OFFSET
0,2
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
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).
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
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)
EXAMPLE
for n=5, a(5) = 1+5+(4*25)/2 = 1+5+100/2 = 1+5+50 = 56.
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 *)
PROG
(PARI) Vec((1 - 2*x + 3*x^2 + x^3) / (1 - x)^4 + O(x^40)) \\ Colin Barker, May 13 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mokhtar Mohamed, Oct 24 2012
EXTENSIONS
Corrected and edited by Mokhtar Mohamed, Nov 17 2012
Missing term 1937 inserted by Alois P. Heinz, Jun 11 2017
STATUS
approved