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 1-5 of 5 results.

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

A193376 T(n,k) = number of ways to place any number of 2 X 1 tiles of k distinguishable colors into an n X 1 grid; array read by descending antidiagonals, with n, k >= 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 5, 5, 1, 5, 7, 11, 8, 1, 6, 9, 19, 21, 13, 1, 7, 11, 29, 40, 43, 21, 1, 8, 13, 41, 65, 97, 85, 34, 1, 9, 15, 55, 96, 181, 217, 171, 55, 1, 10, 17, 71, 133, 301, 441, 508, 341, 89, 1, 11, 19, 89, 176, 463, 781, 1165, 1159, 683, 144, 1, 12, 21, 109, 225, 673
Offset: 1

Views

Author

R. H. Hardin, Jul 24 2011

Keywords

Comments

Transposed variant of A083856. - R. J. Mathar, Aug 23 2011
As to the sequences by columns beginning (1, N, ...), let m = (N-1). The g.f. for the sequence (1, N, ...) is 1/(1 - x - m*x^2). Alternatively, the corresponding matrix generator is [[1,1], [m,0]]. Another equivalency is simply: The sequence beginning (1, N, ...) is the INVERT transform of (1, m, 0, 0, 0, ...). Convergents to the sequences a(n)/a(n-1) are (1 + sqrt(4*m+1))/2. - Gary W. Adamson, Feb 25 2014

Examples

			Array T(n,k) (with rows n >= 1 and column k >= 1) begins as follows:
  ..1...1....1....1.....1.....1.....1......1......1......1......1......1...
  ..2...3....4....5.....6.....7.....8......9.....10.....11.....12.....13...
  ..3...5....7....9....11....13....15.....17.....19.....21.....23.....25...
  ..5..11...19...29....41....55....71.....89....109....131....155....181...
  ..8..21...40...65....96...133...176....225....280....341....408....481...
  .13..43...97..181...301...463...673....937...1261...1651...2113...2653...
  .21..85..217..441...781..1261..1905...2737...3781...5061...6601...8425...
  .34.171..508.1165..2286..4039..6616..10233..15130..21571..29844..40261...
  .55.341.1159.2929..6191.11605.19951..32129..49159..72181.102455.141361...
  .89.683.2683.7589.17621.35839.66263.113993.185329.287891.430739.624493...
  ...
Some solutions for n = 5 and k = 3 with colors = 1, 2, 3 and empty = 0:
..0....2....3....2....0....1....0....0....2....0....0....2....3....0....0....0
..0....2....3....2....2....1....2....3....2....1....0....2....3....1....1....1
..1....0....0....0....2....0....2....3....2....1....0....1....0....1....1....1
..1....2....2....0....3....2....2....3....2....0....3....1....3....3....2....1
..0....2....2....0....3....2....2....3....0....0....3....0....3....3....2....1
		

Crossrefs

Column 1 is A000045(n+1), column 2 is A001045(n+1), column 3 is A006130, column 4 is A006131, column 5 is A015440, column 6 is A015441(n+1), column 7 is A015442(n+1), column 8 is A015443, column 9 is A015445, column 10 is A015446, column 11 is A015447, and column 12 is A053404,
Row 2 is A000027(n+1), row 3 is A004273(n+1), row 4 is A028387, row 5 is A000567(n+1), and row 6 is A106734(n+2).
Diagonal is A171180, superdiagonal 1 is A083859(n+1), and superdiagonal 2 is A083860(n+1).

Programs

  • Maple
    T:= proc(n,k) option remember; `if`(n<0, 0,
          `if`(n<2 or k=0, 1, k*T(n-2, k) +T(n-1, k)))
        end;
    seq(seq(T(n, d+1-n), n=1..d), d=1..12); # Alois P. Heinz, Jul 29 2011
  • Mathematica
    T[n_, k_] := T[n, k] = If[n < 0, 0, If[n < 2 || k == 0, 1, k*T[n-2, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 12}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

Formula

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. Thus, T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n = 0, 1, ..., z-1.
The solution is T(n,k) = Sum_r r^(-n-1)/(1 + z*k*r^(z-1)), where the sum is over the roots r of the polynomial k*x^z + x - 1.
For z = 2, T(n,k) = ((2*k / (sqrt(1 + 4*k) - 1))^(n+1) - (-2*k/(sqrt(1 + 4*k) + 1))^(n+1)) / sqrt(1 + 4*k).
T(n,k) = Sum_{s=0..[n/2]} binomial(n-s,s) * k^s.
For z X 1 tiles, T(n,k,z) = Sum_{s = 0..[n/z]} binomial(n-(z-1)*s, s) * k^s. - R. H. Hardin, Jul 31 2011

Extensions

Formula and proof from Robert Israel in the Sequence Fans mailing list.

