a(n) = (n*(n+1)/2)^2 =
A000217(n)^2 = Sum_{k=1..n}
A000578(k), that is, 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2.
G.f.: (x+4*x^2+x^3)/(1-x)^5. -
Simon Plouffe in his 1992 dissertation
a(n) = Sum ( Sum ( 1 + Sum (6*n) ) ), rephrasing the formula in
A000578. - Xavier Acloque, Jan 21 2003
This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=1. -
Alexander R. Povolotsky, May 17 2008
a(n) = -Sum_{j=1..3} j*Stirling1(n+1,n+1-j)*Stirling2(n+3-j,n). -
Mircea Merca, Jan 25 2014
a(n)*((s-2)*(s-3)/2) = P(3, P(s, n+1)) - P(s, P(3, n+1)), where P(s, m) = ((s-2)*m^2-(s-4)*m)/2 is the m-th s-gonal number. For s=7, 10*a(n) =
A000217(
A000566(n+1)) -
A000566(
A000217(n+1)). -
Bruno Berselli, Aug 04 2015
E.g.f.: x*(4 + 14*x + 8*x^2 + x^3)*exp(x)/4.
Dirichlet g.f.: (zeta(s-4) + 2*zeta(s-3) + zeta(s-2))/4. (End)
a(n) = (Bernoulli(4, n+1) - Bernoulli(4, 1))/4, n >= 0, with the Bernoulli polynomial B(4, x) from row n=4 of
A053382/
A053383. See, e.g., the Ash-Gross reference, p. 62, eq. (6.3) for k=3. -
Wolfdieter Lang, Mar 12 2017
a(n) = n*binomial(n+2, 3) + binomial(n+2, 4) + binomial(n+1, 4). -
Tony Foster III, Nov 14 2017
Another identity: ..., a(3) = (1/2)*(1*(2+4+6)+3*(4+6)+5*6) = 36, a(4) = (1/2)*(1*(2+4+6+8)+3*(4+6+8)+5*(6+8)+7*(8)) = 100, a(5) = (1/2)*(1*(2+4+6+8+10)+3*(4+6+8+10)+5*(6+8+10)+7*(8+10)+9*(10)) = 225, ... -
J. M. Bergot, Aug 27 2022
The previous comment expresses a(n) as the sum of all of the n X n multiplication table array entries.
For example, for n = 4:
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16
This array sum can be split up as follows:
+---+---------------+
| 0 | 1 2 3 4 | (0+1)*(1+2+3+4)
| +---+-----------+
| 0 | 2 | 4 6 8 | (1+2)*(2+3+4)
| | +---+-------+
| 0 | 3 | 6 | 9 12 | (2+3)*(3+4)
| | | +---+---+
| 0 | 4 | 8 |12 |16 | (3+4)*(4)
+---+---+---+---+---+
This kind of row+column sums was used by Ramanujan and others for summing Lambert series. (End)
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