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 1-5 of 5 results.

A262397 a(n) = floor(A261327(n)/9).

Original entry on oeis.org

0, 0, 0, 1, 0, 3, 1, 5, 1, 9, 2, 13, 4, 19, 5, 25, 7, 32, 9, 40, 11, 49, 13, 59, 16, 69, 18, 81, 21, 93, 25, 107, 28, 121, 32, 136, 36, 152, 40, 169, 44, 187, 49, 205, 53, 225, 58, 245, 64, 267, 69, 289, 75, 312, 81, 336, 87, 361, 93, 387, 100, 413, 106, 441
Offset: 0

Views

Author

Paul Curtz, Sep 21 2015

Keywords

Comments

Hexasections:
0, 1, 4, 9, 16, 25, 36, ... = A000290(n)
0, 5, 19, 40, 69, 107, 152, ... = c(n)
0, 1, 5, 11, 18, 28, 40, ... = d(n+1)
1, 9, 25, 49, 81, 121, 169, ... = A016754(n)
0, 2, 7, 13, 21, 32, 44, ... = A240438(n+1)
3, 13, 32, 59, 93, 136, 187, ... = e(n+1).
The six sequences have the signature (2, -1, 1, -2, 1), that is, the signature of a(n) without the 0's.
It appears that d(n+1) and A240438(n+1) are connected via the following scheme.
Let x(n) be the sequence that concatenates terms of d(n+1) in reverse order with terms of A240438(n+1), both without their index_0 term:
..., 18, 11, 5, 1, 0, 0, 2, 7, 13, 21, 32, ...
And consider the first and second differences of this sequence:
..., -7, -6, -4, -1, 0, 2, 5, 6, 8, 11, 12, ...
..., 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, ...
In the first differences, we get A047234(n+1) and A047267(n+1). And in the second differences, we get A010882(n).
In the same way, c(n) and e(n+1) are connected via the first and second differences of this sequence, with both their index_0 term:
..., 69, 40, 19, 5, 0, 3, 13, 32, 59, ...
that are respectively:
..., -29, -21, -14, -5, 3, 10, 19, 27, 34, ...
..., 8, 7, 9, 8, 7, 9, 8, 7, 9, ... .
Is it possible to find a direct definition for a(n)?

Examples

			a(0) = floor(1/9) = 0, a(1)= floor (5/9) = 0, a(2) = floor(2/9) = 0, a(3)= floor (13/9) = 1.

Links

Crossrefs

Cf. A000290, A010882, A016754, A047234, A047267, A240438, A261327.

Programs

  • Mathematica
    LinearRecurrence[{0, 3, 0, -3, 0, 1}, {1, 5, 2, 13, 5, 29}, 70]/9 // Floor (* Jean-François Alcover, Sep 26 2015, after Vincenzo Librandi in A261327 *)
  • PARI
    a(n) = numerator((n^2+4)/4)\9; \\ Michel Marcus, Sep 22 2015
  • PARI
    concat([0,0,0], Vec(-x^3*(x^4 +x^3 +x^2 +x +1)*(x^12 -x^11 +x^10 -x^8 +2*x^6 -x^4 +x^2 -x +1) / ((x -1)^3*(x +1)^3*(x^2 -x +1)*(x^2 +x +1)*(x^6 -x^3 +1)*(x^6 +x^3 +1)) + O(x^100))) \\ Colin Barker, Sep 25 2015
  • PARI
    a(n)=if(n%2,n^2+4,(n/2)^2+1)\9 \\ Charles R Greathouse IV, Oct 16 2015

Formula

a(n) = (A261327(n) - A261327(n) mod 9)/9.
From Colin Barker, Sep 25 2015: (Start)
a(n) = floor((n^2+4)/36) for n even.
a(n) = floor((n^2+4)/9) for n odd.
G.f.: -x^3*(x^4 +x^3 +x^2 +x +1)*(x^12 -x^11 +x^10 -x^8 +2*x^6 -x^4 +x^2 -x +1) / ((x -1)^3*(x +1)^3*(x^2 -x +1)*(x^2 +x +1)*(x^6 -x^3 +1)*(x^6 +x^3 +1)). (End)

Extensions

New name suggested by Michel Marcus, Sep 22 2015

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

Original entry on oeis.org

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

Views

Author

Paul Curtz, Sep 24 2015

Keywords

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

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

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 1-5 of 5 results.

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