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.

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A253145 Triangular numbers (A000217) omitting the term 1.

Original entry on oeis.org

0, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275
Offset: 0

Views

Author

Paul Curtz, Mar 23 2015

Keywords

Comments

The full triangle of the inverse Akiyama-Tanigawa transform applied to (-1)^n*A062510(n)=3*(-1)^n*A001045(n) yielding a(n) is
0, 3, 6, 10, 15, 21, 28, 36, ...
-3, -6, -12, -20, -30, -42, -56, ... essentially -A002378
3, 12, 24, 40, 60, 84, ... essentially A046092
-9, -24, -48, -80, -120, ... essentially -A033996
15, 48, 96, 160, ...
-33, -96, -192, ...
63, 192, ...
-129, ...
etc.
First column: (-1)^n*A062510(n).
The following columns are multiples of A122803(n)=(-2)^n. See A007283(n), A091629(n), A020714(n+1), A110286, A175805(n), 4*A005010(n).
An autosequence of the first kind is a sequence whose main diagonal is A000004 = 0's.
b(n) = 0, 0 followed by a(n) is an autosequence of the first kind.
The successive differences of b(n) are
0, 0, 0, 3, 6, 10, 15, 21, ...
0, 0, 3, 3, 4, 5, 6, 7, ... see A194880(n)
0, 3, 0, 1, 1, 1, 1, 1, ...
3, -3, 1, 0, 0, 0, 0, 0, ...
-6, 4, -1, 0, 0, 0, 0, 0, ...
10, -5, 1, 0, 0, 0, 0, 0, ...
-15, 6, -1, 0, 0, 0, 0, 0, ...
21, -7, 1, 0, 0, 0, 0, 0, ...
The inverse binomial transform (first column) is the signed sequence. This is general.
Also generalized hexagonal numbers without 1. - Omar E. Pol, Mar 23 2015

Crossrefs

Programs

Formula

Inverse Akiyama-Tanigawa transform of (-1)^n*A062510(n).
a(n) = (n+1)*(n+2)/2 for n > 0. - Charles R Greathouse IV, Mar 23 2015
a(n+1) = 3*A001840(n+1) + A022003(n).
a(n) = A161680(n+2) for n >= 1. - Georg Fischer, Oct 30 2018
From Stefano Spezia, May 28 2025: (Start)
G.f.: x*(3 - 3*x + x^2)/(1 - x)^3.
E.g.f.: exp(x)*(2 + 4*x + x^2)/2 - 1. (End)

A206012 Modular recursion: a(0)=a(1)=a(2)=a(3)=1, thereafter: a(n) equals a(n - 2) + a(n - 3) when n = 0 mod 5, a(n - 1) + a(n - 3) when n = 1 mod 5, a(n - 1) + a(n - 2) when n = 2 mod 5, a(n - 1) + a(n - 4) when n = 3 mod 5, and a(n - 1) + a(n - 2) + a(n - 3) otherwise.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 3, 5, 8, 16, 13, 21, 34, 50, 105, 84, 134, 218, 323, 675, 541, 864, 1405, 2080, 4349, 3485, 5565, 9050, 13399, 28014, 22449, 35848, 58297, 86311, 180456, 144608, 230919, 375527, 555983, 1162429, 931510, 1487493, 2419003, 3581432, 7487928
Offset: 0

Views

Author

Roger L. Bagula, Mar 19 2012

Keywords

Comments

This sequence was inspired by the work of Paul Curtz on three part sequences. I did a three part version of this that gave a generating polynomial and got even more variance by adding two more modulo sequences.

Crossrefs

Programs

  • Magma
    m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((x^15-x^13+x^12-2*x^10-2*x^9+2*x^8-x^7-3*x^6-4*x^5+3*x^4+x^3+x^2+x+1)/(x^15-3*x^10-6*x^5+1))); // Bruno Berselli, Mar 20 2012
  • Mathematica
    a[0] = 1; a[1] = 1; a[2] = 1; a[3] = 1;a[n_Integer] := a[n]=If[Mod[n, 5] == 0, a[n - 2] + a[n - 3], If[Mod[n, 5] == 1, a[n - 1] + a[n - 3], If[Mod[n, 5] == 2, a[n - 1] + a[n - 2], If[Mod[n, 5] == 3, a[n - 1] + a[n - 4], a[n - 1] + a[n - 2] + a[n - 3]]]]];b = Table[a[n], {n, 0, 50}];(* FindSequenceFunction gives*);Table[c[n] = b[[n]], {n, 1, 16}];c[n_Integer] := c[n] = -c[-15 + n] + c[-10 + n] + 6 c[-5 + n];d = Table[c[n], {n, 1, Length[b]}]
    CoefficientList[Series[(x^15-x^13+x^12-2*x^10-2*x^9+2*x^8-x^7-3*x^6-4*x^5+3*x^4+x^3+x^2+x+1)/(x^15-3*x^10-6*x^5+1),{x,0,1001}],x] (* Vincenzo Librandi, Apr 01 2012 *)
  • PARI
    Vec((x^15-x^13+x^12-2*x^10-2*x^9+2*x^8-x^7-3*x^6-4*x^5+3*x^4+x^3+x^2+x+1)/(x^15-3*x^10-6*x^5+1)+O(x^99)) \\ Charles R Greathouse IV, Mar 19 2012
    

Formula

G.f.: (x^15 - x^13 + x^12 - 2x^10 - 2x^9 + 2x^8 - x^7 - 3x^6 - 4x^5 + 3x^4 + x^3 + x^2 + x + 1) / (x^15 - 3x^10 - 6x^5 + 1). - Alois P. Heinz, Mar 19 2012
Showing 1-2 of 2 results.