A283394 a(n) = 3*n*(3*n + 7)/2 + 4.

4, 19, 43, 76, 118, 169, 229, 298, 376, 463, 559, 664, 778, 901, 1033, 1174, 1324, 1483, 1651, 1828, 2014, 2209, 2413, 2626, 2848, 3079, 3319, 3568, 3826, 4093, 4369, 4654, 4948, 5251, 5563, 5884, 6214, 6553, 6901, 7258, 7624, 7999, 8383, 8776, 9178, 9589, 10009

Offset: 0

Author

Bruno Berselli, Mar 23 2017

Keywords

• nonn
• easy

Comments

Sum_{k = 0..n} (3*k + r)^3 is divisible by 3*n*(3*n + 2*r + 3)/2 + r^2: the sequence corresponds to the case r = 2 of this formula (other cases are listed in Crossrefs section).

Also, Sum_{k = 0..n} (3*k + 2)^3 / a(n) gives 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, ... (A005449).

a(n) is even if n belongs to A014601. No term is divisible by 3, 5, 7 and 11.

Links

• Bruno Berselli, Table of n, a(n) for n = 0..1000
• Nickolas Arustamyan, Christopher Cox, Erik Lundberg, Sean Perry, and Zvi Rosen, On the Number of Equilibria Balancing Newtonian Point Masses with a Central Force, arXiv:2106.11416 [math-ph], 2021.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A005449, A034856, A095794, A101881, A165157, A186349.

Sequences with formula 3*n*(3*n + 2*r + 3)/2 + r^2: A038764 (r=-1), A027468 (r=0), A081271 (r=1), this sequence (r=2), A027468 (r=3; offset: -1), A080855 (r=4; offset: -2).

Programs

Magma

[3*n*(3*n+7)/2+4: n in [0..50]];

Mathematica

Table[3 n (3 n + 7)/2 + 4, {n, 0, 50}]
LinearRecurrence[{3,-3,1},{4,19,43},50] (* Harvey P. Dale, Mar 02 2019 *)

Maxima

makelist(3*n*(3*n+7)/2+4, n, 0, 50);

PARI

a(n) = 3*n*(3*n + 7)/2 + 4; \\ Indranil Ghosh, Mar 24 2017

Python

[3*n*(3*n+7)/2+4 for n in range(50)]

Sage

[3*n*(3*n+7)/2+4 for n in range(50)]

Formula

O.g.f.: (4 + 7*x - 2*x^2)/(1 - x)^3.

E.g.f.: (8 + 30*x + 9*x^2)*exp(x)/2.

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

a(n) = A081271(-n-2).

a(n) = 3*A095794(n+1) + 1.

a(n) = A034856(3*n+2) = A101881(6*n+2) = A165157(6*n+3) = A186349(6*n+3).

The inverse binomial transform yields 4, 15, 9, 0 (0 continued), therefore:

a(n) = 4*binomial(n,0) + 15*binomial(n,1) + 9*binomial(n,2).