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.

A371758 a(n) = Sum_{k=0..floor(n/3)} binomial(2*n-3*k-1,n-3*k).

Original entry on oeis.org

1, 1, 3, 11, 39, 141, 519, 1933, 7263, 27479, 104543, 399543, 1532779, 5899167, 22766607, 88073091, 341425551, 1326019653, 5158412943, 20096457549, 78396460299, 306190920837, 1197181197567, 4685523856881, 18354865147011, 71962695111841, 282357198103815
Offset: 0

Views

Author

Seiichi Manyama, Apr 05 2024

Keywords

Crossrefs

Cf. A371770, A371771, A371772.
Cf. A144904, A371773, A371777.

Programs

  • PARI
    a(n) = sum(k=0, n\3, binomial(2*n-3*k-1, n-3*k));

Formula

a(n) = [x^n] 1/((1-x^3) * (1-x)^n).
a(n) = binomial(2*n-1, n)*hypergeom([1, (1-n)/3, (2-n)/3, -n/3], [(1-2*n)/3, 2*(1-n)/3, 1-2*n/3], 1). - Stefano Spezia, Apr 06 2024
From Vaclav Kotesovec, Apr 08 2024: (Start)
Recurrence: 3*n*(7*n-11)*a(n) = 6*(2*n-3)*(7*n-4)*a(n-1) - n*(7*n-11)*a(n-2) + 2*(2*n-3)*(7*n-4)*a(n-3).
a(n) ~ 2^(2*n+2) / (7*sqrt(Pi*n)). (End)

A371770 a(n) = Sum_{k=0..floor(n/3)} binomial(3*n-3*k-1,n-3*k).

Original entry on oeis.org

1, 2, 10, 57, 338, 2057, 12741, 79914, 505954, 3226638, 20696685, 133382658, 862978221, 5601919325, 36467212610, 237974911737, 1556281907586, 10196788555859, 66921360130374, 439860632463462, 2895002186799453, 19077000179746293, 125849150650146714
Offset: 0

Views

Author

Seiichi Manyama, Apr 05 2024

Keywords

Links

Crossrefs

Cf. A005809, A165817, A183160.
Cf. A371758, A371771, A371772.

Programs

  • Maple
    f:= proc(n) local k; add(binomial(3*n-3*k-1,n-3*k),k=0..n/3) end proc:
map(f, [$0..30]); # Robert Israel, Feb 28 2025
  • PARI
    a(n) = sum(k=0, n\3, binomial(3*n-3*k-1, n-3*k));

Formula

a(n) = [x^n] 1/((1-x^3) * (1-x)^(2*n)).
a(n) = binomial(3*n-1, n)*hypergeom([1, (1-n)/3, (2-n)/3, -n/3], [1/3-n, 2/3-n, 1-n], 1). - Stefano Spezia, Apr 06 2024
From Vaclav Kotesovec, Apr 08 2024: (Start)
Recurrence: 18*n*(2*n - 1)*(13*n - 22)*(37*n - 51)*a(n) = 3*(40885*n^4 - 165468*n^3 + 229373*n^2 - 125562*n + 22680)*a(n-1) - (40885*n^4 - 165468*n^3 + 229373*n^2 - 125562*n + 22680)*a(n-2) + 3*(3*n - 5)*(3*n - 4)*(13*n - 9)*(37*n - 14)*a(n-3).
a(n) ~ 3^(3*n + 5/2) / (13 * sqrt(Pi*n) * 2^(2*n+1)). (End)

A371772 a(n) = Sum_{k=0..floor(n/3)} binomial(5*n-3*k-1,n-3*k).

Original entry on oeis.org

1, 4, 36, 365, 3892, 42714, 477621, 5411109, 61901268, 713435333, 8271470666, 96361329024, 1127086021461, 13227336997645, 155680966681101, 1836862248992565, 21719923705450260, 257316706385394615, 3053599633736172765, 36292098436808314572, 431918050456887676362
Offset: 0

Views

Author

Seiichi Manyama, Apr 05 2024

Keywords

Crossrefs

Cf. A371758, A371770, A371771.
Cf. A371756.

Programs

  • PARI
    a(n) = sum(k=0, n\3, binomial(5*n-3*k-1, n-3*k));

Formula

a(n) = [x^n] 1/((1-x^3) * (1-x)^(4*n)).
a(n) = binomial(5*n-1, n)*hypergeom([1, (1-n)/3, (2-n)/3, -n/3], [(1-5*n)/3, (2-5*n)/3, 1-5*n/3], 1). - Stefano Spezia, Apr 06 2024
From Vaclav Kotesovec, Apr 08 2024: (Start)
Recurrence: 72*n*(2*n - 1)*(4*n - 3)*(4*n - 1)*(899*n^2 - 2355*n + 1534)*a(n) = (25514519*n^6 - 117751221*n^5 + 212960873*n^4 - 191684487*n^3 + 89835824*n^2 - 20567076*n + 1769040)*a(n-1) - 5*(5*n - 7)*(5*n - 6)*(5*n - 4)*(5*n - 3)*(899*n^2 - 557*n + 78)*a(n-2).
a(n) ~ 5^(5*n + 5/2) / (31 * sqrt(Pi*n) * 2^(8*n + 3/2)). (End)

A371817 a(n) = Sum_{k=0..floor(n/3)} (-1)^k * binomial(4*n-3*k-1,n-3*k).

Original entry on oeis.org

1, 3, 21, 164, 1353, 11508, 99808, 877425, 7790745, 69704921, 627438606, 5675535000, 51546958296, 469764721533, 4293594852225, 39341599326304, 361271345551257, 3323924166943410, 30634431485945569, 282767849049333909, 2613630939017216898
Offset: 0

Views

Author

Seiichi Manyama, Apr 06 2024

Keywords

Crossrefs

Cf. A120305, A262977, A371816.
Cf. A371771.

Programs

  • PARI
    a(n) = sum(k=0, n\3, (-1)^k*binomial(4*n-3*k-1, n-3*k));

Formula

a(n) = [x^n] 1/((1+x^3) * (1-x)^(3*n)).
a(n) = binomial(4*n-1, n)*hypergeom([1, (1-n)/3, (2-n)/3, -n/3], [(1-4*n)/3, (2-4*n)/3, 1-4*n/3], -1). - Stefano Spezia, Apr 07 2024
Showing 1-4 of 4 results.

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