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.

A330298 a(n) is the number of subsets of {1..n} that contain exactly 1 odd and 2 even numbers.

Original entry on oeis.org

0, 0, 0, 0, 2, 3, 9, 12, 24, 30, 50, 60, 90, 105, 147, 168, 224, 252, 324, 360, 450, 495, 605, 660, 792, 858, 1014, 1092, 1274, 1365, 1575, 1680, 1920, 2040, 2312, 2448, 2754, 2907, 3249, 3420, 3800, 3990, 4410, 4620, 5082, 5313, 5819, 6072, 6624, 6900, 7500, 7800, 8450, 8775, 9477
Offset: 0

Views

Author

Enrique Navarrete, Feb 29 2020

Keywords

Comments

The general formula for the number of subsets of {1..n} that contain exactly k odd and j even numbers is binomial(ceiling(n/2), k) * binomial(floor(n/2), j).

Examples

			For n=6, a(6) = 9 and the 9 subsets are: {1,2,4}, {1,2,6}, {1,4,6}, {2,3,4}, {2,3,6}, {2,4,5}, {2,5,6}, {3,4,6}, {4,5,6}.

Links

Crossrefs

Cf. A330299, A330300.
Cf. A000217, A002378.
Interleaves the positive integers of A006002 and A027480.

Programs

  • Mathematica
    a[n_] := Ceiling[n/2] * Binomial[Floor[n/2], 2]; Array[a, 55, 0] (* Amiram Eldar, Mar 01 2020 *)
Table[Length[Select[Subsets[Range[n],{3}],Total[Boole[OddQ[#]]]==1&]],{n,0,60}] (* Harvey P. Dale, Jul 26 2020 *)
  • PARI
    a(n) = ceil(n/2) * binomial(floor(n/2), 2) \\ Andrew Howroyd, Mar 01 2020
  • PARI
    concat([0,0,0,0], Vec(x^4*(2 + x) / ((1 - x)^4*(1 + x)^3) + O(x^40))) \\ Colin Barker, Mar 02 2020

Formula

a(n) = ceiling(n/2) * binomial(floor(n/2), 2).
From Colin Barker, Mar 01 2020: (Start)
G.f.: x^4*(2 + x) / ((1 - x)^4*(1 + x)^3).
a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7) for n>6. (End)
E.g.f.: (x*(-3 + x + x^2)*cosh(x) + (3 - x + x^3)*sinh(x))/16. - Stefano Spezia, Mar 02 2020
a(n) = 3/32-5*n^2/32-n/32+n^3/16+(-)^n*(n-3+n^2)/32. - R. J. Mathar, Mar 31 2025

A330300 a(n) is the number of subsets of {1..n} that contain exactly 2 odd and 3 even numbers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 3, 6, 24, 40, 100, 150, 300, 420, 735, 980, 1568, 2016, 3024, 3780, 5400, 6600, 9075, 10890, 14520, 17160, 22308, 26026, 33124, 38220, 47775, 54600, 67200, 76160, 92480, 104040, 124848, 139536, 165699, 184110, 216600, 239400, 279300, 307230, 355740, 389620, 448063
Offset: 0

Views

Author

Enrique Navarrete, Feb 29 2020

Keywords

Comments

The general formula for the number of subsets of {1..n} that contain exactly k odd and j even numbers is binomial(ceiling(n/2), k) * binomial(floor(n/2), j).

Examples

			a(7) = 6 and the 6 subsets are {1,2,3,4,6}, {1,2,4,5,6}, {1,2,4,6,7}, {2,3,4,5,6}, {2,3,4,6,7}, {2,4,5,6,7}.

Links

Crossrefs

Cf. A330298, A330299.

Programs

  • Mathematica
    a[n_] := Binomial[Ceiling[n/2], 2] * Binomial[Floor[n/2], 3]; Array[a, 47, 0] (* Amiram Eldar, Mar 01 2020 *)
  • PARI
    a(n) = binomial(ceil(n/2), 2) * binomial(floor(n/2), 3) \\ Andrew Howroyd, Mar 01 2020
  • PARI
    concat([0,0,0,0,0,0], Vec(x^6*(3 + 3*x + 3*x^2 + x^3) / ((1 - x)^6*(1 + x)^5) + O(x^40))) \\ Colin Barker, Mar 02 2020

