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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A083856 Square array T(n,k) of generalized Fibonacci numbers, read by antidiagonals upwards (n, k >= 0).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 3, 1, 0, 1, 1, 4, 5, 5, 1, 0, 1, 1, 5, 7, 11, 8, 1, 0, 1, 1, 6, 9, 19, 21, 13, 1, 0, 1, 1, 7, 11, 29, 40, 43, 21, 1, 0, 1, 1, 8, 13, 41, 65, 97, 85, 34, 1, 0, 1, 1, 9, 15, 55, 96, 181, 217, 171, 55, 1
Offset: 0

Views

Author

Paul Barry, May 06 2003

Keywords

Comments

Row n >= 0 of the array gives the solution to the recurrence b(k) = b(k-1) + n*b(k-2) for k >= 2 with b(0) = 0 and b(1) = 1.

Examples

			Array T(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
  0, 1, 1,  1,  1,   1,   1,    1,    1,     1, ... [A057427]
  0, 1, 1,  2,  3,   5,   8,   13,   21,    34, ... [A000045]
  0, 1, 1,  3,  5,  11,  21,   43,   85,   171, ... [A001045]
  0, 1, 1,  4,  7,  19,  40,   97,  217,   508, ... [A006130]
  0, 1, 1,  5,  9,  29,  65,  181,  441,  1165, ... [A006131]
  0, 1, 1,  6, 11,  41,  96,  301,  781,  2286, ... [A015440]
  0, 1, 1,  7, 13,  55, 133,  463, 1261,  4039, ... [A015441]
  0, 1, 1,  8, 15,  71, 176,  673, 1905,  6616, ... [A015442]
  0, 1, 1,  9, 17,  89, 225,  937, 2737, 10233, ... [A015443]
  0, 1, 1, 10, 19, 109, 280, 1261, 3781, 15130, ... [A015445]
  ...
		

Crossrefs

Rows include A057427 (n=0), A000045 (n=1), A001045 (n=2), A006130 (n=3), A006131 (n=4), A015440 (n=5), A015441 (n=6), A015442 (n=7), A015443 (n=8), A015445 (n=9).
Columns include A000012 (k=1,2), A000027 (k=3), A005408 (k=4), A028387 (k=5), A000567 (k=6), A106734 (k=7).
Cf. A083857 (binomial transform), A083859 (main diagonal), A083860 (first subdiagonal), A083861 (second binomial transform), A110112, A110113 (diagonal sums), A193376 (transposed variant), A172237 (transposed variant).

Programs

  • Julia
    function generalized_fibonacci(r, n)
       F = BigInt[1 r; 1 0]
       Fn = F^n
       Fn[2, 1]
    end
    for r in 0:6 println([generalized_fibonacci(r, n) for n in 0:9]) end # Peter Luschny, Mar 06 2017
  • Maple
    A083856_row := proc(r, n) local R; R := proc(n) option remember;
    if n<=1 then n else R(n-1)+r*R(n-2) fi end: R(n) end:
    for r from 0 to 9 do seq(A083856_row(r, n), n=0..9) od; # Peter Luschny, Mar 06 2017
  • Mathematica
    T[, 0] = 0; T[, 1|2] = 1; T[n_, k_] := T[n, k] = T[n, k-1] + n T[n, k-2];
    Table[T[n-k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 22 2018 *)

Formula

T(n, k) = (((1 + sqrt(4*n + 1))/2)^k - ((1 - sqrt(4*n + 1))/2)^k)/sqrt(4*n + 1). [corrected by Michel Marcus, Jun 25 2018]
From Thomas Baruchel, Jun 25 2018: (Start)
The g.f. for row n >= 0 is x/(1 - x - n*x^2).
The g.f. for column k >= 1 is g(k,x) = 1/(1-x) + Sum_{m = 1..floor((k-1)/2)} (1 - x)^(-1 - m) * binomial(k - 1 - m, m) * Sum_{i = 0..m} x^i * Sum_{j = 0..i} (-1)^j * (i - j)^m * binomial(1 + m, j).
The g.f. for column k >= 1 is also g(k,x) = 1 + Sum_{m = 1..floor((k+1)/2)} ((1 - x)^(-m) * binomial(k-m, m-1) * Sum_{j = 0..m} (-1)^j * binomial(m, j) * x^m * Phi(x, -m+1, -j+m)) + Sum_{s = 0..floor((k-1)/2)} binomial(k-s-1, s) * PolyLog(-s, x), where Phi is the Lerch transcendent function. (End)
T(n,k) = Sum_{i = 0..k} (-1)^(k+i) * binomial(k,i) * A083857(n,i). - Petros Hadjicostas, Dec 24 2019

Extensions

Various sections edited by Petros Hadjicostas, Dec 24 2019

A110111 Sequence associated to the recurrence b(n) = b(n-1) + 3*b(n-2).

Original entry on oeis.org

0, 1, 7, 133, 1330, 18430, 210490, 2673223, 31940881, 394918819, 4788779380, 58709030380, 715296121540, 8745656280829, 106717441265323, 1303667366328817, 15915556720909510, 194371775990116810
Offset: 0

Views

Author

Paul Barry, Jul 12 2005

Keywords

Comments

In general, let T(n,k) be the solution to T(n,k) = T(n-1,k) + k*T(n-2,k) for n >= 2 with T(0,k) = 0 and T(1,k) = 1 for all k. Then, for fixed k, S(n,k) = T(n,k) * T(n+1,k) * T(n+2,k)/(k+1) has g.f. x/((1 + k*x - k^3*x^2) * (1 - (3*k + 1)*x - k^3*x^2)) (cf. A110112). For the current sequence, a(n) = S(n,k=3) = T(n,3) * T(n+1,3) * T(n+2,3)/4. [Edited by Petros Hadjicostas, Dec 26 2019]

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[x/((1+3x-27x^2)(1-10x-27x^2)),{x,0,30}],x] (* or *) LinearRecurrence[{7,84,-189,-729},{0,1,7,133},30] (* Harvey P. Dale, Jun 29 2022 *)

Formula

G.f.: x/((1 + 3*x - 27*x^2) * (1 - 10*x - 27*x^2)).
a(n) = 7*a(n-1) + 84*a(n-2) - 189*a(n-3) - 729*a(n-4) for n >= 4.
a(n) = b(n) * b(n+1) * b(n+2)/4, where b(n) = (((1 + sqrt(13))/2 )^n - ((1 - sqrt(13))/2)^n)/sqrt(13). [Corrected by Petros Hadjicostas, Dec 26 2019]
a(n) = A006130(n-1) * A006130(n) * A006130(n+1)/4 for n >= 1.
Showing 1-2 of 2 results.