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URL: https://oeis.org/A024166

⇱ A024166 - OEIS

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A024166
a(n) = Sum_{1 <= i < j <= n} (j-i)^3.
53
0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088, 70584, 93993, 123234, 159334, 203434, 256795, 320804, 396980, 486980, 592605, 715806, 858690, 1023526, 1212751, 1428976, 1674992, 1953776, 2268497, 2622522, 3019422
OFFSET
0,3
COMMENTS
Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006
a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012
In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013
The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022
REFERENCES
Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.
LINKS
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See pp. 13, 15.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 90--99. MR0428678 (55 #1698). See Table 1. - N. J. A. Sloane, Mar 23 2014
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7.
Alexander R. Povolotsky, Problem 1147, Pi Mu Epsilon Fall 2006 Problems.
Alexander R. Povolotsky, Problem, Pi Mu Epsilon Spring 2007 Problems.
FORMULA
From Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999: (Start)
a(n) = Sum_{i=0..n} A000537(i), partial sums of A000537.
a(n) = n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60. (End)
a(A004772(n)) mod 2 = 0; a(A016813(n)) mod 2 = 1. - Reinhard Zumkeller, Oct 14 2001
a(n) = Sum_{k=0..n} k^3*(n+1-k). - Paul Barry, Sep 14 2003; edited by Jon E. Schoenfield, Dec 29 2014
a(n) = 2*n*(n+1)*(n+2)*((n+1)^2 + 2*n*(n+2))/5!. This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=2. - Alexander R. Povolotsky, May 17 2008
O.g.f.: x*(1 + 4*x + x^2)/(-1 + x)^6. - R. J. Mathar, Jun 06 2008
a(n) = (6*n^2 + 12*n + 2)*(n+2)!/(120*(n-1)!), n > 0. - Gary Detlefs, Mar 01 2013
a(n) = A222716(n+1)/10 = A000292(n)*A100536(n+1)/10. - Jonathan Sondow, Mar 04 2013
4*a(n) = Sum_{i=0..n} A000290(i)*A000290(i+1). - Bruno Berselli, Feb 05 2014
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j*(n - max(i, j) + 1). - Melvin Peralta, May 12 2016
a(n) = n*binomial(n+3, 4) + binomial(n+2, 5). - Tony Foster III, Nov 14 2017
a(n) = Sum_{i=1..n} i*A143037(n,n-i+1). - J. M. Bergot, Aug 30 2022
EXAMPLE
4*a(7) = 6384 = (0*1)^2 + (1*2)^2 + (2*3)^2 + (3*4)^2 + (4*5)^2 + (5*6)^2 + (6*7)^2 + (7*8)^2. - Bruno Berselli, Feb 05 2014
MAPLE
A024166:=n->n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: seq(A024166(n), n=0..50); # Wesley Ivan Hurt, Nov 21 2017
MATHEMATICA
Nest[Accumulate, Range[0, 40]^3, 2] (* Harvey P. Dale, Jan 10 2016 *)
Table[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, {n, 0, 30}] (* G. C. Greubel, Nov 21 2017 *)
PROG
(PARI) a(n)=sum(j=1, n, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1)))) \\ Alexander R. Povolotsky, May 17 2008
(Haskell)
a024166 n = sum $ zipWith (*) [n+1, n..0] a000578_list
-- Reinhard Zumkeller, Oct 14 2001
(Magma) [n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: n in [0..30]]; // G. C. Greubel, Nov 21 2017
(PARI) for(n=0, 30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017
KEYWORD
nonn,easy,nice
STATUS
approved