A262997 -id:A262997

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A287893

a(n) = floor(n*(n+2)/9).

+10
1

0, 0, 0, 1, 2, 3, 5, 7, 8, 11, 13, 15, 18, 21, 24, 28, 32, 35, 40, 44, 48, 53, 58, 63, 69, 75, 80, 87, 93, 99, 106, 113, 120, 128, 136, 143, 152, 160, 168, 177, 186, 195, 205, 215, 224, 235, 245, 255, 266, 277, 288, 300, 312, 323, 336, 348, 360, 373, 386

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,5

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,1,-2,1).

FORMULA

a(n) = (A005563(n) - A005563(n) mod 9)/9. Note that A005563(n) mod 9 has period 9: repeat [0, 3, 8, 6, 6, 8, 3, 0, 8].

Interleave A240438(n+1), A262523(n), A005563(n).

From Colin Barker, Jun 02 2017: (Start)

G.f.: x^3*(1 + x^3 - x^5 + 2*x^6 - x^7) / ((1 - x)^3*(1 + x + x^2)*(1 + x^3 + x^6)).

a(n) = 2*a(n-1) - a(n-2) + a(n-9) - 2*a(n-10) + a(n-11) for n>10.

(End)

a(n) = floor(n*(n+2)/9). - Alois P. Heinz, Jun 02 2017

EXAMPLE

a(3) = (15-6)/9 = 1.

MATHEMATICA

Table[Floor[(n(n+2))/9], {n, 0, 60}] (* or *) LinearRecurrence[{2, -1, 0, 0, 0, 0, 0, 0, 1, -2, 1}, {0, 0, 0, 1, 2, 3, 5, 7, 8, 11, 13}, 60] (* Harvey P. Dale, Jan 09 2023 *)

PROG

(PARI) concat(vector(3), Vec(x^3*(1 + x^3 - x^5 + 2*x^6 - x^7) / ((1 - x)^3*(1 + x + x^2)*(1 + x^3 + x^6)) + O(x^100))) \\ Colin Barker, Jun 02 2017

(PARI) a(n)=n*(n+2)\9 \\ Charles R Greathouse IV, Jun 06 2017

CROSSREFS

Cf. A005563, A240438, A262523, A262397, A262997.

KEYWORD

nonn,easy

AUTHOR

Paul Curtz, Jun 02 2017

EXTENSIONS

Definition simplified by Alois P. Heinz, Jun 02 2017

STATUS

approved

A301926

a(n+3) = a(n) + 24*n + 32, a(0)=0, a(1)=3, a(2)=13.

+10
1

0, 3, 13, 32, 59, 93, 136, 187, 245, 312, 387, 469, 560, 659, 765, 880, 1003, 1133, 1272, 1419, 1573, 1736, 1907, 2085, 2272, 2467, 2669, 2880, 3099, 3325, 3560, 3803, 4053, 4312, 4579, 4853, 5136, 5427, 5725

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

Difference table:

0, 3, 13, 32, 59, 93, 136, 187, ...

3, 10, 19, 27, 34, 43, 51, ... = b(n)

7, 9, 8, 7, 9, 8, ... .

The sequence of last decimal digits of a(n) has period 15 and contain no 1's, 4's or 8's.

a(n) is e(n), hexasection, in A262397(n-1).

b(n) mod 9 is of period 9: 3, 1, 1, 0, 7, 7, 6, 4, 4.

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).

FORMULA

a(-n) = A262997(n).

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).

Trisections: a(3n) = 4*n*(9*n-1), a(3n+1) = 3 + 20*n + 36*n^2, a(3n+2) = 13 + 44*n + 36*n^2.

a(n+15) = a(n) + 40*(22+3*n).

G.f.: x*(1 + x)*(3 + 4*x + 5*x^2) / ((1 - x)^3*(1 + x + x^2)). - Colin Barker, Jun 20 2018

MATHEMATICA

CoefficientList[ Series[ -x (5^3 +9x^2 +7x +3)/(x -1)^3 (x^2 +x +1), {x, 0, 40}], x] (* or *)LinearRecurrence[{2, -1, 1, -2, 1}, {0, 3, 13, 32, 59, 93}, 41] (* Robert G. Wilson v, Jun 20 2018 *)

PROG

(PARI) concat(0, Vec(x*(1 + x)*(3 + 4*x + 5*x^2) / ((1 - x)^3*(1 + x + x^2)) + O(x^40))) \\ Colin Barker, Jun 20 2018