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.

Previous Showing 11-19 of 19 results.

A319503 Number of partitions of Fibonacci(n) into exactly n positive Fibonacci numbers.

Original entry on oeis.org

1, 1, 0, 0, 0, 1, 2, 6, 16, 43, 117, 305, 769, 1907, 4686, 11587, 28580, 70451, 172880, 423629, 1036332, 2533559, 6186635, 15092985, 36784586, 89590410, 218069921, 530551804, 1290218120, 3136385254, 7621522229, 18515039477, 44966884766, 109184448962
Offset: 0

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Author

Alois P. Heinz, Sep 20 2018

Keywords

Examples

			a(0) = 1: the empty partition.
a(1) = 1: 1.
a(5) = 1: 11111.
a(6) = 2: 221111, 311111.
a(7) = 6: 2222221, 3222211, 3322111, 3331111, 5221111, 5311111.
		

Crossrefs

Programs

  • Mathematica
    (* Program not suitable for a large number of terms. *)
    a[n_] := a[n] = If[n < 2, 1, IntegerPartitions[Fibonacci[n], {n}, Fibonacci[Range[2, n - 1]]] //Length];
    Table[Print[n, " ", a[n]]; a[n], {n, 0, 24}] (* Jean-François Alcover, Dec 08 2020 *)

Formula

a(n) = [x^A000045(n) y^n] 1/Product_{j>=2} (1-y*x^A000045(j)).
a(n) = A319394(A000045(n),n).

A359511 Number of partitions of n into at most 2 positive Fibonacci numbers (with a single type of 1).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 1, 0, 2, 1, 1, 2, 0, 1, 0, 0, 2, 1, 1, 1, 0, 2, 0, 0, 1, 0, 0, 0, 0, 2, 1, 1, 1, 0, 1, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 1, 1, 0, 1, 0, 0, 1
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 03 2023

Keywords

Crossrefs

Formula

a(n) = Sum_{k=0..2} A319394(n,k). - Alois P. Heinz, Jan 03 2023

A359512 Number of partitions of n into at most 3 positive Fibonacci numbers (with a single type of 1).

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 3, 4, 3, 4, 4, 3, 4, 3, 1, 4, 2, 4, 4, 2, 4, 3, 2, 4, 1, 3, 1, 0, 4, 2, 3, 4, 2, 4, 2, 1, 4, 2, 3, 2, 0, 4, 1, 1, 3, 0, 1, 0, 0, 4, 2, 3, 3, 2, 4, 2, 1, 4, 1, 2, 1, 0, 4, 2, 2, 3, 0, 2, 0, 0, 4, 1, 1, 1, 0, 3, 0, 0, 1, 0, 0, 0, 0, 4, 2, 3, 3
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 03 2023

Keywords

Crossrefs

Formula

a(n) = Sum_{k=0..3} A319394(n,k). - Alois P. Heinz, Jan 03 2023

A319402 Number of partitions of n into exactly nine positive Fibonacci numbers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 4, 4, 6, 7, 10, 11, 13, 14, 18, 18, 22, 23, 27, 26, 31, 30, 36, 36, 39, 39, 45, 43, 49, 49, 55, 52, 58, 56, 63, 62, 65, 64, 71, 68, 73, 72, 79, 77, 82, 81, 87, 86, 90, 89, 96, 93, 96, 96, 101, 100, 101, 100, 107, 103, 108
Offset: 0

Views

Author

Alois P. Heinz, Sep 18 2018

Keywords

Crossrefs

Column k=9 of A319394.
Cf. A000045.

Programs

  • Maple
    h:= proc(n) option remember; `if`(n<1, 0, `if`((t->
          issqr(t+4) or issqr(t-4))(5*n^2), n, h(n-1)))
        end:
    b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(i<1 or
          t<1, 0, b(n, h(i-1), t)+b(n-i, h(min(n-i, i)), t-1)))
        end:
    a:= n-> (k-> b(n, h(n), k)-b(n, h(n), k-1))(9):
    seq(a(n), n=0..120);

Formula

a(n) = [x^n y^9] 1/Product_{j>=2} (1-y*x^A000045(j)).

A319403 Number of partitions of n into exactly ten positive Fibonacci numbers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 4, 4, 6, 7, 10, 11, 14, 15, 19, 20, 24, 25, 31, 30, 35, 36, 42, 42, 48, 47, 54, 54, 59, 60, 69, 66, 73, 72, 80, 79, 86, 85, 92, 91, 97, 96, 107, 103, 110, 110, 118, 117, 123, 123, 132, 130, 135, 134, 142, 141, 146, 145
Offset: 0

Views

Author

Alois P. Heinz, Sep 18 2018

Keywords

Crossrefs

Column k=10 of A319394.
Cf. A000045.

