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 1-7 of 7 results.

A211210 a(n) = Sum_{k=0..n} binomial(n, k)*|S1(n, k)|.

Original entry on oeis.org

1, 1, 3, 16, 115, 1021, 10696, 128472, 1734447, 25937683, 424852351, 7554471156, 144767131444, 2971727661124, 65013102375404, 1509186299410896, 37032678328740751, 957376811266995031, 25999194631060525009, 739741591417352081464, 22000132609456951524051
Offset: 0

Views

Author

Olivier Gérard, Oct 23 2012

Keywords

Comments

Binomial convolution of the unsigned Stirling numbers of the first kind.
Row sums of triangle A187555.

Crossrefs

Cf. A317274 (signed S1), A187555, A134090, A211211.
Cf. A122455 (second kind), A271702, A134094, A343841 (second kind inverse).

Programs

  • Mathematica
    Table[Sum[Binomial[n, k] Abs[StirlingS1[n, k]], {k, 0, n}], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*abs(stirling(n, k, 1))); \\ Michel Marcus, May 10 2021

A134090 Triangle, read by rows, where T(n,k) = [(I + D*C)^n](n,k); that is, row n of T = row n of (I + D*C)^n for n>=0 where C denotes Pascal's triangle, I the identity matrix and D a matrix where D(n+1,n)=1 and zeros elsewhere.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 13, 9, 3, 1, 71, 46, 18, 4, 1, 456, 285, 110, 30, 5, 1, 3337, 2021, 780, 215, 45, 6, 1, 27203, 16023, 6167, 1729, 371, 63, 7, 1, 243203, 139812, 53494, 15176, 3346, 588, 84, 8, 1, 2357356, 1326111, 504030, 143814, 32376, 5886, 876, 108, 9, 1
Offset: 0

Views

Author

Paul D. Hanna, Oct 07 2007

Keywords

Comments

Column 0 equals A122455 if we define A122455(0)=1.

Examples

			Triangle T begins:
1;
1, 1;
3, 2, 1;
13, 9, 3, 1;
71, 46, 18, 4, 1;
456, 285, 110, 30, 5, 1;
3337, 2021, 780, 215, 45, 6, 1;
27203, 16023, 6167, 1729, 371, 63, 7, 1;
243203, 139812, 53494, 15176, 3346, 588, 84, 8, 1;
2357356, 1326111, 504030, 143814, 32376, 5886, 876, 108, 9, 1; ...
Let P denote the matrix equal to Pascal's triangle shift down 1 row:
P(n,k) = C(n+1,k) for n>k>=0, with P(n,n)=1 for n>=0.
Illustrate row n of T = row n of P^n as follows.
Matrix P = I + D*C begins:
1;
1, 1;
1, 1, 1;
1, 2, 1, 1;
1, 3, 3, 1, 1;
1, 4, 6, 4, 1, 1; ...
Matrix cube P^3 begins:
1;
3, 1;
6, 3, 1;
13, 9, 3, 1; <== row 3 of P^3 = row 3 of T
30, 25, 12, 3, 1;
73, 72, 40, 15, 3, 1; ...
Matrix 4th power P^4 begins:
1;
4, 1;
10, 4, 1;
26, 14, 4, 1;
71, 46, 18, 4, 1; <== row 4 of P^4 = row 4 of T
204, 155, 70, 22, 4, 1; ...
Matrix 5th power P^5 begins:
1;
5, 1;
15, 5, 1;
45, 20, 5, 1;
140, 75, 25, 5, 1;
456, 285, 110, 30, 5, 1; <== row 5 of P^5 = row 5 of T.

Crossrefs

Cf. columns: A134091, A134092, A134093; A134094 (row sums).

Programs

  • PARI
    \\ As generated by the g.f.
{T(n,k)=polcoeff(sum(j=0,n,binomial(n,j)*x^j/(1-j*x)^k/prod(i=0,j,1-i*x+x*O(x^(n-k)))),n-k)}
  • PARI
    \\ As generated by matrix power: row n of T equals row n of P^n
{T(n,k)=local(P=matrix(n+1,n+1,r,c,if(r==c,1,if(r>c,binomial(r-2,c-1)))));(P^n)[n+1,k+1]}

Formula

T(n,k) = [x^(n-k)] Sum_{j=0..n} C(n,j)*x^j/(1-j*x)^k /[Product_{i=0..j}(1-i*x)].

A134091 Column 1 of triangle A134090.

Original entry on oeis.org

1, 2, 9, 46, 285, 2021, 16023, 139812, 1326111, 13544857, 147880458, 1715413558, 21036674321, 271585117428, 3677831536291, 52081368845176, 769123715337395, 11816582501728389, 188470925178659344, 3114771205613655362
Offset: 0

Views

Author

Paul D. Hanna, Oct 07 2007

Keywords

Comments

Row n of triangle T=A134090 = row n of (I + D*C)^n for n>=0 where C denotes Pascal's triangle, I the identity matrix and D a matrix where D(n+1,n)=1 and zeros elsewhere.

Crossrefs

Cf. A134090; columns: A122455, A134092, A134093; A134094 (row sums); A048993 (S2).

