a(n) = n*(n-1)*(n-2)*(n-3)/24.
a(n) = Sum_{k=1..n-3} Sum_{i=1..k} i*(i+1)/2. -
Benoit Cloitre, Jun 15 2003
Convolution of natural numbers {1, 2, 3, 4, ...} and
A000217, the triangular numbers {1, 3, 6, 10, ...}. -
Jon Perry, Jun 25 2003
a(n+1) = ((n^5-(n-1)^5) - (n^3-(n-1)^3))/24 - (n^5-(n-1)^5-1)/30; a(n) =
A006322(n-2)-
A006325(n-1). - Xavier Acloque, Oct 20 2003;
R. J. Mathar, Jul 07 2009
a(4*n+2) = Pyr(n+4, 4*n+2) where the polygonal pyramidal numbers are defined for integers A>2 and B>=0 by Pyr(A, B) = B-th A-gonal pyramid number = ((A-2)*B^3 + 3*B^2 - (A-5)*B)/6; For all positive integers i and the pentagonal number function P(x) = x*(3*x-1)/2: a(3*i-2) = P(P(i)) and a(3*i-1) = P(P(i) + i); 1 + 24*a(n) = (n^2 + 3*n + 1)^2. -
Jonathan Vos Post, Nov 15 2004
For n > 3, the sum of the first n-2 tetrahedral numbers (
A000292). - Martin Steven McCormick (mathseq(AT)wazer.net), Apr 06 2005 [Corrected by
Doug Bell, Jun 25 2017]
Starting (1, 5, 15, 35, ...), = binomial transform of [1, 4, 6, 4, 1, 0, 0, 0, ...]. -
Gary W. Adamson, Dec 28 2007
Sum_{n>=4} 1/a(n) = 4/3, from the Taylor expansion of (1-x)^3*log(1-x) in the limit x->1. -
R. J. Mathar, Jan 27 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=1. -
Harvey P. Dale, Aug 22 2011
a(n) = (binomial(n-1,2)^2 - binomial(n-1,2))/6. -
Gary Detlefs, Nov 20 2011
Sum_{n>=0} a(n)/n! = e/24. Sum_{n>=3} a(n)/(n-3)! = 73*e/24. See
A067764 regarding the second ratio. -
Richard R. Forberg, Dec 26 2013
a(n+3) = C(n,1) + 3*C(n,2) + 3*C(n,3) + C(n,4). Each term indicates the number of ways to use n colors to color a tetrahedron with exactly 1, 2, 3, or 4 colors.
G.f.: Starting (1, 5, 14, ...), x/(1-x)^5 can be written
as (x * r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = (1+x)^5;
as (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...) where r(x) = (1+x+x^2)^5;
as (x * r(x) * r(x^4) * r(x^16) * r(x^64) * ...) where r(x) = (1+x+x^2+x^3)^5;
... (as a conjectured infinite set). (End)
Product_{n>=5} (1 - 1/a(n)) = cosh(sqrt(15)*Pi/2)/(100*Pi). -
Amiram Eldar, Jan 21 2021
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