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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A131820 Row sums of triangle A131819.

Original entry on oeis.org

1, 6, 16, 33, 59, 96, 146, 211, 293, 394, 516, 661, 831, 1028, 1254, 1511, 1801, 2126, 2488, 2889, 3331, 3816, 4346, 4923, 5549, 6226, 6956, 7741, 8583, 9484, 10446, 11471, 12561, 13718, 14944, 16241, 17611, 19056, 20578, 22179, 23861, 25626
Offset: 1

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Author

Gary W. Adamson, Jul 18 2007

Keywords

Comments

Let M(n) be the n-th square matrix whose (i,j)-entry equals i^2/(i^2+1) if i=j and equals 1 otherwise. Then a(n) = (-1)^(n+1) * gamma(1-i+n) * gamma(1+i+n) * sinh(Pi)/Pi times the determinant of M(n). - John M. Campbell, Sep 07 2011

Examples

			a(4) = 33 = (1, 3, 3, 1) dot (1, 5, 5, 2) = (1 + 15 + 15 + 2).
a(4) = 33 = sum of row 4 terms of triangle A131819: (13 + 9 + 7 + 4).

Crossrefs

Cf. A131819, A000330, A081489, A005563.

Programs

  • Maple
    a:= n-> (7+(3+2*n)*n)*n/6-1:
seq(a(n), n=1..40);  # Alois P. Heinz, May 04 2009
  • Mathematica
    Table[n^3/3 + n^2/2 + 7*n/6 - 1, {n, 100}]

Formula

Binomial transform of (1, 5, 5, 2, 0, 0, 0, ...).
From Alois P. Heinz, May 04 2009: (Start)
a(n) = n^3/3 + n^2/2 + (7/6)*n - 1.
a(n) = -1 + Sum_{k=1..n} (k^2+1).
a(n) = A000330(n) + A000027(n) - A000012(n).
G.f.: x*(1 + 2*x - 2*x^2 + x^3)/(1 - x)^4. (End)
a(n) = n^2 + a(n-1) + 1, n > 1. - Gary Detlefs, Jun 29 2010
From Gary Detlefs, Jun 30 2010: (Start)
a(n) = (2n^3 + 3n^2 + 7n - 6)/6, n > 0.
a(n) = A081489(n) + A005563(n-1), n > 0. (End)
E.g.f.: 1 + exp(x)*(2*x^3 + 9*x^2 + 12*x - 6)/6. - Stefano Spezia, Mar 02 2025

Extensions

More terms from Alois P. Heinz, May 04 2009

A329854 Triangle read by rows: T(n,k) = ((n - k)*(n + k - 1) + 2)/2, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 4, 3, 1, 7, 7, 6, 4, 1, 11, 11, 10, 8, 5, 1, 16, 16, 15, 13, 10, 6, 1, 22, 22, 21, 19, 16, 12, 7, 1, 29, 29, 28, 26, 23, 19, 14, 8, 1, 37, 37, 36, 34, 31, 27, 22, 16, 9, 1, 46, 46, 45, 43, 40, 36, 31, 25, 18, 10, 1, 56, 56, 55, 53, 50, 46, 41, 35, 28, 20, 11, 1
Offset: 0

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Author

Werner Schulte, Nov 22 2019

Keywords

Comments

This triangle equals A309559 with reversed rows and supplemented main diagonal (all terms are 1).
There are two lower triangular matrices M and N so that the matrix product M * N equals T (seen as a matrix).
/ 1 \ / 1 \
| 0 1 | | 1 1 |
| 0 1 1 | | 1 1 1 |
M(n,k) = | 0 1 2 1 | N(n,k) = | 1 1 1 1 |
| 0 1 2 3 1 | | 1 1 1 1 1 |
| 0 1 2 3 4 1 | | 1 1 1 1 1 1 |
\ . . . . . . . / \ . . . . . . . /
The matrix product N * M equals the rascal triangle A077028 (seen as a matrix).

