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 1-4 of 4 results.

A363622 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with weighted alternating sum k (leading and trailing 0's omitted).

Original entry on oeis.org

1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 2, 0, 0, 1, 0, 1, 1, 2, 0, 1, 1, 1, 1, 1, 3, 0, 0, 2, 0, 1, 1, 2, 1, 1, 3, 0, 2, 2, 1, 1, 2, 2, 1, 1, 5, 0, 0, 3, 0, 2, 2, 2, 1, 3, 2, 1, 1, 5, 0, 3, 3, 2, 2, 3, 2, 2, 4, 2, 1, 1, 7, 0, 0, 5, 0, 3, 3, 4, 2, 4, 2, 4, 4, 2, 1, 1
Offset: 0

Views

Author

Gus Wiseman, Jun 15 2023

Keywords

Comments

We define the weighted alternating sum of a sequence (y_1,...,y_k) to be Sum_{i=1..k} (-1)^(i-1) i * y_i. For example:
- (3,3,2,1,1) has weighted alternating sum 1*3 - 2*3 + 3*2 - 4*1 + 5*1 = 4.
- (1,2,2,3) has weighted alternating sum 1*1 - 2*2 + 3*2 - 4*3 = -9.

Examples

			Triangle begins:
  1
  1
  1  0  0  1
  1  0  1  1
  2  0  0  1  0  1  1
  2  0  1  1  1  1  1
  3  0  0  2  0  1  1  2  1  1
  3  0  2  2  1  1  2  2  1  1
  5  0  0  3  0  2  2  2  1  3  2  1  1
  5  0  3  3  2  2  3  2  2  4  2  1  1
  7  0  0  5  0  3  3  4  2  4  2  4  4  2  1  1
  7  0  5  5  3  3  5  4  3  5  3  5  4  2  1  1
Row n = 6 counts the following partitions:
  k=-3            k=0        k=2    k=3   k=4      k=5    k=6
  -----------------------------------------------------------
  (33)      .  .  (42)    .  (321)  (51)  (222)    (411)  (6)
  (2211)          (3111)                  (21111)
  (111111)
		

Crossrefs

Row sums are A000041.
The unweighted version is A103919 with leading zeros removed.
Row-lengths appear to be A168233.
Central column T(n,0) is A363532, ranks A363621.
The corresponding rank statistic is A363619, reverse A363620.
The reverse version is A363623.
A053632 counts compositions by weighted sum.
A264034 counts partitions by weighted sum, reverse A358194.
A316524 gives alternating sum of prime indices, reverse A344616.
A363624 gives weighted alternating sum of Heinz partition, reverse A363625.

Programs

  • Mathematica
    altwtsum[y_]:=Sum[(-1)^(k-1)*k*y[[k]],{k,1,Length[y]}];
    Table[Length[Select[IntegerPartitions[n],altwtsum[#]==k&]],{n,0,15},{k,Min[altwtsum/@IntegerPartitions[n]], Max[altwtsum/@IntegerPartitions[n]]}]

A178977 a(n) = (3*n+2)*(3*n+5)/2.

Original entry on oeis.org

5, 20, 44, 77, 119, 170, 230, 299, 377, 464, 560, 665, 779, 902, 1034, 1175, 1325, 1484, 1652, 1829, 2015, 2210, 2414, 2627, 2849, 3080, 3320, 3569, 3827, 4094, 4370, 4655, 4949, 5252, 5564, 5885, 6215, 6554, 6902, 7259, 7625, 8000, 8384, 8777, 9179, 9590, 10010
Offset: 0

Views

Author

Paul Curtz, Jan 02 2011

Keywords

Comments

Companion to A145910.

