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

1,2

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

Links

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

Formula

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

Example

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

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 *)

Crossrefs

Cf. A000292, A006472.

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

Sequence in context: A283767 A174142 A091158 * A284684 A175008 A336408

Adjacent sequences: A175550 A175551 A175552 * A175554 A175555 A175556

Keyword

nonn

Author

Ctibor O. Zizka, Jun 26 2010

Extensions

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

Status

approved