Examples

			The triangle T(n,k) starts:
n \ k :   0    1    2    3    4    5    6    7    8    9   10   11
==================================================================
   0  :   1
   1  :   1    1
   2  :   2    2    1
   3  :   4    4    3    1
   4  :   7    7    6    4    1
   5  :  11   11   10    8    5    1
   6  :  16   16   15   13   10    6    1
   7  :  22   22   21   19   16   12    7    1
   8  :  29   29   28   26   23   19   14    8    1
   9  :  37   37   36   34   31   27   22   16    9    1
  10  :  46   46   45   43   40   36   31   25   18   10    1
  11  :  56   56   55   53   50   46   41   35   28   20   11    1
etc.

Crossrefs

Row sums equal A116731(n+1).
Row sums apart from column 0 equal A081489.
Cf. A077028, A309559.

Formula

O.g.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n = ((t^2+(1-t)^2) * (1-x*t) + x * t^2 * (1-t)) / ((1-t)^3 * (1-x*t)^2).
G.f. of column k: Sum_{n>=k} T(n,k) * t^n = t^k * (t^2/(1-t)^3 + 1/(1-t) + k*t/(1-t)^2) for k >= 0.
T(n,k) = 1 + T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) for 0 < k < n with initial values T(n,0) = (n*(n-1)+2)/2 and T(n,n) = 1 for n >= 0.
T(n,k) = (2 + T(n-1,k-1) * T(n-1,k+1)) / T(n-2,k) for 0 < k < n-1 with initial values given above and T(n,n-1) = n for n > 0.
Referring to the triangle M(n,k) (see comments), we get:
(1) Sum_{k=0..n} (k+1) * M(n,k) = A116731(n+1) for n >= 0;
(2) Sum_{k=1..n} k * M(n,k) = A081489(n) for n >= 1.
T(n,k) = T(n-1,k-1) + n-k for 0 < k <= n with initial values T(n,0) = (n*(n-1)+2)/2 for n >= 0.
T(n,k) = 2 * T(n-1,k-1) - T(n-2,k-2) for 1 < k <= n with initial values T(0,0) = 1 and T(n,0) = T(n,1) = (n*(n-1)+2)/2 for n > 0.

A349118 Row sums of a triangle based on A261327.

Original entry on oeis.org

1, 5, 3, 18, 8, 47, 18, 100, 35, 185, 61, 310, 98, 483, 148, 712, 213, 1005, 295, 1370, 396, 1815, 518, 2348, 663, 2977, 833, 3710, 1030, 4555, 1256, 5520, 1513, 6613, 1803, 7842, 2128, 9215, 2490, 10740, 2891, 12425, 3333, 14278, 3818, 16307, 4348, 18520, 4925
Offset: 2

Views

Author

Paul Curtz, Nov 08 2021

Keywords

Comments

The following triangle has A261327 as its diagonals:
1
5
1 2
5 13
1 2 5
5 13 29
1 2 5 10
5 13 29 53
1 2 5 10 17
5 13 29 53 85
...
a(0) = a(1) = 0.
a(n)'s final digit: neither 4 nor 9.
First full bisection difference table:
0, 1, 3, 8, 18, 35, 61, 98, ... = 0, A081489 = b(n)
1, 2, 5, 10, 17, 26, 37, 50, ... = A002522
1, 3, 5, 7, 9, 11, 13, 15, ... = A005408
2, 2, 2, 2, 2, 2, 2, 2, ... = A007395
0, 0, 0, 0, 0, 0, 0, 0, ... = A000004
Second full bisection difference table:
0, 5, 18, 47, 100, 185, 310, 483, ... = c(n)
5, 13, 29, 53, 85, 125, 173, 229, ... = A078370
8, 16, 24, 32, 40, 48, 56, 64, ... = A008590(n+1)
8, 8, 8, 8, 8, 8, 8, 8, ... = A010731
0, 0, 0, 0, 0, 0, 0, 0, ... = A000004
Both bisections are cubic polynomials.
c(-n) = -c(n).

Links

Crossrefs

Cf. A002522, A005408, A007395, A078370, A081489 (first bisection).
Cf. also A008590, A010731, A261327.

Programs

  • Mathematica
    LinearRecurrence[{0, 4, 0, -6, 0, 4, 0, -1}, {1, 5, 3, 18, 8, 47, 18, 100}, 50] (* Amiram Eldar, Nov 08 2021 *)

Formula

G.f.: (5*x^5+2*x^4-2*x^3-x^2+5*x+1)/((x-1)^4*(x+1)^4).
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