Programs

  • PARI
    a(n)=polcoeff(sum(k=0,n+1,binomial(n+1,k)*x^k/(1-k*x)/prod(i=0,k,1-i*x +x*O(x^n))),n)

Formula

a(n) = [x^n] Sum_{k=0..n+1} C(n+1,k)*x^k/(1-k*x) / [Product_{i=1..k}(1-i*x)].

A134092 Column 2 of triangle A134090.

Original entry on oeis.org

1, 3, 18, 110, 780, 6167, 53494, 504030, 5112090, 55411697, 638154165, 7770348170, 99618149267, 1339889000543, 18848892749144, 276573551651632, 4222814264496510, 66947348027905977, 1099955438013660173
Offset: 0

Views

Author

Paul D. Hanna, Oct 08 2007

Keywords

Comments

Row n of triangle T=A134090 = row n of (I + D*C)^n for n>=0 where C denotes Pascal's triangle, I the identity matrix and D a matrix where D(n+1,n)=1 and zeros elsewhere.

Crossrefs

Cf. A134090; columns: A122455, A134091, A134093; A134094 (row sums); A048993 (S2).

Programs

  • PARI
    {a(n)= polcoeff(sum(k=0,n+2,binomial(n+2,k)*x^k/(1-k*x)^2/prod(i=0,k,1-i*x +x*O(x^n))),n)}

Formula

a(n) = [x^n] Sum_{k=0..n+2} C(n+2,k)*x^k/(1-k*x)^2 / [Product_{i=1..k}(1-i*x)].

A134093 Column 3 of triangle A134090.

Original entry on oeis.org

1, 4, 30, 215, 1729, 15176, 143814, 1462995, 15876410, 182811992, 2223580281, 28458251185, 381943459065, 5359649816728, 78430018675440, 1194057733357517, 18873870914263424, 309154787519651284, 5238840625331179517
Offset: 0

Views

Author

Paul D. Hanna, Oct 08 2007

Keywords

Comments

Row n of triangle T=A134090 = row n of (I + D*C)^n for n>=0 where C denotes Pascal's triangle, I the identity matrix and D a matrix where D(n+1,n)=1 and zeros elsewhere.

Crossrefs

Cf. A134090; columns: A122455, A134091, A134092; A134094 (row sums); A048993 (S2).

Programs

  • PARI
    {a(n)= polcoeff(sum(k=0,n+3,binomial(n+3,k)*x^k/(1-k*x)^3/prod(i=0,k,1-i*x +x*O(x^n))),n)}

Formula

a(n) = [x^n] Sum_{k=0..n+3} C(n+3,k)*x^k/(1-k*x)^3 / [Product_{i=1..k}(1-i*x)].

A211211 sum( C(n+1,k)*|S1(n,k)|, k=0..n). Binomial convolution of the Stirling numbers of the first kind.

Original entry on oeis.org

1, 2, 6, 30, 205, 1750, 17766, 207942, 2746815, 40315858, 649688072, 11387466948, 215440517656, 4371810051908, 94649397546302, 2176321870192342, 52938365091640943, 1357592080006964806, 36593629200726397630, 1033979281229140895582, 30552322294916306960625
Offset: 0

Views

Author

Olivier Gérard, Oct 23 2012

Keywords

Crossrefs

Shifted version of A211210. S1 Analog of A134094.

Programs

  • Mathematica
    Table[Sum[Binomial[n + 1, k] Abs[StirlingS1[n, k]], {k, 0, n}], {n, 0, 20}]

A271702 Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)*C(-j-1,-n-1)*S2(k,j), S2 the Stirling set numbers A048993, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 13, 1, 4, 10, 26, 71, 1, 5, 15, 45, 140, 456, 1, 6, 21, 71, 246, 887, 3337, 1, 7, 28, 105, 399, 1568, 6405, 27203, 1, 8, 36, 148, 610, 2584, 11334, 51564, 243203, 1, 9, 45, 201, 891, 4035, 18849, 91101, 455712, 2357356
Offset: 0

Views

Author

Peter Luschny, Apr 14 2016

Keywords

Examples

			Triangle starts:
[1]
[1, 1]
[1, 2, 3]
[1, 3, 6, 13]
[1, 4, 10, 26, 71]
[1, 5, 15, 45, 140, 456]
[1, 6, 21, 71, 246, 887, 3337]
[1, 7, 28, 105, 399, 1568, 6405, 27203]

Crossrefs

A000012 (col. 0), A000027 (col. 1), A000217 (col. 2), A008778 (col. 3), A122455 (diag. n,n), A134094 (diag. n,n-1).
Cf. A048993.

Programs

  • Maple
    T := (n,k) -> add(Stirling2(k,j)*binomial(-j-1,-n-1)*(-1)^(n-j),j=0..n):
seq(seq(T(n,k), k=0..n), n=0..9);
  • Mathematica
    Flatten[Table[Sum[(-1)^(n-j) Binomial[-j-1,-n-1] StirlingS2[k,j], {j,0,n}], {n,0,9}, {k,0,n}]]

Formula

T(n,k) = Sum_{j=0..k} C(n,j) * S2(k,j). - Alois P. Heinz, Sep 03 2019
Showing 1-7 of 7 results.

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