80,82,83,84,85,86,88,89,90,91,92,93,94,96
%N Product of first k triangular numbers divided by the sum of first k triangular numbers is an integer.
%C Numbers k such that (1*3*6*10* ... *(k*(k+1)/2)) / (1+3+6+10+ ... +(k*(k+1)/2)) is an integer. What if, instead of triangular numbers, we use squares, 1*4*...*(k*k) / (1+4+...+k*k); odd numbers, 1*3*...*(2*k-1) / (1+3+...+(2*k-1)); or Fibonacci numbers, F(1)* ... *F(k) / (F(1)+ ... + F(k))?
%C It appears that the corresponding sequence for the Fibonacci numbers is given in A133653. - _John W. Layman_, Jul 10 2010
%C k > 6 is in this sequence if and only if k+2 is composite. - _Robert Israel_, Nov 04 2021
%H Robert Israel, <a href="/A175553/b175553.txt">Table of n, a(n) for n = 1..10000</a>
%F {k: A006472(k+1)/A000292(k) in Z}. - _R. J. Mathar_, Jun 28 2010
%e For k=4 we have 1*3*6*10 /(1+3+6+10) = 9 so k=4 belongs to the sequence.
%p A006472 := proc(n) n!*(n-1)!/2^(n-1) ; end proc:
%p A000292 := proc(n) binomial(n+2,3) ; end proc:
%p for n from 1 to 200 do a := A006472(n+1)/A000292(n) ; if type(a,'integer') then printf("%d,",n) ; end if; end do: # _R. J. Mathar_, Jun 28 2010
%t fQ[n_] := Mod[6n!(n - 1)!, (n + 2)2^n ] == 0; Select[Range@ 96, fQ@# &] (* _Robert G. Wilson v_, Jun 29 2010 *)
%Y Cf. A000292, A006472.
%Y Cf. A133653. - _John W. Layman_, Jul 10 2010
%K nonn
%O 1,2
%A _Ctibor O. Zizka_, Jun 26 2010
%E More terms from _R. J. Mathar_ and _Robert G. Wilson v_, Jun 28 2010
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