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.

A108479 Antidiagonal sums of number triangle A086645.

Original entry on oeis.org

1, 1, 2, 7, 17, 44, 117, 305, 798, 2091, 5473, 14328, 37513, 98209, 257114, 673135, 1762289, 4613732, 12078909, 31622993, 82790070, 216747219, 567451585, 1485607536, 3889371025, 10182505537, 26658145586, 69791931223, 182717648081
Offset: 0

Views

Author

Paul Barry, Jun 04 2005

Keywords

Links

Crossrefs

Cf. A005252, A051286, A086645, A182890.
Cf. A381421, A382230, A382470, A382471, A382472, A382473, A382474.
Cf. A376729, A376730, A376731.

Programs

  • Mathematica
    LinearRecurrence[{2,1,2,-1},{1,1,2,7},30] (* Harvey P. Dale, Jun 01 2021 *)

Formula

G.f.: (1 - x - x^2)/(1 - 2*x - x^2 - 2*x^3 + x^4).
a(n) = 2*(n-1) + a(n-2) + 2*a(n-3) - a(n-4).
a(n) = Sum_{k=0..floor(n/2)} C(2*(n-k), 2*k).
a(n) = Sum_{k=0..floor(n/2)} Sum_{j=0..n-k} C(2*(n-2*k), j) * C(2*k, j).
a(n) = A005252(2*n). - Seiichi Manyama, Aug 11 2024

A376729 Expansion of (1 - x^2 - x^3)/((1 - x^2 - x^3)^2 - 4*x^5).

Original entry on oeis.org

1, 0, 1, 1, 1, 6, 2, 15, 16, 29, 71, 73, 212, 276, 541, 1016, 1497, 3189, 4825, 9162, 16022, 26763, 50424, 82869, 151851, 262705, 456520, 820328, 1401913, 2511824, 4361521, 7657481, 13528913, 23509678, 41633002, 72630919, 127709888, 224418509, 392539055, 691382201
Offset: 0

Views

Author

Seiichi Manyama, Oct 03 2024

Keywords

Links

Crossrefs

Cf. A108479, A376730, A376731.
Cf. A376723, A376726.

Programs

  • PARI
    my(N=40, x='x+O('x^N)); Vec((1-x^2-x^3)/((1-x^2-x^3)^2-4*x^5))
  • PARI
    a(n) = sum(k=0, n\2, binomial(2*k, 2*n-4*k));

Formula

a(n) = 2*a(n-2) + 2*a(n-3) - a(n-4) + 2*a(n-5) - a(n-6).
a(n) = Sum_{k=0..floor(n/2)} binomial(2*k,2*n-4*k).

A376730 Expansion of (1 - x^3 - x^4)/((1 - x^3 - x^4)^2 - 4*x^7).

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 1, 6, 1, 1, 15, 15, 2, 28, 70, 29, 46, 210, 211, 111, 496, 925, 586, 1067, 3005, 3123, 2821, 8100, 13024, 11068, 20385, 44068, 48604, 57325, 129261, 192224, 200585, 358806, 662117, 781433, 1055567, 2050819, 2941702, 3524140, 6067682, 10169037
Offset: 0

Views

Author

Seiichi Manyama, Oct 03 2024

Keywords

Links

Crossrefs

Cf. A108479, A376729, A376731.
Cf. A376724, A376727.

Programs

  • PARI
    my(N=50, x='x+O('x^N)); Vec((1-x^3-x^4)/((1-x^3-x^4)^2-4*x^7))
  • PARI
    a(n) = sum(k=0, n\3, binomial(2*k, 2*n-6*k));

Formula

a(n) = 2*a(n-3) + 2*a(n-4) - a(n-6) + 2*a(n-7) - a(n-8).
a(n) = Sum_{k=0..floor(n/3)} binomial(2*k,2*n-6*k).

A376725 Expansion of 1/((1 - x^4 - x^5)^2 - 4*x^9).

Original entry on oeis.org

1, 0, 0, 0, 2, 2, 0, 0, 3, 10, 3, 0, 4, 28, 28, 4, 5, 60, 126, 60, 11, 110, 396, 396, 117, 188, 1001, 1716, 1009, 462, 2191, 5720, 5729, 2592, 4564, 15920, 24320, 16482, 12036, 39168, 84000, 84750, 51927, 93024, 249292, 353738, 269962, 258324, 666932, 1250142
Offset: 0

Views

Author

Seiichi Manyama, Oct 02 2024

Keywords

Links

Crossrefs

Cf. A182890, A376723, A376724.
Cf. A376728, A376731.

Programs

  • PARI
    my(N=50, x='x+O('x^N)); Vec(1/((1-x^4-x^5)^2-4*x^9))
  • PARI
    a(n) = sum(k=0, n\4, binomial(2*k+2, 2*n-8*k+1))/2;

Formula

a(n) = 2*a(n-4) + 2*a(n-5) - a(n-8) + 2*a(n-9) - a(n-10).
a(n) = (1/2) * Sum_{k=0..floor(n/4)} binomial(2*k+2,2*n-8*k+1).

A376728 Expansion of (1 + x^4 - x^5)/((1 + x^4 - x^5)^2 - 4*x^4).

Original entry on oeis.org

1, 0, 0, 0, 3, 1, 0, 0, 5, 10, 1, 0, 7, 35, 21, 1, 9, 84, 126, 36, 12, 165, 462, 330, 68, 287, 1287, 1716, 730, 533, 3004, 6435, 5022, 2045, 6293, 19449, 24329, 13345, 14008, 50524, 92400, 76912, 47481, 120156, 294124, 354488, 237139, 299421, 823200, 1354588
Offset: 0

Views

Author

Seiichi Manyama, Oct 03 2024

Keywords

Links

Crossrefs

Cf. A099511, A376726, A376727.
Cf. A376725, A376731.

Programs

  • PARI
    my(N=50, x='x+O('x^N)); Vec((1+x^4-x^5)/((1+x^4-x^5)^2-4*x^4))
  • PARI
    a(n) = sum(k=0, n\4, binomial(2*k+1, 2*n-8*k+1));

Formula

a(n) = 2*a(n-4) + 2*a(n-5) - a(n-8) + 2*a(n-9) - a(n-10).
a(n) = Sum_{k=0..floor(n/4)} binomial(2*k+1,2*n-8*k+1).
Showing 1-5 of 5 results.

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