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.

Previous Showing 21-25 of 25 results.

A355725 Row 5 of table A355721.

Original entry on oeis.org

1, 2, 26, 426, 8178, 176802, 4206618, 108577674, 3011332338, 89141101506, 2802596567706, 93232011912426, 3271729161905010, 120810104634555234, 4683805718871051162, 190294015841923438026, 8087576641287426829170, 358981130096398432055682, 16615841072836741527510810
Offset: 0

Views

Author

Peter Bala, Jul 15 2022

Keywords

Links

Crossrefs

Cf. A001147, A355721 (table), A112934 (row 0), A000698 (row 1), A355722 (row 2), A355723 (row 3), A355724 (row 4).

Programs

  • Maple
    n := 5: seq(coeff(series( hypergeom([n+1/2, 1], [], 2*x)/hypergeom([n-1/2, 1], [], 2*x ), x, 21), x, k), k = 0..20);

Formula

O.g.f: A(x) = ( Sum_{k >= 0} d(k+5)/d(5)*x^k )/( Sum_{k >= 0} d(k+4)/d(4)*x^k ), where d(n) = Product_{k = 1..n} (2*k-1) = A001147(n).
A(x) = 1/(1 + 9*x - 11*x/(1 + 11*x - 13*x/(1 + 13*x - 15*x/(1 + 15*x - ... )))).
The o.g.f. satisfies the Riccati differential equation 2*x^2*A'(x) + 9*x*A(x)^2 - (1 + 7*x)*A(x) + 1 = 0 with A(0) = 1.
Applying Stokes 1982 gives A(x) = 1/(1 - 2*x/(1 - 11*x/(1 - 4*x/(1 - 13*x/(1 - 6*x/(1 - 15*x/(1 - ... - 2*n*x/(1 - (2*n+9)*x )))))))), a continued fraction of Stieltjes-type.

A303943 Expansion of 1/(1 - x/(1 - 1^2*x/(1 - 2^2*x/(1 - 3^2*x/(1 - 4^2*x/(1 - ...)))))), a continued fraction.

Original entry on oeis.org

1, 1, 2, 8, 76, 1540, 53684, 2812148, 205054036, 19805016628, 2444724910292, 375282530128052, 70102075181928148, 15655136160745164340, 4118456236678107528404, 1260512820941791994429876, 444069171743010266366969044, 178408825363590577961830752052
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 04 2018

Keywords

Comments

Invert transform of Euler (or secant) numbers (A000364), shifted right one place.

Links

Crossrefs

Cf. A000290, A000364, A112934, A305532.

Programs

  • Maple
    b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
a:= proc(n) option remember; `if`(n<1, 1,
      add(a(n-i)*b((i-1)*2, 0), i=1..n))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 13 2018
# Alternative:
T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1)
else (n - k)^2 * T(n, k - 1) + T(n - 1, k) fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..17);  # Peter Luschny, Oct 02 2023
  • Mathematica
    nmax = 17; CoefficientList[Series[1/(1 - x/(1 + ContinuedFractionK[-k^2 x, 1, {k, 1, nmax}])), {x, 0, nmax}], x]
nmax = 17; CoefficientList[Series[1/(1 - x Sum[Abs[EulerE[2 k]] x^k, {k, 0, nmax}]), {x, 0, nmax}], x]
a[0] = 1; a[n_] := a[n] = Sum[Abs[EulerE[2 (k - 1)]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 17}]

Formula

a(n) ~ 2^(4*n - 1) * n^(2*n - 3/2) / (Pi^(2*n - 3/2) * exp(2*n)). - Vaclav Kotesovec, Jun 08 2019

A137592 Duplicate of A098694.

Original entry on oeis.org

1, 2, 48, 34560, 1393459200, 5056584744960000, 2422112183371431936000000, 211155601241022491077587763200000000, 4417964278440225627098723475313498521600000000000
Offset: 0

Views

Author

Paul Barry, Jan 28 2008

Keywords

Comments

Hankel transform of A112934(n+1). [From Paul Barry, Dec 04 2009]

Formula

a(n)=Product{k=0..n, (2(k+1)(2k+1))^(n-k)}; a(n)=A000178(n)*A057863(n)*A006125(n+1)=A121835(n)*A006125(n+1);
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 1/(1 + 1/(2*k+2)!/x/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 14 2013

A302699 G.f. A(x) satisfies: A(x) = 1 + x * A(x) * (A(x) + 3*x*A'(x)) / (A(x) + x*A'(x)).

