A218152 a(n) = 1 + n + ((n-1)*n^2)/2.

1, 2, 5, 13, 29, 56, 97, 155, 233, 334, 461, 617, 805, 1028, 1289, 1591, 1937, 2330, 2773, 3269, 3821, 4432, 5105, 5843, 6649, 7526, 8477, 9505, 10613, 11804, 13081, 14447, 15905, 17458, 19109, 20861, 22717, 24680, 26753, 28939, 31241, 33662, 36205, 38873

Offset: 0

Author

Mokhtar Mohamed, Oct 24 2012

Keywords

• nonn
• easy

Comments

a(n) = sum(i=0,1,2,...k) d(i)*C(n,i), d(0)=a(0), C(n,i)=0 for all i > n. I would introduce the arithmetic-arithmetic sequence which is defined as the sequence of finite differences, that is, with k consecutive rows of differences, whose first terms are d(1), d(2), d(3),..., d(k), the last row (k-th row) being of a constant difference. Here, it is submitted a special case of the above mentioned sequence with k=3, d(0)=d(1)=1, d(2)=2, d(3)=3.

This sequence is not in Comtet. - T. D. Noe, Nov 16 2012

a(n) appears to be the number of configurations of n equilateral triangles that are allowed to have common vertices, where A002061(n) gives the number of connected configurations and A060354(n) is the number of configurations consisting of several pieces. - Anton Zakharov, May 13 2018

Examples

			for n=5, a(5) = 1+5+(4*25)/2 = 1+5+100/2 = 1+5+50 = 56.
		

References

• L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Anton Zakharov, Illustration of initial terms
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A000027, A000124, A000125, A060354, A002061.

Programs

Mathematica

Table[1+n+((n-1)n^2)/2,{n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,2,5,13},50] (* Harvey P. Dale, May 04 2023 *)

PARI

Vec((1 - 2*x + 3*x^2 + x^3) / (1 - x)^4 + O(x^40)) \\ Colin Barker, May 13 2018

Formula

a(n) = a(n-1)+(4-5*n+3*n^2)/2 for n > 0 and a(0)=1.

a(n) = A006000(n-1)+1 for n > 0. - Antti Karttunen, Oct 24 2012

a(n) = A060354(n) + A002061(n). - Anton Zakharov, May 13 2018

G.f.: (x^3+3*x^2-2*x+1)/(x-1)^4. - Alois P. Heinz, May 13 2018

From Colin Barker, May 13 2018: (Start)

a(n) = (2 + 2*n - n^2 + n^3) / 2.

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

(End)

Extensions

Corrected and edited by Mokhtar Mohamed, Nov 17 2012

Missing term 1937 inserted by Alois P. Heinz, Jun 11 2017