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 1-4 of 4 results.

A028412 Rectangular array of numbers Fibonacci(m(n+1))/Fibonacci(m), m >= 1, n >= 0, read by downward antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 4, 8, 3, 1, 7, 17, 21, 5, 1, 11, 48, 72, 55, 8, 1, 18, 122, 329, 305, 144, 13, 1, 29, 323, 1353, 2255, 1292, 377, 21, 1, 47, 842, 5796, 15005, 15456, 5473, 987, 34, 1, 76, 2208, 24447, 104005, 166408, 105937, 23184, 2584, 55, 1, 123, 5777
Offset: 0

Views

Author

Keywords

Comments

Every integer-valued quotient of two Fibonacci numbers is in this array. - Clark Kimberling, Aug 28 2008
Not only does 5 divide row 5, but 50 divides (-5 + row 5), as in A214984. - Clark Kimberling, Nov 02 2012

Examples

			   1   1    1      1       1        1
   1   3    4      7      11       18
   2   8   17     48     122      323
   3  21   72    329    1353     5796
   5  55  305   2255   15005   104005
   8 144 1292  15456  166408  1866294
  13 377 5473 105937 1845493 33489287
  ...
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 142.

Crossrefs

Rows include (essentially) A000032, A047946, A083564, A103226.
Main diagonal is A051294.
Transpose is A214978.

Programs

  • Mathematica
    max = 11; col[m_] := CoefficientList[ Series[ 1/(1 - LucasL[m]*x + (-1)^m*x^2), {x, 0, max}], x]; t = Transpose[ Table[ col[m], {m, 1, max}]] ; Flatten[ Table[ t[[n - m + 1, m]], {n, 1, max }, {m, n, 1, -1}]] (* Jean-François Alcover, Feb 21 2012, after Paul D. Hanna *)
    f[n_] := Fibonacci[n]; t[m_, n_] := f[m*n]/f[n]
    TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]] (* array *)
    t = Flatten[Table[t[k, n + 1 - k], {n, 1, 120}, {k, 1, n}]] (* sequence *) (* Clark Kimberling, Nov 02 2012 *)
  • PARI
    {T(n,m)=polcoeff(1/(1 - Lucas(m)*x + (-1)^m*x^2 +x*O(x^n)),n)}

Formula

T(n, m) = Sum_{i_1>=0} Sum_{i_2>=0} ... Sum_{i_m>=0} C(n-i_m, i_1)*C(n-i_1, i_2)*C(n-i_2, i_3)*...*C(n-i_{m-1}, i_m).
G.f. for column m >= 1: 1/(1 - Lucas(m)*x + (-1)^m*x^2), where Lucas(m) = A000204(m). - Paul D. Hanna, Jan 28 2012

Extensions

More terms from Erich Friedman, Jun 03 2001
Edited by Ralf Stephan, Feb 03 2005
Better description from Clark Kimberling, Aug 28 2008

A056567 Fibonomial coefficients.

Original entry on oeis.org

1, 55, 4895, 352440, 27372840, 2063912136, 157373300370, 11948265189630, 908637119420910, 69056421075989160, 5249543573067466872, 399024295188779925720, 30331388438447118520355, 2305576054220330112077285, 175254358052498673797874685, 13321629629800423781409595728
Offset: 0

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Crossrefs

Cf. A010048, A000045, A001654-8, A056565-6, A001076 (signed), A049666, A049667 (signed), A049669.

Programs

  • Mathematica
    a[n_] := (1/2227680) Times @@ Fibonacci[n + Range[9]]; Array[a, 20, 0] (* Giovanni Resta, May 08 2016 *)
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
    vector(20, n, b(n-1, 9))  \\ Joerg Arndt, May 08 2016

Formula

a(n) = A010048(n+9, 9) = Fibonomial(n+9, 9).
G.f.: 1/p(10, n) with p(10, n)= 1 - 55*x - 1870*x^2 + 19635*x^3 + 85085*x^4 - 136136*x^5 - 85085*x^6 + 19635*x^7 + 1870*x^8 - 55*x^9 - x^10 = (1 - x - x^2)*(1 + 4*x - x^2)*(1 - 11*x - x^2)*(1 + 29*x - x^2)*(1 - 76*x - x^2) (n=10 row polynomial of signed Fibonomial triangle A055870; see this entry for Knuth and Riordan references).
Recursion: a(n) = 76*a(n-1) + a(n-2)+((-1)^n)*A056565(n), n >= 2, a(0)=1, a(1)=55.
G.f.: exp( Sum_{k>=1} F(10*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

A138590 a(n) = Fibonacci(9*n).

Original entry on oeis.org

0, 34, 2584, 196418, 14930352, 1134903170, 86267571272, 6557470319842, 498454011879264, 37889062373143906, 2880067194370816120, 218922995834555169026, 16641027750620563662096, 1264937032042997393488322, 96151855463018422468774568
Offset: 0

Views

Author

Zerinvary Lajos, May 13 2008

Keywords

Crossrefs

Programs

Formula

a(n) = 34*A049669(n). G.f.: -34*x / ( -1+76*x+x^2 ). - R. J. Mathar, Sep 30 2013
From Colin Barker, Apr 06 2017: (Start)
a(n) = ((-38 - 17*sqrt(5))^(-n)*(-1 + (-2889 - 1292*sqrt(5))^n)) / sqrt(5).
a(n) = 76*a(n-1) + a(n-2) for n > 1.
(End)

A305413 a(n) = Fibonacci(11*n)/89.

Original entry on oeis.org

0, 1, 199, 39602, 7880997, 1568358005, 312111123992, 62111682032413, 12360536835574179, 2459808941961294034, 489514339987133086945, 97415813466381445596089, 19386236394149894806708656, 3857958458249295447980618633, 767753119428003944042949816623
Offset: 0

Views

Author

Vincenzo Librandi, Jun 05 2018

Keywords

Crossrefs

Cf. similar sequences: F(3*n)/2 (A001076), F(4*n)/3 (A004187), F(5*n)/5 (A049666), F(6*n)/8 (A049660), F(7*n)/13 (A049667), F(8*n)/21 (A049668), F(9*n)/34 (A049669), F(10*n)/55 (A049670), F(11*n)/89 (this sequence), F(12*n)/144 (A253368).

Programs

  • Magma
    [Fibonacci(11*n)/89: n in [0..30]];
    
  • Mathematica
    Fibonacci[11 Range[0, 20]]/89
    LinearRecurrence[{199,1},{0,1},20] (* Harvey P. Dale, Aug 03 2024 *)
  • PARI
    a(n) = fibonacci(11*n)/89 \\ Felix Fröhlich, Jul 30 2019

Formula

G.f.: x/(1 - 199*x - x^2).
a(n) = 199*a(n-1) + a(n-2) for n>1, a(0)=0, a(1)=1.
a(n) = A167398(n)/89.
For n >= 1, a(n) equals the denominator of the continued fraction [199, 199, ..., 199] (with n copies of 199). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 29 2019
Showing 1-4 of 4 results.