Formula

a(n) = binomial(ceiling(n/2), 2) * binomial(floor(n/2), 3).
From Colin Barker, Mar 01 2020: (Start)
G.f.: x^6*(3 + 3*x + 3*x^2 + x^3) / ((1 - x)^6*(1 + x)^5).
a(n) = a(n-1) + 5*a(n-2) - 5*a(n-3) - 10*a(n-4) + 10*a(n-5) + 10*a(n-6) - 10*a(n-7) - 5*a(n-8) + 5*a(n-9) + a(n-10) - a(n-11) for n>10.
(End)
E.g.f.: (x*(-15 + 3*x - 7*x^2 + 2*x^3 + x^4)*cosh(x) + (15 - 3*x + 12*x^2 - 3*x^3 + x^4 + x^5)*sinh(x))/384. - Stefano Spezia, Mar 02 2020

A333321 a(n) is the number of subsets of {1..n} that contain exactly 1 odd and 4 even numbers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 4, 5, 25, 30, 90, 105, 245, 280, 560, 630, 1134, 1260, 2100, 2310, 3630, 3960, 5940, 6435, 9295, 10010, 14014, 15015, 20475, 21840, 29120, 30940, 40460, 42840, 55080, 58140, 73644, 77520, 96900, 101745, 125685, 131670, 160930, 168245, 203665, 212520
Offset: 0

Views

Author

Enrique Navarrete, Mar 14 2020

Keywords

Comments

The general formula for the number of subsets of {1..n} that contain exactly k odd and j even numbers is binomial(ceiling(n/2), k) * binomial(floor(n/2), j).

Examples

			a(9)=5 and the 5 subsets are {1,2,4,6,8}, {2,3,4,6,8}, {2,4,5,6,8}, {2,4,6,7,8}, {2,4,6,8,9}.

Links

Crossrefs

Cf. A330298, A330299, A333320.

Programs

  • Mathematica
    Array[Binomial[Ceiling[#], 1] Binomial[Floor[#], 4] &[#/2] &, 48, 0] (* Michael De Vlieger, Mar 14 2020 *)
  • PARI
    concat([0,0,0,0,0,0,0,0], Vec(x^8*(4 + x) / ((1 - x)^6*(1 + x)^5) + O(x^45))) \\ Colin Barker, Mar 17 2020

Formula

a(n) = ceiling(n/2) * binomial(floor(n/2), 4).
From Colin Barker, Mar 17 2020: (Start)
G.f.: x^8*(4 + x) / ((1 - x)^6*(1 + x)^5).
a(n) = a(n-1) + 5*a(n-2) - 5*a(n-3) - 10*a(n-4) + 10*a(n-5) + 10*a(n-6) - 10*a(n-7) - 5*a(n-8) + 5*a(n-9) + a(n-10) - a(n-11) for n>10.
(End)

A333319 a(n) is the number of subsets of {1..n} that contain exactly 3 odd and 1 even numbers.

Original entry on oeis.org

0, 0, 0, 0, 0, 2, 3, 12, 16, 40, 50, 100, 120, 210, 245, 392, 448, 672, 756, 1080, 1200, 1650, 1815, 2420, 2640, 3432, 3718, 4732, 5096, 6370, 6825, 8400, 8960, 10880, 11560, 13872, 14688, 17442, 18411, 21660, 22800, 26600, 27930, 32340, 33880, 38962, 40733, 46552
Offset: 0

Views

Author

Enrique Navarrete, Mar 14 2020

Keywords

Comments

The general formula for the number of subsets of {1..n} that contain exactly k odd and j even numbers is binomial(ceiling(n/2), k) * binomial(floor(n/2), j).

Examples

			a(7)=12 and the 12 subsets are {1,2,3,5}, {1,2,3,7}, {1,2,5,7}, {1,3,4,5}, {1,3,4,7}, {1,3,5,6}, {1,3,6,7}, {1,4,5,7}, {1,5,6,7}, {2,3,5,7}, {3,4,5,7}, {3,5,6,7}.

Links

Crossrefs

Cf. A330299.

Programs

  • Mathematica
    Array[Binomial[Ceiling[#], 3] Binomial[Floor[#], 1] &[#/2] &, 48, 0] (* Michael De Vlieger, Mar 14 2020 *)
  • PARI
    concat([0,0,0,0,0], Vec(x^5*(2 + x + x^2) / ((1 - x)^5*(1 + x)^4) + O(x^40))) \\ Colin Barker, Mar 15 2020

Formula

a(n) = binomial(ceiling(n/2),3) * binomial(floor(n/2), 1).
From Colin Barker, Mar 15 2020: (Start)
G.f.: x^5*(2 + x + x^2) / ((1 - x)^5*(1 + x)^4).
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) - 6*a(n-4) + 6*a(n-5) + 4*a(n-6) - 4*a(n-7) - a(n-8) + a(n-9) for n>8.
(End)
Showing 1-4 of 4 results.

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