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 1-6 of 6 results.

A384950 a(n) = Sum_{k=0..n} 3^k * binomial(2*n+k-1,k).

Original entry on oeis.org

1, 7, 103, 1720, 30319, 550867, 10204660, 191606380, 3633593071, 69434167357, 1334845289023, 25787841299392, 500217562201348, 9736067678711524, 190051513661403112, 3719197868485767940, 72942019051301120239, 1433317465944902210161, 28212929859612197439829
Offset: 0

Views

Author

Seiichi Manyama, Aug 01 2025

Keywords

Links

Crossrefs

Cf. A383888, A385438.
Cf. A385474.

Programs

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

Formula

a(n) = Sum_{k=0..n} 2^(n-k) * binomial(3*n,k) * binomial(3*n-k-1,n-k).
a(n) = [x^n] ( (1+x)^3/(1-2*x)^2 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-2*x)^2 / (1+x)^3 ). See A385474.
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(3*n,k).
a(n) = 4^(-n) - 3^n*binomial(3*n-1, n)*(hypergeom([1, 3*n], [1+n], 3) - 1). - Stefano Spezia, Aug 02 2025
a(n) ~ 3^(4*n + 3/2) / (2^(2*n+3) * sqrt(Pi*n)). - Vaclav Kotesovec, Aug 04 2025

A385438 a(n) = Sum_{k=0..n} 3^k * binomial(3*n+k-1,k).

Original entry on oeis.org

1, 10, 208, 4888, 121132, 3092950, 80506684, 2123780536, 56581885468, 1518936682888, 41021505946468, 1113273696074968, 30335161535834212, 829405495046080612, 22742967214283811976, 625193974445825554408, 17223870801864911429404, 475423918887141016417144
Offset: 0

Views

Author

Seiichi Manyama, Aug 01 2025

Keywords

Links

Crossrefs

Cf. A383888, A384950.
Cf. A385475.

Programs

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

Formula

a(n) = Sum_{k=0..n} 2^(n-k) * binomial(4*n,k) * binomial(4*n-k-1,n-k).
a(n) = [x^n] ( (1+x)^4/(1-2*x)^3 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-2*x)^3 / (1+x)^4 ). See A385475.
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(4*n,k).
a(n) = (-2)^(-3*n) - 3^n*binomial(4*n-1, n)*(hypergeom([1, 4*n], [1+n], 3) - 1). - Stefano Spezia, Aug 02 2025
a(n) ~ 2^(8*n + 1/2) / (11 * 3^(2*n - 3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Aug 04 2025

A386763 a(n) = Sum_{k=0..n} 5^k * 3^(n-k) * binomial(n+k-1,k).

Original entry on oeis.org

1, 8, 114, 1862, 32246, 576768, 10529544, 194960802, 3647285766, 68772760928, 1304858513324, 24882531221292, 476462691535436, 9155397868559288, 176447193966483204, 3409285356643013082, 66020593061854488006, 1280989373915746600848, 24897996624141835608684
Offset: 0

Views

Author

Seiichi Manyama, Aug 02 2025

Keywords

Crossrefs

Cf. A386764, A386765.
Cf. A383888, A386769.

Programs

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

Formula

a(n) = Sum_{k=0..n} 3^k * 2^(n-k) * binomial(2*n,k) * binomial(2*n-k-1,n-k).
a(n) = [x^n] ( (1+3*x)^2/(1-2*x) )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-2*x) / (1+3*x)^2 ). See A386769.
a(n) = Sum_{k=0..n} 5^k * (-2)^(n-k) * binomial(2*n,k).
a(n) = (-9/2)^n*(1 - (-10/9)^n*binomial(2*n-1, n)*(hypergeom([1, 2*n], [1+n], 5/3) - 1)). - Stefano Spezia, Aug 02 2025

A385514 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(2*n+1,k) * binomial(2*n-k,n-k).

Original entry on oeis.org

1, 7, 64, 643, 6766, 73162, 805414, 8979523, 101060326, 1145704162, 13064219224, 149674343518, 1721537039236, 19866626222632, 229912254620434, 2667252458378083, 31009548579437446, 361198085246048602, 4214267651960927824, 49243413868632029338, 576179701092650156356
Offset: 0

Views

Author

Seiichi Manyama, Aug 04 2025

Keywords

Links

Crossrefs

Cf. A385667, A385668.
Cf. A383888.

Programs

  • Magma
    [&+[2^(n-k) * Binomial(2*n+1,k) * Binomial(2*n-k,n-k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 05 2025
  • Mathematica
    Table[Sum[2^(n-k)*Binomial[2*n+1,k]*Binomial[2*n-k,n-k],{k,0,n}],{n,0,25}] (* Vincenzo Librandi, Aug 05 2025 *)
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n+1, k)*binomial(2*n-k, n-k));

Formula

a(n) = [x^n] (1+x)^(2*n+1)/(1-2*x)^(n+1).
a(n) = [x^n] 1/((1-x) * (1-3*x)^(n+1)).
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(2*n+1,k).
a(n) = Sum_{k=0..n} 3^k * binomial(n+k,k).
a(n) ~ 2^(2*n+1) * 3^(n+1) / (5*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 06 2025

A386862 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(2*n+1,k) * binomial(2*n-k-1,n-k).

Original entry on oeis.org

1, 5, 42, 409, 4238, 45414, 496996, 5517929, 61909878, 700189606, 7968994124, 91158632250, 1047156227068, 12071222381456, 139569181458552, 1617879480097129, 18796461329347238, 218806784598226926, 2551538498649588892, 29800118958422522414, 348529038403155280548
Offset: 0

Views

Author

Seiichi Manyama, Aug 06 2025

Keywords

Crossrefs

Cf. A383888, A386865.
Cf. A116881.

Programs

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

Formula

a(n) = [x^n] (1+x)^(2*n+1)/(1-2*x)^n.
a(n) = [x^n] 1/((1-x)^2 * (1-3*x)^n).
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * (n-k+1) * binomial(2*n+1,k).
a(n) = Sum_{k=0..n} 3^k * (n-k+1) * binomial(n+k-1,k).

A386865 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(2*n+2,k) * binomial(2*n-k-1,n-k).

Original entry on oeis.org

1, 6, 51, 496, 5130, 54894, 600103, 6657312, 74646702, 843819580, 9599776494, 109776491664, 1260666279964, 14528980409454, 167951183468655, 1946529575164864, 22611104963042646, 263175370423429428, 3068541416792813338, 35834296592951011680, 419059482092284948908
Offset: 0

Views

Author

Seiichi Manyama, Aug 06 2025

Keywords

Crossrefs

Cf. A383888, A386862.
Cf. A386835.

Programs

  • Mathematica
    Table[Sum[(-1)^k*(k+1)*(k+2)*2^(k-1)*3^(n-k)* Binomial[2*n+2, n+k+2], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Aug 06 2025 *)
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n+2, k)*binomial(2*n-k-1, n-k));

Formula

a(n) = [x^n] (1+x)^(2*n+2)/(1-2*x)^n.
a(n) = [x^n] 1/((1-x)^3 * (1-3*x)^n).
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(2*n+2,k) * binomial(n-k+2,n-k).
a(n) = Sum_{k=0..n} 3^k * binomial(n+k-1,k) * binomial(n-k+2,n-k).
a(n) ~ 2^(2*n+2) * 3^(n+3) / (125*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 06 2025
Showing 1-6 of 6 results.

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