A164845 a(n) = (6 + 10*n + 5*n^2 + n^3)/2.

3, 11, 27, 54, 95, 153, 231, 332, 459, 615, 803, 1026, 1287, 1589, 1935, 2328, 2771, 3267, 3819, 4430, 5103, 5841, 6647, 7524, 8475, 9503, 10611, 11802, 13079, 14445, 15903, 17456, 19107, 20859, 22715, 24678, 26751, 28937, 31239, 33660, 36203, 38871

Offset: 0

Author

Paul Curtz, Aug 28 2009

Keywords

• nonn
• easy

Comments

Row sums of the triangle defined by non-interrupted runs in A080036.

If the sequence of integers is split at positions defined by A000124 we obtain A080036. Its runs of consecutive integers can be placed into rows of a triangle:

3;

5, 6;

8, 9, 10;

12, 13, 14, 15;

17, 18, 19, 20, 21;

...

The a(n) are the row sums of this triangle.

The a(n) are also the binomial transform of the quasi-finite sequence 3, 8, 8, 3, 0 (0 continued).

An associated integer sequence could be defined by a(n)/A026741(n+1) = 3, 11, 9, 27, ...

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A135278.

Programs

Magma

[3+5*n+5*n^2/2+n^3/2: n in [0..50]]; // Vincenzo Librandi, Aug 07 2011

Mathematica

Table[(6 + 10*n + 5*n^2 + n^3)/2, {n,0,50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {3, 11, 27, 54}, 50] (* G. C. Greubel, Apr 21 2018 *)

PARI

for(n=0, 50, print1((6+10*n+5*n^2+n^3)/2, ", ")) \\ G. C. Greubel, Apr 21 2018

Formula

a(n) = A162607(n+3) + n.

First differences: a(n+1) - a(n) = A104249(n+2), i.e., a(n) = a(n-1) + 3*n^2/2 + 7*n/2 +3.

Second differences: a(n+2) - 2*a(n+1) + a(n) = A016789(n+2).

a(n) = 2*a(n-1) - a(n-2) + 3*n + 5, n>1.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3, n>2.

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.

G.f.: (3-x+x^2)/(x-1)^4.

E.g.f.: (6 + 16*x + 8*x^2 + x^3)*exp(x)/2. - G. C. Greubel, Apr 21 2018

Extensions

Edited and extended by R. J. Mathar, Aug 31 2009

Corrected typo in recurrence, observed by Paul Curtz - R. J. Mathar, Sep 25 2009