A083860 First subdiagonal of generalized Fibonacci array A083856.

Original entry on oeis.org

0, 1, 1, 5, 11, 55, 176, 937, 3781, 21571, 102455, 624493, 3356640, 21752431, 129055681, 884773585, 5696734715, 41129090011, 283908657880, 2149818248341, 15765656131765, 124759995175751, 965186517474191, 7956847444317049, 64577172850366176, 553048437381116275
Offset: 0

Views

Author

Paul Barry, May 06 2003

Keywords

Crossrefs

Programs

  • Maple
    T := proc(n, k) local v; option remember; if 0 <= n and k = 0 then v := 0; end if; if 0 <= n and k = 1 then v := 1; end if; if 0 <= n and 2 <= k then v := T(n, k - 1) + n*T(n, k - 2); end if; v; end proc;
    seq(T(n + 1, n), n = 0 .. 40); # Petros Hadjicostas, Dec 25 2019
  • Mathematica
    T[, 0] = 0; T[, 1|2] = 1;
    T[n_, k_] := T[n, k] = T[n, k-1] + n T[n, k-2];
    a[n_] := T[n+1, n];
    Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Sep 26 2022 *)

Formula

a(n) = (((1 + sqrt(4*n + 5))/2)^n - ((1 - sqrt(4*n + 5))/2)^n)/sqrt(4*n + 5).
a(n) = A193376(n-1, n+1) for n >= 2. - R. J. Mathar, Aug 23 2011
a(n) = Sum_{s = 0..floor((n-1)/2)} binomial(n-1-s, s) * (n+1)^s. - Petros Hadjicostas, Dec 25 2019

A083862 Main diagonal of array A083857.

Original entry on oeis.org

0, 1, 3, 10, 39, 171, 819, 4229, 23247, 135088, 824163, 5255361, 34876647, 240130801, 1710323667, 12572548582, 95183606559, 740843916471, 5918533467075, 48463830575837, 406239439253175, 3481899731310196
Offset: 0

Views

Author

Paul Barry, May 06 2003

Keywords

Crossrefs

Programs

  • Maple
    T := proc(n, k) local v; option remember; if 0 <= n and k = 0 then v := 0; end if; if 0 <= n and k = 1 then v := 1; end if; if 0 <= n and 2 <= k then v := 3*T(n, k - 1) + (n - 2)*T(n, k - 2); end if; v; end proc;
    seq(T(n, n), n = 0 .. 40); # Petros Hadjicostas, Dec 25 2019

Formula

a(n) = (((3 + sqrt(4*n + 1))/2)^n - ((3 - sqrt(4*n + 1))/2)^n)/sqrt(4*n + 1).

A082297 Main diagonal of array A083861.

Original entry on oeis.org

0, 1, 5, 22, 105, 551, 3125, 18901, 120785, 809956, 5669525, 41253125, 310968905, 2421594449, 19434554165, 160416509794, 1359425390625, 11809173506531, 105013666399445, 954781355707373, 8865807366284825, 83996265859375000, 811222567266570805, 7980019803318567233
Offset: 0

Views

Author

Paul Barry, May 06 2003

Keywords

Crossrefs

Programs

  • Maple
    T := proc(n, k) local v; option remember; if 0 <= n and k = 0 then v := 0; end if; if 0 <= n and k = 1 then v := 1; end if; if 0 <= n and 2 <= k then v := 5*T(n, k - 1) + (n - 6)*T(n, k - 2); end if; v; end proc;
    seq(T(n, n), n = 0..40); # Petros Hadjicostas, Dec 25 2019

Formula

a(n) = (((5 + sqrt(4*n + 1))/2)^n - ((5 - sqrt(4*n + 1))/2)^n)/sqrt(4*n + 1).
Showing 1-5 of 5 results.