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 1-3 of 3 results.

A329268 Positions of 1 in A309797.

Original entry on oeis.org

1, 2, 3, 4, 8, 9, 10, 11, 22, 23, 24, 25, 29, 58, 59, 60, 61, 65, 66, 67, 68, 88, 142, 143, 144, 145, 149, 150, 209, 300, 301, 302, 303, 307, 315, 317, 319, 322, 325, 650, 651, 652, 653, 657, 658, 659, 660, 671, 673, 675, 677, 680, 705, 706, 711, 712, 713, 716
Offset: 1

Views

Author

Rémy Sigrist, Nov 11 2019

Keywords

Comments

This sequence is infinite.

Examples

			A309797(67) = 1, hence 67 appears in the sequence.

Links

Crossrefs

Cf. A309797.

Programs

  • PARI
    See Links section.

Formula

a(A309797(n)) = 1.
a(n+1) <= 2*a(n).

A309793 Number of odd parts appearing among the second largest parts of the partitions of n into 4 parts.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 3, 5, 6, 8, 9, 11, 13, 17, 20, 24, 27, 32, 36, 42, 47, 54, 60, 68, 75, 85, 93, 103, 112, 124, 135, 149, 161, 176, 189, 205, 220, 239, 256, 276, 294, 316, 336, 360, 382, 408, 432, 460, 486, 517, 545, 577, 607, 642, 675, 713, 748
Offset: 0

Views

Author

Wesley Ivan Hurt, Aug 17 2019

Keywords

Examples

			Figure 1: The partitions of n into 4 parts for n = 8, 9, ..
                                                         1+1+1+9
                                                         1+1+2+8
                                                         1+1+3+7
                                                         1+1+4+6
                                             1+1+1+8     1+1+5+5
                                             1+1+2+7     1+2+2+7
                                 1+1+1+7     1+1+3+6     1+2+3+6
                                 1+1+2+6     1+1+4+5     1+2+4+5
                                 1+1+3+5     1+2+2+6     1+3+3+5
                     1+1+1+6     1+1+4+4     1+2+3+5     1+3+4+4
         1+1+1+5     1+1+2+5     1+2+2+5     1+2+4+4     2+2+2+6
         1+1+2+4     1+1+3+4     1+2+3+4     1+3+3+4     2+2+3+5
         1+1+3+3     1+2+2+4     1+3+3+3     2+2+2+5     2+2+4+4
         1+2+2+3     1+2+3+3     2+2+2+4     2+2+3+4     2+3+3+4
         2+2+2+2     2+2+2+3     2+2+3+3     2+3+3+3     3+3+3+3
--------------------------------------------------------------------------
  n  |      8           9          10          11          12        ...
--------------------------------------------------------------------------
a(n) |      2           3           5           6           8        ...
--------------------------------------------------------------------------

Links

Crossrefs

Cf. A309795, A309797, A026928.

Programs

  • Mathematica
    LinearRecurrence[{2, -1, 0, 0, 0, 1, -2, 2, -2, 1, 0, 0, 0, -1, 2, -1}, {0, 0, 0, 0, 1, 1, 1, 1, 2, 3, 5, 6, 8, 9, 11, 13}, 50]
  • PARI
    concat([0,0,0,0], Vec(x^4*(1 - x + x^4) / ((1 - x)^4*(1 + x)^2*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)*(1 + x^4)) + O(x^50))) \\ Colin Barker, Oct 10 2019

Formula

a(n) = Sum_{k=1..floor(n/4)} Sum_{j=k..floor((n-k)/3)} Sum_{i=j..floor((n-j-k)/2)} (i mod 2).
From Colin Barker, Aug 18 2019: (Start)
G.f.: x^4*(1 - x + x^4) / ((1 - x)^4*(1 + x)^2*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)*(1 + x^4)).
a(n) = 2*a(n-1) - a(n-2) + a(n-6) - 2*a(n-7) + 2*a(n-8) - 2*a(n-9) + a(n-10) - a(n-14) + 2*a(n-15) - a(n-16) for n>15.
(End) [Recurrence verified by Wesley Ivan Hurt, Aug 24 2019]

A309795 Number of even parts appearing among the second largest parts of the partitions of n into 4 parts.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 2, 3, 3, 4, 5, 7, 9, 12, 14, 17, 19, 23, 27, 32, 36, 42, 47, 54, 60, 68, 75, 84, 92, 103, 113, 125, 135, 148, 160, 175, 189, 206, 221, 239, 255, 275, 294, 316, 336, 360, 382, 408, 432, 460, 486, 516, 544, 577, 608, 643, 675, 712, 747
Offset: 0

Views

Author

Wesley Ivan Hurt, Aug 17 2019

Keywords

Examples

			Figure 1: The partitions of n into 4 parts for n = 8, 9, ..
                                                         1+1+1+9
                                                         1+1+2+8
                                                         1+1+3+7
                                                         1+1+4+6
                                             1+1+1+8     1+1+5+5
                                             1+1+2+7     1+2+2+7
                                 1+1+1+7     1+1+3+6     1+2+3+6
                                 1+1+2+6     1+1+4+5     1+2+4+5
                                 1+1+3+5     1+2+2+6     1+3+3+5
                     1+1+1+6     1+1+4+4     1+2+3+5     1+3+4+4
         1+1+1+5     1+1+2+5     1+2+2+5     1+2+4+4     2+2+2+6
         1+1+2+4     1+1+3+4     1+2+3+4     1+3+3+4     2+2+3+5
         1+1+3+3     1+2+2+4     1+3+3+3     2+2+2+5     2+2+4+4
         1+2+2+3     1+2+3+3     2+2+2+4     2+2+3+4     2+3+3+4
         2+2+2+2     2+2+2+3     2+2+3+3     2+3+3+3     3+3+3+3
--------------------------------------------------------------------------
  n  |      8           9          10          11          12        ...
--------------------------------------------------------------------------
a(n) |      3           3           4           5           7        ...
--------------------------------------------------------------------------

Links

Crossrefs

Cf. A309793, A309797, A026928.

Programs

  • Mathematica
    LinearRecurrence[{2, -1, 0, 0, 0, 1, -2, 2, -2, 1, 0, 0, 0, -1, 2, -1}, {0, 0, 0, 0, 0, 0, 1, 2, 3, 3, 4, 5, 7, 9, 12, 14}, 50]
  • PARI
    concat([0,0,0,0,0,0], Vec(x^6*(1 - x^3 + x^4) / ((1 - x)^4*(1 + x)^2*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)*(1 + x^4)) + O(x^70))) \\ Colin Barker, Oct 10 2019

Formula

a(n) = Sum_{k=1..floor(n/4)} Sum_{j=k..floor((n-k)/3)} Sum_{i=j..floor((n-j-k)/2)} ((i-1) mod 2).
From Colin Barker, Aug 18 2019: (Start)
G.f.: x^6*(1 - x^3 + x^4) / ((1 - x)^4*(1 + x)^2*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)*(1 + x^4)).
a(n) = 2*a(n-1) - a(n-2) + a(n-6) - 2*a(n-7) + 2*a(n-8) - 2*a(n-9) + a(n-10) - a(n-14) + 2*a(n-15) - a(n-16) for n>15.
(End) [Recurrence verified by Wesley Ivan Hurt, Aug 25 2019]
Showing 1-3 of 3 results.

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