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A087047
a(n) = n*(n+1)*(n+2)*a(n-1)/6 for n >= 1; a(0) = 1.
3
1, 1, 4, 40, 800, 28000, 1568000, 131712000, 15805440000, 2607897600000, 573737472000000, 164088916992000000, 59728365785088000000, 27176406432215040000000, 15218787602040422400000000, 10348775569387487232000000000, 8444600864620189581312000000000, 8182818237816963704291328000000000
OFFSET
0,3
COMMENTS
Product of the first n tetrahedral (or pyramidal) numbers. See 2nd formula. - Alexander Adamchuk, May 19 2006
LINKS
Karl Dienger, Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung, Jahres-Bericht Ludwig-Wilhelm-Gymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]
Ana Luzón, Manuel A. Morón, and José L. Ramírez, Differential Equations in Ward's Calculus, ResearchGate, September 2023.
Eric Weisstein's World of Mathematics, Tetrahedral Number.
FORMULA
a(n) = 2^(-n-1)*3^(-n)*n!*(n+1)!*(n+2)!.
From Alexander Adamchuk, May 19 2006: (Start)
a(n) = Product_{k=1..n} k*(k+1)*(k+2)/6.
a(n) = Product_{k=1..n} A000292(k). (End)
a(n) = denominator( [x^n] 1F3([1], [1, 2, 3], 6*x) ), where 1F3 is the hypergeometric function (see Luzón et al. at page 19). - Stefano Spezia, Oct 13 2023
EXAMPLE
a(4) = (1/32)*(1/81)*24*120*720 = 800.
MAPLE
a[0]:=1: for n from 1 to 20 do a[n]:=n*(n+1)*(n+2)*a[n-1]/6 od: seq(a[n], n=0..17); # Emeric Deutsch, Mar 06 2005
seq(mul(binomial(k+2, 3), k=1..n), n=0..16); # Zerinvary Lajos, Aug 07 2007
MATHEMATICA
Table[Product[k*(k+1)*(k+2)/6, {k, 1, n}], {n, 0, 16}] (* Alexander Adamchuk, May 19 2006 *)
a[n_]:=Denominator[SeriesCoefficient[HypergeometricPFQ[{1}, {1, 2, 3}, 6x], {x, 0, n}]]; Array[a, 18, 0] (* Stefano Spezia, Oct 13 2023 *)
PROG
(Sage)
q=50 # change q for more terms
[2^(-n-1)*3^(-n)*factorial(n)*factorial(n+1)*factorial(n+2) for n in [0..q]] # Tom Edgar, Mar 15 2014
CROSSREFS
Sequence in context: A277748 A278590 A370086 * A211035 A053514 A325293
KEYWORD
nonn,easy
AUTHOR
Enrico T. Federighi (rico125162(AT)aol.com), Aug 08 2003
EXTENSIONS
More terms from Emeric Deutsch, Mar 06 2005
Example and formula corrected by Tom Edgar, Mar 15 2014
a(0)=1 prepended by and a(15)-a(17) from Stefano Spezia, Oct 13 2023
STATUS
approved