Programs

  • Maple
    h:= proc(n) option remember; `if`(n<1, 0, `if`((t->
          issqr(t+4) or issqr(t-4))(5*n^2), n, h(n-1)))
        end:
    b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(i<1 or
          t<1, 0, b(n, h(i-1), t)+b(n-i, h(min(n-i, i)), t-1)))
        end:
    a:= n-> (k-> b(n, h(n), k)-b(n, h(n), k-1))(10):
    seq(a(n), n=0..120);

Formula

a(n) = [x^n y^10] 1/Product_{j>=2} (1-y*x^A000045(j)).

A136343 Number of partitions of n such that each summand is greater than or equal to the sum of the next two summands.

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 6, 7, 10, 11, 14, 16, 21, 23, 29, 32, 40, 43, 52, 57, 69, 75, 88, 96, 113, 122, 141, 153, 177, 190, 216, 233, 265, 285, 320, 345, 387, 415, 461, 495, 551, 589, 650, 695, 767, 818, 896, 957, 1048, 1116, 1214, 1293, 1407, 1495, 1620, 1721, 1864
Offset: 0

Views

Author

David S. Newman, May 11 2008

Keywords

Comments

This sequence was suggested by Moshe Shmuel Newman. The idea came to him while reading a paper of Lev Shneerson.
Number of partitions of 2n into exactly n positive Fibonacci numbers. a(8) = 10: 82111111, 55111111, 53311111, 53221111, 52222111, 33331111, 33322111, 33222211, 32222221, 22222222. - Alois P. Heinz, Sep 18 2018

Examples

			a(5) = 4 because 4 of the 7 partitions of 5 have the required property: {5} {4,1} {3,2} {3,1,1}. The other 3 partitions of 5: {2,2,1} {2,1,1,1} and {1,1,1,1,1} each have an element which is < the sum of next two.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, j) option remember; `if`(n=0, 1,
          `if`(i<1, 0, b(n-i, min(n-i, i,
          `if`(j=0, i, j-i)), i) +b(n, i-1, j)))
        end:
    a:= n-> b(n$2, 0):
    seq(a(n), n=0..60);  # Alois P. Heinz, Jul 29 2017
  • Mathematica
    b[n_, i_, j_]:=b[n, i, j]=If[n==0, 1, If[i<1, 0, b[n - i, Min[n - i, i, If[j==0, i, j - i]], i] + b[n, i - 1, j]]]; Table[b[n, n, 0], {n, 0, 60}] (* Indranil Ghosh, Aug 01 2017, after Maple code *)

Formula

From Alois P. Heinz, Sep 18 2018: (Start)
a(n) = [x^(2n) y^n] 1/Product_{j>=2} (1-y*x^A000045(j)).
a(n) = A319394(2n,n). (End)

Extensions

Conjectured g.f. removed and a(0), a(35)-a(56) added by Alois P. Heinz, Jul 29 2017

A281689 Expansion of Sum_{k>=2} x^Fibonacci(k)/(1 - x^Fibonacci(k)) / Product_{k>=2} (1 - x^Fibonacci(k)).

Original entry on oeis.org

1, 3, 6, 11, 18, 29, 42, 62, 86, 119, 159, 211, 273, 352, 446, 562, 697, 864, 1054, 1284, 1550, 1860, 2220, 2639, 3114, 3669, 4293, 5011, 5823, 6745, 7783, 8956, 10268, 11747, 13390, 15237, 17281, 19561, 22089, 24889, 27979, 31405, 35157, 39309, 43856, 48849, 54319, 60309, 66840, 73992, 81760, 90243
Offset: 1

Views

Author

Ilya Gutkovskiy, Jan 27 2017

Keywords

Comments

Total number of parts in all partitions of n into Fibonacci parts (with a single type of 1).
Convolution of A003107 and A005086.

Examples

			a(5) = 18 because we have [5], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1], [1, 1, 1, 1, 1] and 1 + 2 + 3 + 3 + 4 + 5 = 18.
		

Crossrefs

Programs

  • Maple
    h:= proc(n) option remember; `if`(n<1, 0, `if`((t->
          issqr(t+4) or issqr(t-4))(5*n^2), n, h(n-1)))
        end:
    b:= proc(n, i) option remember; `if`(n=0 or i=1, [1, n],
           b(n, h(i-1))+(p->p+[0, p[1]])(b(n-i, h(min(n-i, i)))))
        end:
    a:= n-> b(n, h(n))[2]:
    seq(a(n), n=1..70);  # Alois P. Heinz, Sep 18 2018
  • Mathematica
    Rest[CoefficientList[Series[Sum[x^Fibonacci[k]/(1 - x^Fibonacci[k]), {k, 2, 20}]/Product[1 - x^Fibonacci[k], {k, 2, 20}], {x, 0, 52}], x]]

