A087047 a(n) = n*(n+1)*(n+2)*a(n-1)/6 for n >= 1; a(0) = 1.

1, 1, 4, 40, 800, 28000, 1568000, 131712000, 15805440000, 2607897600000, 573737472000000, 164088916992000000, 59728365785088000000, 27176406432215040000000, 15218787602040422400000000, 10348775569387487232000000000, 8444600864620189581312000000000, 8182818237816963704291328000000000

Offset: 0

Author

Enrico T. Federighi (rico125162(AT)aol.com), Aug 08 2003

Keywords

• nonn
• easy

Comments

Product of the first n tetrahedral (or pyramidal) numbers. See 2nd formula. - Alexander Adamchuk, May 19 2006

From Peter Bala, Nov 28 2024: (Start)

For n >= 5, a(n-3) == 9 (mod n) if and only if n is a prime (adapt the proof of the Main Theorem in Himane).

The list of primes p such that a(p-3) == 9 (mod p^2) (analog of A007540 - Wilson primes) begins [11, 31, 47, ...]. (End)

Examples

			a(4) = (1/32)*(1/81)*24*120*720 = 800.
		

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]
• Djamel Himane, A simple proof of Werner Schulte's conjecture, arXiv:2404.08646 [math.GM], 2024
• 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.

Crossrefs

Cf. A000292, A006472.

Programs

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 *)

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

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

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