A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

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

Author

Clark Kimberling

Keywords

• nonn
• easy
• nice

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

Examples

			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
		

References

• Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• 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.
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Crossrefs

Cf. A000292, A000332, A000389, A000579, A000580, A024166, A027555, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086025, A086026, A086027, A086028, A086029, A086030, A087127.

Cf. A000330, A000537 (first diffs), A001286, A003215, A100536, A101094 (partial sums), A101097, A101102, A222716.

Programs

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

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 *)

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

PARI

for(n=0,30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017

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