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A175553
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Product of first k triangular numbers divided by the sum of first k triangular numbers is an integer.
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6
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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
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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For k=4 we have 1*3*6*10 /(1+3+6+10) = 9 so k=4 belongs to the sequence.
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MAPLE
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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
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MATHEMATICA
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fQ[n_] := Mod[6n!(n - 1)!, (n + 2)2^n ] == 0; Select[Range@ 96, fQ@# &] (* Robert G. Wilson v, Jun 29 2010 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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