Formula

G.f.: Sum_{k>=2} x^Fibonacci(k)/(1 - x^Fibonacci(k)) / Product_{k>=2} (1 - x^Fibonacci(k)).
a(n) = Sum_{k=1..n} k * A319394(n,k). - Alois P. Heinz, Sep 18 2018

A343537 Number of partitions of the n-th Fibonacci number into a Fibonacci number of Fibonacci parts.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 7, 16, 41, 135, 632, 4091, 37020, 478852, 8897512, 240133480, 9489055662, 552854898873, 47794151866058, 6165361571608551, 1192709563056788508, 347571453153709529743, 153189847887607116894958
Offset: 0

Views

Author

Alois P. Heinz, May 26 2021

Keywords

Examples

			a(5) = 5: [5], [3,2], [3,1,1], [2,2,1], [1,1,1,1,1].  Partition [2,1,1,1] is not counted because 4 (the number of parts) is not a Fibonacci number.
a(6) = 7: [8], [5,3], [5,2,1], [3,3,2], [3,2,1,1,1], [2,2,2,1,1], [1,1,1,1,1,1,1,1].
a(7) = 16: [13], [8,5], [8,3,2], [8,2,1,1,1], [5,5,3], [5,5,1,1,1], [5,3,3,1,1], [5,3,2,2,1], [5,2,2,2,2], [5,2,1,1,1,1,1,1], [3,3,3,3,1], [3,3,3,2,2], [3,3,2,1,1,1,1,1], [3,2,2,2,1,1,1,1], [2,2,2,2,2,1,1,1], [1,1,1,1,1,1,1,1,1,1,1,1,1].
		

Crossrefs

Programs

  • Maple
    f:= n-> (t-> issqr(t+4) or issqr(t-4))(5*n^2):
    h:= proc(n) option remember; `if`(f(n), n, h(n-1)) end:
    b:= proc(n, i, c) option remember; `if`(n=0 or i=1, `if`(
          f(c+n), 1, 0), b(n-i, h(min(n-i, i)), c+1)+b(n, h(i-1), c))
        end:
    a:= n-> b((<<0|1>, <1|1>>^n)[1, 2]$2, 0):
    seq(a(n), n=0..17);
  • Mathematica
    $RecursionLimit = 10000;
    f[n_] := With[{t = 5 n^2}, IntegerQ@Sqrt[t+4] || IntegerQ@Sqrt[t-4]];
    h[n_] := h[n] = If[f[n], n, h[n - 1]] ;
    b[n_, i_, c_] := b[n, i, c] = If[n == 0 || i == 1, If[f[c+n], 1, 0], b[n-i, h[Min[n-i, i]], c+1] + b[n, h[i-1], c]];
    a[n_] := a[n] = With[{m = MatrixPower[{{0, 1}, {1, 1}}, n][[1, 2]]}, b[m, m, 0]];
    Table[Print[n, " ", a[n]]; a[n], {n, 0, 17}] (* Jean-François Alcover, Sep 09 2022, after Alois P. Heinz *)

Formula

a(n) = Sum_{k in {A000045}} A319394(A000045(n),k).

A356977 a(n) is the number of solutions, j >= 0 and 2 <= m_1 <= ... <= m_n, of the equation Sum_{k=1..n} F(m_k) = 2^j where F(i) is the i-th Fibonacci number.

Original entry on oeis.org

0, 3, 6, 10, 36, 66
Offset: 0

Views

Author

Peter Munn and Jon E. Schoenfield, Sep 07 2022

Keywords

Comments

The difficulty of this sequence comes in determining which is the largest j in a solution for a(n), equivalently the last nonzero term in each sum from the A319394-based formula in the formula section.
a(2) derives from Bravo and Luca, a(3) from Bravo and Bravo, a(4) from Pagdame Tiebekabe and Diouf. Pagdame has indicated A356928(5), from which a(5) is derived, has been determined.
a(6) >= 178, a(7) >= 478.

Examples

			For n = 2, the a(2) = 6 solutions are j = 1 with (2,2), j = 2 with (2,4) and (3,3), j = 3 with (4,5), j = 4 with (4,7) and (6,6) according to the paper of Bravo and Luca. [That is, 2 = 1+1, 4 = 1+3 = 2+2, 8 = 3+5, 16 = 3+13 = 8+8.]
		

References

  • J. J. Bravo, and F. Luca, On the Diophantine equation F_n+F_m=2^a, Quaest. Math. 39 (2016) 391-400.
  • P. Tiebekabe and I. Diouf, On solutions of Diophantine equation F_{n_1}+F_{n_2}+F_{n_3}+F_{n_4}=2^a, Journal of Algebra and Related Topics, Volume 9, Issue 2 (2021), 131-148.

Crossrefs

Formula

a(n) = Sum_{i >= 0} A319394(2^i, k).
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