A090902 a(n) = floor((product of first n triangular numbers)/(sum of first n factorials)).

1, 1, 2, 5, 17, 64, 268, 1236, 6286, 35031, 212401, 1392489, 9817398, 74078419, 595722994, 5086611025, 45961503660, 438168680119, 4395396953168, 46281082011630, 510378647082537, 5882795810558767, 70740717281280862, 885944239324190839, 11537420341016416104

Offset

1,3

Comments

Floor(Sum (prod(n))/(Product(Sum (n)). Asymptotic formula?

Links

Harvey P. Dale, Table of n, a(n) for n = 1..505

Formula

a(n) = floor(A000178(n)^(1/n)).

Example

a(4) = floor(((1)*(1+2)*(1+2+3)*(1+2+3+4))/((1)+(1*2)+(1*2*3)+(1*2*3*4))) = 5.

Mathematica

Do[Print[Floor[ Product[k*(k+1)/2, {k, 1, n}] / Sum[k!, {k, 1, n}] ]], {n, 1, 20}] (* Ryan Propper, Jun 18 2005 *)
Module[{nn=20, trs, facs}, trs=Rest[FoldList[Times, 1, Accumulate[Range[ nn]]]]; facs = Accumulate[Range[nn]!]; Floor/@(trs/facs)] (* Harvey P. Dale, Jul 23 2014 *)

Prog

(PARI) {a(n) = (prod(j=1, n, binomial(j+1, 2))/sum(k=1, n, k!))\1};
for(n=1, 25, print1(a(n), ", ")) \\ G. C. Greubel, Feb 05 2019
(Magma) [Floor((&*[Binomial(j+1, 2): j in [1..n]])/(&+[Factorial(k): k in [1..n]])): n in [1..25]]; // G. C. Greubel, Feb 05 2019
(Sage) [floor(prod(binomial(j+1, 2) for j in (1..n))/sum(factorial(k) for k in (1..n))) for n in (1..25)] # G. C. Greubel, Feb 05 2019

Crossrefs

Cf. A090901.

Sequence in context: A003456 A109084 A217596 * A150012 A150013 A052539

Adjacent sequences: A090899 A090900 A090901 * A090903 A090904 A090905

Keyword

nonn

Author

Amarnath Murthy, Dec 12 2003

Extensions

More terms from Ryan Propper, Jun 18 2005

More terms from Harvey P. Dale, Jul 23 2014

Status

approved