A106734 -id:A106734 - cp's OEIS frontend

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

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

Paul Barry, May 06 2003

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.

			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]
  ...

A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, 21(2) (2015), 35-42.

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).

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

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

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 *)

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

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

R. H. Hardin, Jul 24 2011

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

			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

R. H. Hardin, Table of n, a(n) for n = 1..9999
Ron H. Hardin, Re: A193376 Tabl = 20 existing sequences, Sequence Fans mailing list, 2011.
Robert Israel, Re: A193376 Tabl = 20 existing sequences, Sequence Fans mailing list, 2011.

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).

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

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 *)

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

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

A030440 Values of Newton-Gregory forward interpolating polynomial (1/3)(n-1)(2n+3)(2*n-1).

Original entry on oeis.org

1, 0, 7, 30, 77, 156, 275, 442, 665, 952, 1311, 1750, 2277, 2900, 3627, 4466, 5425, 6512, 7735, 9102, 10621, 12300, 14147, 16170, 18377, 20776, 23375, 26182, 29205, 32452, 35931, 39650, 43617, 47840, 52327, 57086, 62125, 67452, 73075, 79002, 85241, 91800, 98687, 105910

Offset: 0

Ilias.Kotsireas(AT)lip6.fr (Ilias Kotsireas)

Starting at a(2)=7, partial sums of A073577. - J. M. Bergot, Apr 20 2016

S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234; http://www.scirp.org/journal/am; http://dx.doi.org/10.4236/am.2014.515216

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A073577, A000384 (trinomial k=2 column), A106734, A127672, A027907.

[(1/3)*(n-1)*(2*n-1)*(2*n+3):n in [0..50]]; // Vincenzo Librandi, Apr 20 2018

LinearRecurrence[{4, -6, 4, -1}, {1, 0, 7, 30}, 40] (* Vincenzo Librandi, Apr 20 2018 *)

a(n) = (n-1)*(2*n-1)*(2*n+3)/3; \\ Altug Alkan, Apr 19 2018

G.f.: (1+13*x^2-2*x^3-4*x)/(1-x)^4. - R. J. Mathar, May 18 2014
a(n) = (1/6) * (A106734(2n) - 1), n > 0. - Mathew Englander, Jun 06 2014
E.g.f.: (3 - 3*x + 12*x^2 + 4*x^3)*exp(x)/3. - Ilya Gutkovskiy, Apr 20 2016
a(n+1) = trinomial(2*n+1, 3) = binomial(2*n+1, 3) + (2*n+1)*(2*n) = n*(2*n+1)*(2*n+5)/3, for n >= 0, with the trinomial irregular triangle A027907. a(n+1) = trinomial(2*n+1, 4*n-1), for n >= 1 (symmetry). a(n+1) = Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n+1)*R(4*(n-1), x)/Pi with the R polynomial coefficients given in A127672. [Comtet, p. 77, the integral formula for q=3, n -> 2*n+1, k = 3, rewritten with x = 2*cos(phi)]. The g.f. of {a(n+1)}{n>= 0} is x*(7 + 2*x - x^2)/(1 - x)^4. - _Wolfdieter Lang, Apr 19 2018

cp's OEIS Frontend

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

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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.

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A030440 Values of Newton-Gregory forward interpolating polynomial (1/3)*(n-1)*(2*n+3)*(2*n-1).

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A030440 Values of Newton-Gregory forward interpolating polynomial (1/3)(n-1)(2n+3)(2*n-1).