cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A262997 a(n+3) = a(n) + 24*n + 40, a(0)=0, a(1)=5, a(2)=19.

Original entry on oeis.org

0, 5, 19, 40, 69, 107, 152, 205, 267, 336, 413, 499, 592, 693, 803, 920, 1045, 1179, 1320, 1469, 1627, 1792, 1965, 2147, 2336, 2533, 2739, 2952, 3173, 3403, 3640, 3885, 4139, 4400, 4669, 4947, 5232, 5525, 5827, 6136, 6453, 6779, 7112, 7453, 7803, 8160, 8525
Offset: 0

Views

Author

Paul Curtz, Oct 07 2015

Keywords

Comments

The hexasections of A262397(n) are
0, 1, 4, 9, 16, 25, 36, ... = A000290(n)
0, 5, 19, 40, 69, 107, 152, ... = a(n)
0, 1, 5, 11, 18, 28, 40, ... = A240438(n+1)
1, 9, 25, 49, 81, 121, 169, ... = A016754(n)
0, 2, 7, 13, 21, 32, 44, ... = A262523(n)
3, 13, 32, 59, 93, 136, 187, ... = e(n+1).
The five-step recurrence in FORMULA is valuable for the six sequences.
Consider a(n) extended from right to left with their first two differences:
..., 59, 32, 13, 3, 0, 5, 19, 40, 69, ...
..., -27, -19, -10, -3, 5, 14, 21, 29, 38, ...
..., 8, 9, 7, 8, 9, 7, 8, 9, 7, ... .
From 0,the first row is
1) from right to left: e(n)
2) from left to right: a(n).
a(n) and e(n) are companions.
The third row is of period 3.
The last digit of a(n) is of period 15; the same is true of e(n).

Links

Crossrefs

Cf. A000290, A008586, A016742, A016754, A016789, A016921, A016945, A022267, A042965, A240438, A262397, A262523.

Programs

  • Mathematica
    a[0] = 0; a[1] = 5; a[2] = 19; a[n_] := a[n] = a[n - 3] + 24 (n - 3) + 40; Table[a@ n, {n, 0, 46}] (* Michael De Vlieger, Oct 09 2015 *)
LinearRecurrence[{2,-1,1,-2,1},{0,5,19,40,69},60] (* Harvey P. Dale, Dec 16 2024 *)
  • PARI
    vector(100, n, n--; 4*n^2 + (4*(n+1)-3)\3) \\ Altug Alkan, Oct 07 2015
  • PARI
    concat(0, Vec(-x*(x+1)*(3*x^2+4*x+5)/((x-1)^3*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Oct 08 2015

Formula

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) - a(n-5), n> 4.
a(n) = A016742(n) + A042965(n).
a(-n) = e(n).
a(-n) + a(n) = 8*n^2.
a(n+2) - 2*a(n+1) + a(n) = period 3:repeat 9, 7, 8.
a(n+3) - a(n-3) = 8*(1 + 6*n).
a(n+7) - a(n-7) = 40*(2 + 3*n).
a(2n+1) = -a(2n) + 6*n + 3.
a(2n+2) = -a(2n+1) + 4*(n+1).
a(3n) = 4*n*(9*n+1) = 8*A022267(n), a(3n+1) = 36*n^2 +28*n +5, a(3n+2) = 36*n^2 +52*n +19.
G.f.: -x*(x+1)*(3*x^2+4*x+5) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Oct 08 2015

A287893 a(n) = floor(n*(n+2)/9).

Original entry on oeis.org

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
Offset: 0

Views

Author

Paul Curtz, Jun 02 2017

Keywords

Examples

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

Links

Crossrefs

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

Programs

  • 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 *)
  • 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

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

Extensions

Definition simplified by Alois P. Heinz, Jun 02 2017

A301926 a(n+3) = a(n) + 24*n + 32, a(0)=0, a(1)=3, a(2)=13.

Original entry on oeis.org

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
Offset: 0

Views

Author

Paul Curtz, Jun 20 2018

Keywords

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

Crossrefs

Cf. A262997, A262397. A000290, A240438, A016754, A262523 (hexasections). Cf. A130518.

Programs

  • 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 *)
  • 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

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
Showing 1-3 of 3 results.

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