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.

A156308 Inverse of triangle S(n,m) defined by sequence A156290, n >= 1, 1 <= m <= n.

Original entry on oeis.org

1, 4, 1, 9, 6, 1, 16, 20, 8, 1, 25, 50, 35, 10, 1, 36, 105, 112, 54, 12, 1, 49, 196, 294, 210, 77, 14, 1, 64, 336, 672, 660, 352, 104, 16, 1, 81, 540, 1386, 1782, 1287, 546, 135, 18, 1, 100, 825, 2640, 4290, 4004, 2275, 800, 170, 20, 1
Offset: 1

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Author

Hartmut F. W. Hoft, Feb 07 2009

Keywords

Comments

From Wolfdieter Lang, Jun 26 2011: (Start)
This triangle S(n,m) appears as U_m(n) in the Knuth reference on p. 285. It is related to the Riordan triangle T_m(n) = A111125(n,m) by S(n,m) = A111125(n,m) - A111125(n-1,m), n >= m >= 1 (identity on p. 286).
Also, S(n,m)-S(n-1,m) = A111125(n-1,m-1), n >= 2, m >= 1 (identity on p. 286). (End)
These polynomials may be expressed in terms of the Faber polynomials of A263916 and are embedded in A127677 and A208513. - Tom Copeland, Nov 06 2015

Examples

			Triangle starts:
  n=1:  1;
  n=2:  4,  1;
  n=3:  9,  6,  1;
  n=4: 16, 20,  8,  1;
  ...
		

Crossrefs

Same as triangle A208513 with the first column truncated.
Columns: A000290 (m=1), A002415 (m=2), A040977 (m=3), A053347 (m=4), A054334 (m=5).

Programs

  • Magma
    [(n/k)*Binomial(n+k-1, 2*k-1): k in [1..n], n in [1..12]]; // G. C. Greubel, Feb 01 2022
  • Mathematica
    S[m_] := Flatten[Table[k/j Binomial[k + j - 1, 2 j - 1], {k, 1, m}, {j, 1, k}]]
  • Sage
    flatten([[(n/k)*binomial(n+k-1, 2*k-1) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Feb 01 2022
    

Formula

S(n, m) = (n/m) * binomial(n + m - 1, 2*m - 1).
From Peter Bala, May 01 2014: (Start)
The n-th row o.g.f. is polynomial R(n,x) = 2/x*( T(n,(x + 2)/2) - 1 ), where T(n,x) is Chebyshev polynomial of the first kind. They form a divisibility sequence: if n divides m then R(n,x) divides R(m,x) in the ring Z[x].
R(2*n,x) = (x + 4)*U(n-1,(x + 2)/2)^2;
R(2*n + 1,x) = ( U(n,(x + 2)/2) + U(n-1,(x + 2)/2) )^2.
O.g.f.: Sum_{n >= 0} R(n,x)*z^n = z*(1 + z)/( (1 - z)*(1 - (x + 2)*z + z^2) ). (End)
The polynomial R(n,x) defined above satisfies (x + 1/x - 2) * R(n, x + 1/x - 2) = x^n + 1/x^n - 2. - Alexander Burstein, May 23 2021

Extensions

Edited by Max Alekseyev, Mar 05 2018