Original entry on oeis.org

1, 1, 3, 13, 71, 469, 3711, 35181, 398791, 5352149, 83650687, 1494274301, 29988083447, 666634964197, 16233361360559, 429237520044813, 12237655701598503, 374023408217062261, 12195222470567359071, 422440153967133458205, 15490152522612488256855, 599350023954941335582725, 24401304036660493806643215
Offset: 0

Views

Author

Paul D. Hanna, Apr 11 2018

Keywords

Comments

Compare to: C(x) = 1 + x*C(x) * (C(x) + 2*x*C'(x)) / (C(x) + x*C'(x)) holds when C(x) = 1 + x*C(x)^2 is a g.f. of the Catalan numbers (A000108).

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 71*x^4 + 469*x^5 + 3711*x^6 + 35181*x^7 + 398791*x^8 + 5352149*x^9 + 83650687*x^10 + ...
RELATED SERIES.
A'(x)/A(x) = 1 + 5*x + 31*x^2 + 225*x^3 + 1891*x^4 + 18473*x^5 + 210939*x^6 + 2815137*x^7 + 43551715*x^8 + 770297385*x^9 + ...
A(x) + x*A'(x) = 1 + 2*x + 9*x^2 + 52*x^3 + 355*x^4 + 2814*x^5 + 25977*x^6 + 281448*x^7 + 3589119*x^8 + 53521490*x^9 + ...

Links

Crossrefs

Cf. A001147, A112934.

Programs

  • PARI
    /* Differential equation: */
{a(n) = my(A=1); for(i=0, n, A = 1 + x*A*(A + 3*x*A')/(x*A +x^2*O(x^n))'); polcoeff(G=A, n)}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    /* Continued fraction: */
{a(n) = my(A=1, CF = 1+x +x*O(x^n)); for(i=1, n, for(k=0, n, CF = 1/(1 - (n-k+1)*x*A*CF ) ); A=1/(1 - x*A*CF) ); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = 1 + x*A(x) * (A(x) + 3*x*A'(x)) / (A(x) + x*A'(x)).
(2) A(x) = 1/(1 - x*A(x)/(1 - x*A(x)/(1 - 2*x*A(x)/(1 - 3*x*A(x)/(1 - 4*x*A(x)/(1 - 5*x*A(x)/(1 - ...))))))), a continued fraction.
(3) A(x) = Series_Reversion( x - x^2*F(x) ) where F(x) = Sum_{n>=0} (2*n)!/(n!*2^n)*x^n (g.f. of the odd double factorials A001147).
a(n) ~ 2^(n - 1/2) * n^(n-1) / exp(n - 1/2). - Vaclav Kotesovec, Aug 11 2021

A305535 Expansion of 1/(1 - x/(1 - 2*x/(1 - 2*x/(1 - 4*x/(1 - 4*x/(1 - 6*x/(1 - 6*x/(1 - ...)))))))), a continued fraction.

Original entry on oeis.org

1, 1, 3, 13, 75, 557, 5179, 58589, 784715, 12154061, 213593563, 4195613373, 91031201643, 2160916171181, 55687501548539, 1547866851663261, 46150908197995403, 1469089501918434957, 49722765216242122267, 1782934051704982201469, 67514992620138056010667
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 04 2018

Keywords

Comments

Invert transform of A000165, shifted right one place.

Links

Crossrefs

Cf. A000165, A051295, A051296, A112934, A141307, A292778.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[1/(1 - x/(1 + ContinuedFractionK[-2 Floor[(k + 1)/2] x, 1, {k, 1, nmax}])), {x, 0, nmax}], x]
nmax = 20; CoefficientList[Series[1/(1 - Sum[2^(k - 1) (k - 1)! x^k, {k, 1, nmax}]), {x, 0, nmax}], x]
a[0] = 1; a[n_] := a[n] = Sum[2^(k - 1) (k - 1)! a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]

Formula

a(n) ~ 2^(n-1) * (n-1)!. - Vaclav Kotesovec, Sep 18 2021
Previous Showing 21-25 of 25 results.

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