A262523 a(n+3) = a(n) + 6*n + 13, a(0)=0, a(1)=2, a(2)=7.

0, 2, 7, 13, 21, 32, 44, 58, 75, 93, 113, 136, 160, 186, 215, 245, 277, 312, 348, 386, 427, 469, 513, 560, 608, 658, 711, 765, 821, 880, 940, 1002, 1067, 1133, 1201, 1272, 1344, 1418, 1495, 1573, 1653, 1736, 1820, 1906, 1995, 2085, 2177, 2272, 2368, 2466

Offset: 0

Author

Paul Curtz, Sep 24 2015

Keywords

• nonn
• easy

Comments

Companion of A240438 extended from right to left:

..., 21, 13, 7, 2, 0, 0, 1, 5, 11, 18, ...

..., -8, -6, -5, -2, 0, 1, 4, 6, 7, 10, ... see A047267 and A047234

..., 2, 1, 3, 2, 1, 3, 2, 1, 3, 2, ... .

The last digit of a(n) is of period 30. Like A240438.

Is there a definition equivalent to the NAME of A240438?

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Crossrefs

Cf. A000290, A004523, A010882, A016789, A042965, A047234, A047267, A047395, A240438, A262397.

Programs

Mathematica

LinearRecurrence[{2, -1, 1, -2, 1}, {0, 2, 7, 13, 21}, 101] (* Ray Chandler, Sep 24 2015 *)
RecurrenceTable[{a[n+3] == a[n] + 6 n + 13, a[0]==0, a[1]==2, a[2]==7}, a, {n, 0, 500}] (* G. C. Greubel, Sep 28 2015 *)
Table[n (n + 1) + Floor[(n + 1)/3], {n, 0, 50}] (* Bruno Berselli, Jun 06 2017 *)

PARI

concat(0, Vec(-x*(x+1)*(x+2)/ ((x-1)^3*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Sep 25 2015

PARI

A262523(n)=(2+[9,3]*n=divrem(n,6))*4*n[1]+[0,2,7,13,21,32][n[2]+1] \\ M. F. Hasler, Jun 06 2017

Formula

a(n) = A000290(n+1) - A004523(n+2).

a(n) = A240438(n+1) + A004523(n+1).

a(n) = A240438(n) + A047395(n+1).

a(n+2) - 2*a(n+1) + a(n) = period 3: repeat (3, 1, 2).

a(n+3) = a(n-3) + 4*(2 + 3*n). [Thus, a(n+3m) = a(n-3m) + 4m*(2 + 3n), and a(6m+k) = 4m*(9m + 3k + 2) + a(k): explicit formula for a(n) in terms of a(k), 0 <= k <= 5. - M. F. Hasler, Jun 06 2017]

O.g.f.: -x*(x+1)*(x+2) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Sep 25 2015

E.g.f.: (x/3)*(3*x+7)*exp(x) - (2/(3*sqrt(3)))*exp(-x/2)*sin((sqrt(3)*x)/2). - G. C. Greubel, Sep 28 2015

(a(n+3) - a(n)) mod 2 = 1; (a(n+6) - a(n)) mod 2 = 0. - Altug Alkan, Sep 28 2015

(a(n) mod 2) = (0, 0, 1, 1, 1, 0) repeated. (a(n) mod 3) = (0, 2, 1, 1, 0, 2, 2, 1, 0) repeated. (a(n) mod 4) = (0, 2, 3, 1, 1, 0) repeated. (a(n) mod m) has a period of length 3*m, but for m = 4, 8, 12, ... also of length 3*m/2. - M. F. Hasler, Jun 06 2017

a(n) = n*(n+1) + floor((n+1)/3). - Bruno Berselli, Jun 06 2017