Crossrefs

Programs

Formula

a(n) = a(n-1) + 6 + 9*n.
a(n) = A178971(3*n+2).
a(n) = A145910(n) + 3 + 3*n = A145910(n) + A008585(n+1).
a(n) = A168233(n+1)*A168300(n+1).
G.f.: (-5-5*x+x^2)/(x-1)^3. [Adapted to the offset by Bruno Berselli, Apr 14 2011]
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Apr 19 2013
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=0} 1/a(n) = 1/3.
Sum_{n>=0} (-1)^n/a(n) = 4*Pi/(9*sqrt(3)) - 1/3 - 4*log(2)/9. (End)
From Elmo R. Oliveira, Oct 30 2024: (Start)
E.g.f.: exp(x)*exp(x)*(5 + 15*x + 9*x^2/2).
a(n) = A016789(n)*A016789(n+1)/2. (End)

A168286 a(n) = (6*n + 3*(-1)^n + 1)/2.

Original entry on oeis.org

2, 8, 8, 14, 14, 20, 20, 26, 26, 32, 32, 38, 38, 44, 44, 50, 50, 56, 56, 62, 62, 68, 68, 74, 74, 80, 80, 86, 86, 92, 92, 98, 98, 104, 104, 110, 110, 116, 116, 122, 122, 128, 128, 134, 134, 140, 140, 146, 146, 152, 152, 158, 158, 164, 164, 170, 170, 176, 176, 182, 182
Offset: 1

Views

Author

Vincenzo Librandi, Nov 22 2009

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select n+1 else 6*n-Self(n-1)-2: n in [1..70]]; // Vincenzo Librandi, Sep 17 2013
  • Mathematica
    Table[3 n + 3 (-1)^n/2 + 1/2, {n, 70}] (* Bruno Berselli, Sep 17 2013 *)
    CoefficientList[Series[(2 + 6 x - 2 x^2)/((1 + x) (1 - x)^2), {x, 0, 70}], x] (* Vincenzo Librandi, Sep 17 2013 *)

Formula

a(n) = 6*n - a(n-1) - 2, with n>1, a(1)=2.
From Vincenzo Librandi, Sep 17 2013: (Start)
a(n) = a(n-1) +a(n-2) -a(n-3).
G.f.: 2*x*(1 + 3*x - x^2)/((1+x)*(1-x)^2).
a(n) = 2*A168233(n) = A168301(n) + 1. (End)
E.g.f.: (1/2)*(3 - 4*exp(x) + (6*x + 1)*exp(2*x))*exp(-x). - G. C. Greubel, Jul 17 2016

Extensions

New definition by Bruno Berselli, Sep 17 2013

A168410 a(n) = 3 + 9*floor(n/2).

Original entry on oeis.org

3, 12, 12, 21, 21, 30, 30, 39, 39, 48, 48, 57, 57, 66, 66, 75, 75, 84, 84, 93, 93, 102, 102, 111, 111, 120, 120, 129, 129, 138, 138, 147, 147, 156, 156, 165, 165, 174, 174, 183, 183, 192, 192, 201, 201, 210, 210, 219, 219, 228, 228, 237, 237, 246, 246, 255, 255
Offset: 1

Views

Author

Vincenzo Librandi, Nov 25 2009

Keywords

Crossrefs

Programs

  • Magma
    [3+9*Floor(n/2): n in [1..70]]; // Vincenzo Librandi, Sep 19 2013
  • Mathematica
    Table[3 + 9 Floor[n/2], {n, 70}] (* or *) CoefficientList[Series[3 (1 + 3 x - x^2)/((1 + x) (x - 1)^2), {x, 0, 70}], x] (* Vincenzo Librandi, Sep 19 2013 *)

Formula

a(n) = 9*n - a(n-1) - 3, with n>1, a(1)=3.
G.f.: 3*x*(1 + 3*x - x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 10 2011
a(n) = 3*A168233(n). - R. J. Mathar, Jul 10 2011
a(n) = a(n-1) +a(n-2) -a(n-3). - Vincenzo Librandi, Sep 19 2013
E.g.f.: (3/4)*(3 - 4*exp(x) + (6*x + 1)*exp(2*x))*exp(-x). - G. C. Greubel, Jul 21 2016

Extensions

New definition by Vincenzo Librandi, Sep 19 2013
Showing 1-4 of 4 results.