A175553 Product of first k triangular numbers divided by the sum of first k triangular numbers is an integer.

1, 4, 7, 8, 10, 12, 13, 14, 16, 18, 19, 20, 22, 23, 24, 25, 26, 28, 30, 31, 32, 33, 34, 36, 37, 38, 40, 42, 43, 44, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 58, 60, 61, 62, 63, 64, 66, 67, 68, 70, 72, 73, 74, 75, 76, 78, 79, 80, 82, 83, 84, 85, 86, 88, 89, 90, 91, 92, 93, 94, 96

Offset: 1

Author

Ctibor O. Zizka, Jun 26 2010

Keywords

• nonn

Comments

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))?

It appears that the corresponding sequence for the Fibonacci numbers is given in A133653. - John W. Layman, Jul 10 2010

k > 6 is in this sequence if and only if k+2 is composite. - Robert Israel, Nov 04 2021

Examples

			For k=4 we have 1*3*6*10 /(1+3+6+10) = 9 so k=4 belongs to the sequence.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000292, A006472.

Cf. A133653. - John W. Layman, Jul 10 2010

Programs

Maple

A006472 := proc(n) n!*(n-1)!/2^(n-1) ; end proc:
A000292 := proc(n) binomial(n+2,3) ; end proc:
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

Mathematica

fQ[n_] := Mod[6n!(n - 1)!, (n + 2)2^n ] == 0; Select[Range@ 96, fQ@# &] (* Robert G. Wilson v, Jun 29 2010 *)

Formula

{k: A006472(k+1)/A000292(k) in Z}. - R. J. Mathar, Jun 28 2010

Extensions

More terms from R. J. Mathar and Robert G. Wilson v, Jun 28 2010