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A175553 Product of first n triangular numbers divided by the sum of first n triangular numbers is an integer. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Numbers n such that (1*3*6*10* ... *(n*(n+1)/2)) / (1+3+6+10+ ... +(n*(n+1)/2)) is an integer. What if, instead of triangular numbers, we use squares, 1*4*...*(n*n) / (1+4+...+n*n); odd numbers, 1*3*...*(2*n-1) / (1+3+...+(2*n-1)); or Fibonacci numbers, F(1)* ... *F(n) / (F(1)+ ... + F(n))?

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

LINKS

Table of n, a(n) for n=1..71.

FORMULA

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

EXAMPLE

For n=4 we have 1*3*6*10 /(1+3+6+10) = 9 so n=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 A286401

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

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Last modified April 7 04:20 EDT 2020. Contains 333292 sequences. (Running on oeis4.)