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.

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

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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

A214978 Array T(m,n) = Fibonacci(m*n)/Fibonacci(m), by antidiagonals; transpose of A028412.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 8, 4, 1, 5, 21, 17, 7, 1, 8, 55, 72, 48, 11, 1, 13, 144, 305, 329, 122, 18, 1, 21, 377, 1292, 2255, 1353, 323, 29, 1, 34, 987, 5473, 15456, 15005, 5796, 842, 47, 1, 55, 2584, 23184, 105937, 166408, 104005, 24447, 2208, 76, 1, 89
Offset: 1

Views

Author

Clark Kimberling, Oct 27 2012

Keywords

Comments

The main entry is the transpose, A028412. In the present format, the array can be compared directly with A214984 and A214986.

Examples

			Northwest corner:
  1  1   2    3      5       8
  1  3   8   21     55     144
  1  4  17   72    305    1292
  1  7  48  329   2255   15456
  1 11 122 1353  15005  166408
  1 18 323 5796 104005 1866294
		

Crossrefs

Programs

  • Mathematica
    F[n_] := Fibonacci[n]; t[m_, n_] := F[m*n]/F[m]
    TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]]
    u = Table[t[k, n + 1 - k], {n, 1, 12}, {k, 1, n}];
    v[n_] := Sum[F[m*(n + 1 - m)]/F[m], {m, 1, n}];
    Flatten[u]                           (* A213978 *)
    Flatten[Table[t[n, n], {n, 1, 20}]]  (* A051294 *)
    Table[(t[n, 5] - 5)/50, {n, 1, 20}]  (* A214982 *)
    Table[v[n], {n, 1, 30}]              (* A214983 *)

Formula

T(m,n) = Fibonacci(m*n)/Fibonacci(m).

A103624 Quotient F(n(n+1))/{F(n)*F(n+1)}, where F(n) is the n-th Fibonacci number A000045(n).

Original entry on oeis.org

1, 4, 24, 451, 20801, 2576099, 827294629, 698114862576, 1540142884690276, 8900260727038783399, 134626641626040936157699, 5331733869372412024840703621, 552800277142057127306392295957801
Offset: 1

Views

Author

Lekraj Beedassy, Mar 25 2005

Keywords

Crossrefs

Programs

  • Maple
    seq(combinat:-fibonacci(n*(n+1))/combinat:-fibonacci(n)/combinat:-fibonacci(n+1),n=1..25); # Robert Israel, Jan 11 2018
  • Mathematica
    Table[Fibonacci[n(n+1)]/(Fibonacci[n]Fibonacci[n+1]),{n,20}] (* Harvey P. Dale, Aug 04 2012 *)
  • PARI
    {a(n)=fibonacci(n*(n+1))/(fibonacci(n)*fibonacci(n+1))}  \\ Paul D. Hanna, Jan 28 2012

Formula

a(n) = phi^(n^2 - n - 1) + O(phi^(n^2 - 3n)). [Charles R Greathouse IV, Feb 01 2012]

Extensions

a(2) corrected by Paul D. Hanna, Jan 28 2012

A205505 Fibonacci(n*(n+1)) / Fibonacci(n).

Original entry on oeis.org

1, 8, 72, 2255, 166408, 33489287, 17373187209, 23735905327584, 84707858657965180, 792123204706451722511, 19386236394149894806708656, 1242293991563772001787883943693, 208405704482555536994509895576090977, 91533085042008706066658193727853843719640
Offset: 1

Views

Author

Paul D. Hanna, Jan 28 2012

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Fibonacci[n(n+1)]/Fibonacci[n],{n,20}] (* Harvey P. Dale, Mar 30 2012 *)
  • PARI
    {a(n)=fibonacci(n*(n+1))/fibonacci(n)}
    
  • PARI
    {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
    {a(n)=polcoeff(1/(1-Lucas(n)*x+(-1)^n*x^2+x*O(x^n)), n)}

Formula

a(n) = [x^n] 1/(1 - Lucas(n)*x + (-1)^n*x^2), where Lucas(n) = A000204(n).
Forms a diagonal in table A028412.

A380083 a(n) = Pell(n^2)/Pell(n).

Original entry on oeis.org

1, 6, 197, 39236, 45232349, 304285766994, 11928254138546089, 2725453049877127789064, 3629520789795568149638626009, 28171611459441395148628640333550174, 1274457582507820938168220698796661580252461, 336039604487720392926819615640785342048933644491212
Offset: 1

Views

Author

Seiichi Manyama, May 08 2025

Keywords

Crossrefs

Main diagonal of A383742.

Programs

  • Maple
    a:= n-> (f->f(n^2)/f(n))(k->(<<2|1>, <1|0>>^k)[1, 2]):
    seq(a(n), n=1..12);  # Alois P. Heinz, May 08 2025
  • Mathematica
    a[n_] := Fibonacci[n^2, 2]/Fibonacci[n, 2]; Array[a, 12] (* Amiram Eldar, May 08 2025 *)
  • PARI
    pell(n) = ([2, 1; 1, 0]^n)[2, 1];
    a(n) = pell(n^2)/pell(n);

Formula

a(n) = A204327(n)/A000129(n).
a(n) = [x^n] x/(1 - A002203(n)*x + (-1)^n*x^2).
Showing 1-5 of 5 results.