A083857 -id:A083857 - cp's OEIS frontend

A083862 Main diagonal of array A083857.

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

Offset: 0

Paul Barry, May 06 2003

Cf. A083857, A083859, A083863.

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

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

A083858 Expansion of x/(1 - 3x - 6x^2).

Original entry on oeis.org

0, 1, 3, 15, 63, 279, 1215, 5319, 23247, 101655, 444447, 1943271, 8496495, 37149111, 162426303, 710173575, 3105078543, 13576277079, 59359302495, 259535569959, 1134762524847, 4961500994295, 21693078131967, 94848240361671, 414703189876815, 1813199011800471

Offset: 0

Paul Barry, May 06 2003

Binomial transform of A015443. A row of array A083857.

Pisano period lengths: 1, 1, 1, 1, 12, 1, 8, 1, 1, 12, 110, 1, 168, 8, 12, 2, 16, 1, 360, 12, ... - R. J. Mathar, Aug 10 2012

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,6).

Cf. A015523, A015524.

I:=[0,1]; [n le 2 select I[n] else 3*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 16 2018

a[n_]:=(MatrixPower[{{1,2},{1,-4}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{3,6}, {0,1}, 30] (* G. C. Greubel, Jan 16 2018 *)

x='x+O('x^30); concat([0], Vec(x/(1-3*x-6*x^2))) \\ G. C. Greubel, Jan 16 2018

[lucas_number1(n,3,-6) for n in range(0, 24)] # Zerinvary Lajos, Apr 22 2009

a(n) = 3*a(n-1) + 6*a(n-2), a(0)=0, a(1)=1.
a(n) = (3*sqrt(33)/2 + 21/2)^(n/2)/sqrt(33) - (21/2 - 3*sqrt(33)/2)^(n/2)*(-1)^n/sqrt(33).
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(6*k+3 + 6*x )/( x*(6*k+6 + 6*x ) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 21 2013
a(n) = B(n, k + 2^(n-1)) - B(n,k) where B(n,k) is formed by the family of recursions b(n) = 3*(b(n-1) + b(n-2))/2, with b(0) = 1 and b(1) = k, as explained further in A249861. - Richard R. Forberg, Nov 04 2014
a(n) = Sum_{k=0..n-1} 3^k * 2^(n-1-k) * binomial(k,n-1-k). - Seiichi Manyama, Aug 31 2025

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

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

A083861 Square array T(n,k) of second binomial transforms of generalized Fibonacci numbers, read by ascending antidiagonals, with n, k >= 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 5, 0, 1, 5, 19, 0, 1, 5, 20, 65, 0, 1, 5, 21, 75, 211, 0, 1, 5, 22, 85, 275, 665, 0, 1, 5, 23, 95, 341, 1000, 2059, 0, 1, 5, 24, 105, 409, 1365, 3625, 6305, 0, 1, 5, 25, 115, 479, 1760, 5461, 13125, 19171, 0, 1, 5, 26, 125, 551, 2185, 7573, 21845, 47500, 58025

Offset: 0

Paul Barry, May 06 2003

Row n >= 0 of the array gives the solution to the recurrence b(k) = 5*b(k-1) + (n - 6)*b(k-2) for k >= 2 with b(0) = 0 and b(1) = 1. The rows are the binomial transforms of the rows of array A083857. The rows are the second binomial transforms of the generalized Fibonacci numbers in array A083856.

			Array T(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
  0, 1, 5, 19,  65, 211,  665,  2059,  6305,  19171, ...
  0, 1, 5, 20,  75, 275, 1000,  3625, 13125,  47500, ...
  0, 1, 5, 21,  85, 341, 1365,  5461, 21845,  87381, ...
  0, 1, 5, 22,  95, 409, 1760,  7573, 32585, 140206, ...
  0, 1, 5, 23, 105, 479, 2185,  9967, 45465, 207391, ...
  0, 1, 5, 24, 115, 551, 2640, 12649, 60605, 290376, ...
  0, 1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, ...
  ...

G. C. Greubel, Antidiagonals n = 0..100, flattened
OEIS, Transformations of integer sequences.

Rows include A001047 (n=0), A093131 (n=1), A002450 (n=2), A004254 (n=5), A000351 (n=6), A052918 (n=7), A015535 (n=8), A015536 (n=9), A015537 (n=10).

Cf. A083856 (second inverse binomial transform), A083856 (first inverse binomial transform), A082297 (main diagonal).

T:= func< n,k | Round( (((5+Sqrt(4*n+1))/2)^k - ((5-Sqrt(4*n+1))/2)^k)/Sqrt(4*n + 1) ) >;
[T(n-k,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 27 2019

seq(seq(round( (((5+sqrt(4*(n-k)+1))/2)^k - ((5-sqrt(4*(n-k)+1))/2)^k)/sqrt(4*(n-k)+1) ), k=0..n), n=0..10); # G. C. Greubel, Dec 27 2019

T[n_, k_]:= Round[(((5 +Sqrt[4*n+1])/2)^k - ((5 -Sqrt[4*n+1])/2)^k)/Sqrt[4*n+1]]; Table[T[n-k, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 27 2019 *)

T(n, k) = round( (((5+sqrt(4*n+1))/2)^k - ((5-sqrt(4*n+1))/2)^k)/sqrt(4*n + 1) );
for(n=0,10, for(k=0,n, print1(T(n-k,k), ", "))) \\ G. C. Greubel, Dec 27 2019

[[round( (((5+sqrt(4*(n-k)+1))/2)^k - ((5-sqrt(4*(n-k)+1))/2)^k)/sqrt(4*(n-k)+1) ) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Dec 27 2019

T(n, k) = (((5 + sqrt(4*n + 1))/2)^k - ((5 - sqrt(4*n + 1))/2)^k)/sqrt(4*n + 1).
O.g.f. for row n >= 0: -x/(-1 + 5*x + (n-6)*x^2) . - R. J. Mathar, Dec 02 2007
From Petros Hadjicostas, Dec 25 2019: (Start)
T(n,k) = 5*T(n,k-1) + (n - 6)*T(n,k-2) for k >= 2 with T(n,0) = 0 and T(n,1) = 1 for all n >= 0.
T(n,k) = Sum_{i = 0..k} binomial(k,i) * A083857(n,i).
T(n,k) = Sum_{i = 0..k} Sum_{j = 0..i} binomial(k,i) * binomial(i,j) * A083856(n,j). (End)

Name and various sections edited by Petros Hadjicostas, Dec 25 2019

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A083862 Main diagonal of array A083857.

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A083858 Expansion of x/(1 - 3*x - 6*x^2).

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

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A083861 Square array T(n,k) of second binomial transforms of generalized Fibonacci numbers, read by ascending antidiagonals, with n, k >= 0.

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Magma

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A083858 Expansion of x/(1 - 3x - 6x^2).