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.

A335095 Square array T(n,k), n>=0, k>=0, read by antidiagonals: T(n,k) = ((2n+1)!!)^k * Sum_{j=1..n} 1/(2*j+1)^k.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 8, 3, 0, 1, 34, 71, 4, 0, 1, 152, 1891, 744, 5, 0, 1, 706, 55511, 164196, 9129, 6, 0, 1, 3368, 1745731, 41625144, 20760741, 129072, 7, 0, 1, 16354, 57365351, 11575291716, 56246975289, 3616621254, 2071215, 8
Offset: 0

Views

Author

Seiichi Manyama, Sep 12 2020

Keywords

Examples

			Square array begins:
  0,   0,      0,        0,           0, ...
  1,   1,      1,        1,           1, ...
  2,   8,     34,      152,         706, ...
  3,  71,   1891,    55511,     1745731, ...
  4, 744, 164196, 41625144, 11575291716, ...
		

Crossrefs

Column k=0..4 give A001477, A334670, A335090, A335091, A335092.
Rows n=0-2 give: A000004, A000012, A074606.
Main diagonal gives A335096.
Cf. A291656.

Programs

  • Mathematica
    T[n_, k_] := ((2*n + 1)!!)^k * Sum[1/(2*j + 1)^k, {j, 1, n}]; Table[T[k, n - k], {n, 0, 8}, {k, 0, n}] // Flatten (* Amiram Eldar, Apr 29 2021 *)
  • PARI
    {T(n, k) = prod(j=1, n, 2*j+1)^k*sum(j=1, n, 1/(2*j+1)^k)}

Formula

T(0,k) = 0, T(1,k) = 1 and T(n,k) = ((2*n-1)^k+(2*n+1)^k) * T(n-1,k) - (2*n-1)^(2*k) * T(n-2, k) for n>1.

A335090 a(n) = ((2*n+1)!!)^2 * (Sum_{k=1..n} 1/(2*k+1)^2).

Original entry on oeis.org

0, 1, 34, 1891, 164196, 20760741, 3616621254, 832001250375, 244557191709000, 89472598178279625, 39886085958271670250, 21288783013213520392875, 13405493416599700058947500, 9835107221539462476348118125, 8316889511005794888839427108750, 8030850428074789829954674314399375
Offset: 0

Views

Author

Seiichi Manyama, Sep 11 2020

Keywords

Crossrefs

Column k=2 of A335095.

Programs

  • Mathematica
    a[n_] := ((2*n + 1)!!)^2 * Sum[1/(2*k + 1)^2, {k, 1, n}]; Array[a, 16, 0] (* Amiram Eldar, Apr 29 2021 *)
  • PARI
    {a(n) = prod(k=1, n, 2*k+1)^2*sum(k=1, n, 1/(2*k+1)^2)}
    
  • PARI
    {a(n) = if(n<2, n, (8*n^2+2)*a(n-1)-(2*n-1)^4*a(n-2))}

Formula

a(n) = (8*n^2+2) * a(n-1) - (2*n-1)^4 * a(n-2) for n>1.
a(n) ~ (Pi^2/8 - 1) * 2^(2*n + 3) * n^(2*n + 2) / exp(2*n). - Vaclav Kotesovec, Sep 25 2020

A335091 a(n) = ((2*n+1)!!)^3 * (Sum_{k=1..n} 1/(2*k+1)^3).

Original entry on oeis.org

0, 1, 152, 55511, 41625144, 56246975289, 124697847089808, 423322997436687375, 2088114588247920714000, 14363296872939657999716625, 133299155158711610547152961000, 1624450039177408057102079622846375, 25413656551949715361011431877529125000, 500711137690193661025654228810320074015625
Offset: 0

Views

Author

Seiichi Manyama, Sep 11 2020

Keywords

Crossrefs

Column k=3 of A335095.

Programs

  • Mathematica
    a[n_] := ((2*n + 1)!!)^3 * Sum[1/(2*k + 1)^3, {k, 1, n}]; Array[a, 14, 0] (* Amiram Eldar, Apr 29 2021 *)
  • PARI
    {a(n) = prod(k=1, n, 2*k+1)^3*sum(k=1, n, 1/(2*k+1)^3)}
    
  • PARI
    {a(n) = if(n<2, n, ((2*n-1)^3+(2*n+1)^3)*a(n-1)-(2*n-1)^6*a(n-2))}

Formula

a(n) = ((2*n-1)^3+(2*n+1)^3) * a(n-1) - (2*n-1)^6 * a(n-2) for n>1.
a(n) ~ (7*zeta(3)/8 - 1) * 2^(3*n + 9/2) * n^(3*n + 3) / exp(3*n). - Vaclav Kotesovec, Sep 25 2020

A335092 a(n) = ((2*n+1)!!)^4 * (Sum_{k=1..n} 1/(2*k+1)^4).

Original entry on oeis.org

0, 1, 706, 1745731, 11575291716, 170271339664581, 4874795836698898566, 247120020454614424554375, 20656593715240068513643845000, 2693397991748017956223512587135625, 523998492940635622166679925147692626250
Offset: 0

Views

Author

Seiichi Manyama, Sep 12 2020

Keywords

Crossrefs

Column k=4 of A335095.

Programs

  • Mathematica
    a[n_] := ((2*n + 1)!!)^4 * Sum[1/(2*k + 1)^4, {k, 1, n}]; Array[a, 11, 0] (* Amiram Eldar, Apr 28 2021 *)
  • PARI
    {a(n) = prod(k=1, n, 2*k+1)^4*sum(k=1, n, 1/(2*k+1)^4)}
    
  • PARI
    {a(n) = if(n<2, n, ((2*n-1)^4+(2*n+1)^4)*a(n-1)-(2*n-1)^8*a(n-2))}

Formula

a(n) = ((2*n-1)^4+(2*n+1)^4) * a(n-1) - (2*n-1)^8 * a(n-2) for n>1.
a(n) ~ (Pi^4/96 - 1) * 2^(4*n + 6) * n^(4*n + 4) / exp(4*n). - Vaclav Kotesovec, Sep 25 2020
Showing 1-4 of 4 results.