A033430 a(n) = 4*n^3.

0, 4, 32, 108, 256, 500, 864, 1372, 2048, 2916, 4000, 5324, 6912, 8788, 10976, 13500, 16384, 19652, 23328, 27436, 32000, 37044, 42592, 48668, 55296, 62500, 70304, 78732, 87808, 97556, 108000, 119164, 131072, 143748, 157216, 171500, 186624, 202612, 219488

Offset: 0

Author

Jeff Burch

Keywords

• nonn
• easy

Comments

2*a(n) = (2*n)^3 is a perfect cube.

Number of edges of the product of two complete bipartite graphs, each of order 2n, K_n,n x K_n,n - Roberto E. Martinez II, Jan 07 2002

This sequence is related to A049451 by a(n) = n*A049451(n) + sum( A049451(i), i=0..n-1 ) for n>0. - Bruno Berselli, Dec 19 2013

For n>=3, also the detour index of the n-gear graph. - Eric W. Weisstein, Dec 20 2017

For n > 0, this sequence can be obtained by summing consecutive blocks of odd numbers where the n-th block contains the next 2n odd numbers. - Marco Zárate, Jun 15 2025

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..750
• Frank Ellermann, Illustration of binomial transforms
• Eric Weisstein's World of Mathematics, Detour Index
• Eric Weisstein's World of Mathematics, Gear Graph
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A000578, A049451.

Cf. A000292, A007588.

Cf. A001844.

Programs

Magma

[4*n^3: n in [0..40]]; // Vincenzo Librandi, Apr 27 2011

Mathematica

4 Range[0, 40]^3 (* Harvey P. Dale, Sep 07 2016 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 4, 32, 108}, 40] (* Harvey P. Dale, Sep 07 2016 *)
Table[4 n^3, {n, 0, 20}] (* Eric W. Weisstein, Dec 20 2017 *)
CoefficientList[Series[(4 x (1 + 4 x + x^2))/(-1 + x)^4, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 20 2017 *)

PARI

a(n)=4*n^3 \\ Charles R Greathouse IV, Oct 07 2015

Formula

G.f. 4*x*(1+4*x+x^2)/ (x-1)^4. - R. J. Mathar, Feb 01 2011

From Ilya Gutkovskiy, May 25 2016: (Start)

E.g.f.: 4*x*(1 + 3*x + x^2)*exp(x).

Sum_{n>=1} 1/a(n) = zeta(3)/4. (End)

Product_{n>=1} a(n)/A280089(n) = Pi. - Daniel Suteu, Dec 26 2016

From Bruce J. Nicholson, Dec 07 2019: (Start)

a(n) = 24*A000292(n-1) + 4*n.

a(n) = 2*A007588(n) + 2*n. (End)

a(n) = Sum_{k=0..2*n-1} (2*n*(n-1)-2*k+1). - Sean A. Irvine, Jun 19 2025