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.

A199475 G.f. satisfies A(x) = Sum_{n>=0} x^n * (1 - A(x)^(2*n+2))/(1 - A(x)^2).

Original entry on oeis.org

1, 2, 7, 34, 195, 1225, 8146, 56336, 401005, 2918308, 21614216, 162385693, 1234515922, 9479336440, 73410868630, 572719097908, 4496923141241, 35509806367132, 281814387290431, 2246608404455588, 17982234787607464, 144458551104066553, 1164342291135424494
Offset: 0

Views

Author

Paul D. Hanna, Nov 08 2011

Keywords

Comments

Compare to g.f. B(x) of A007317 (binomial transform of Catalan numbers):
B(x) = Sum_{n>=0} x^n * (1 - B(x)^(n+1))/(1 - B(x)).

Examples

			G.f.: A(x) = 1 + 2*x + 7*x^2 + 34*x^3 + 195*x^4 + 1225*x^5 +...
where g.f. A = A(x) satisfies the equivalent expressions:
A = 1 + x*(1-A^4)/(1-A^2) + x^2*(1-A^6)/(1-A^2) + x^3*(1-A^8)/(1-A^2) +...
A = 1 + x*(1 + A^2) + x^2*(1 + A^2 + A^4) + x^3*(1 + A^2 + A^4 + A^6) +...

Links

Crossrefs

Cf. A007317, A171199, A199548, A349289, A349290, A349291, A349292, A349293.

Programs

  • Mathematica
    Rest[CoefficientList[InverseSeries[Series[(2*x^2)/(1 + x^2 - Sqrt[1 - 4*x - 2*x^2 + x^4]), {x, 0, 30}], x], x]] (* Vaclav Kotesovec, Jul 30 2021 *)
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, x^m*sum(k=0, m, A^(2*k))+x*O(x^n))); polcoeff(A, n)}
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1/((1-x)*(1 - x*A^2+x*O(x^n))));polcoeff(A,n)}
  • PARI
    {a(n)=polcoeff(1/x*serreverse(2*x^2/(1+x^2-sqrt(1-4*x-2*x^2+x^4+x^3*O(x^n)))),n)}

Formula

G.f. satisfies: A(x) = 1/((1-x)*(1 - x*A(x)^2)).
G.f.: A(x) = (1/x)*Series_Reversion( 2*x^2/(1+x^2 - sqrt(1-4*x-2*x^2+x^4)) ).
G.f. satisfies: A(x) = G(x*A(x)) and G(x) = A(x/G(x)) = g.f. of A171199, where G(x) = exp( Sum_{n>=1} [G(x)^n + G(x)^-n]*x^n/n ).
a(n) = 1 + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1). - Ilya Gutkovskiy, Jul 25 2021
a(n) ~ sqrt(387 + 35*sqrt(129)) * (35 + 3*sqrt(129))^n / (9 * sqrt(Pi) * n^(3/2) * 2^(3*n + 5/2)). - Vaclav Kotesovec, Jul 30 2021
a(n) = Sum_{k=0..n} binomial(n+k,n-k) * binomial(3*k,k)/(2*k+1). - Seiichi Manyama, Oct 03 2023
D-finite with recurrence 2*n*(2*n+1)*a(n) +3*(-13*n^2+11*n-2)*a(n-1) +(35*n^2-23*n-42)*a(n-2) +(35*n^2-257*n+426)*a(n-3) +3*(-13*n^2+93*n-166)*a(n-4) +2*(n-4)*(2*n-9)*a(n-5)=0. - R. J. Mathar, Feb 10 2024

A171190 G.f. satisfies: A(x) = exp( Sum_{n>=1} (A(x)^n + A(-x)^n) * x^n/n ).

Original entry on oeis.org

1, 2, 3, 10, 27, 112, 336, 1490, 4791, 22138, 74079, 351288, 1207620, 5831208, 20436516, 100004994, 355610367, 1758044950, 6322608561, 31511387450, 114359284515, 573713781760, 2097612975456, 10580600244664, 38925304968612
Offset: 0

Views

Author

Paul D. Hanna, Dec 05 2009

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 3*x^2 + 10*x^3 + 27*x^4 + 112*x^5 + 336*x^6 + 1490*x^7 + 4791*x^8 + 22138*x^9 + 74079*x^10 + 351288*x^11 + 1207620*x^12 + ...
where the logarithm of A(x) may be written as
log(A(x)) = (A(x) + A(-x))*x + (A(x)^2 + A(-x)^2)*x^2/2 + (A(x)^3 + A(-x)^3)*x^3/3 + (A(x)^4 + A(-x)^4)*x^4/4 + (A(x)^5 + A(-x)^5)*x^5/5 + ...
Incidentally, the square root of g.f. A(x) is an integer series starting
A(x)^(1/2) = 1 + x + x^2 + 4*x^3 + 9*x^4 + 43*x^5 + 108*x^6 + 558*x^7 + 1517*x^8 + 8175*x^9 + 23219*x^10 + 128516*x^11 + 375896*x^12 + ...

Links

Crossrefs

Cf. A171191, A171199.

Programs

  • PARI
    {a(n) = my(A=1+x+x*O(x^n)); for(i=1,n, A=exp(sum(m=1,n,(A^m+subst(A^m,x,-x)+x*O(x^n))*x^m/m)));polcoeff(A,n)}
  • PARI
    {a(n) = my(A=1+x); for(i=1,n, A=(1-x*A+x*O(x^n))^-1*(1-x*subst(A,x,-x)+x*O(x^n))^-1);polcoeff(A,n)} \\ Paul D. Hanna, Dec 06 2009

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x) = exp( Sum_{n>=1} (A(x)^n + A(-x)^n) * x^n/n ).
(2) A(x) = 1/((1 - x*A(x)) * (1 - x*A(-x))). - Paul D. Hanna, Dec 06 2009
(3) 0 = 1 - (3-x)*A(x) + (2-x)*A(x)^2 - (2-5*x)*x*A(x)^3 - (2+x)*x^2*A(x)^4 + 2*x^3*A(x)^5. - Paul D. Hanna, Feb 11 2024

A171191 G.f. satisfies: A(x) = exp( Sum_{n>=1} [A(x)^n + 1/A(-x)^n]*x^n/n ).

Original entry on oeis.org

1, 2, 7, 20, 73, 263, 1111, 4451, 20161, 85304, 401401, 1755593, 8465311, 37866818, 185756605, 844627115, 4196759383, 19321634594, 96962969047, 450810982796, 2280344734891, 10686378006479, 54406554842287, 256637809742444
Offset: 0

Views

Author

Paul D. Hanna, Dec 05 2009

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 7*x^2 + 20*x^3 + 73*x^4 + 263*x^5 + 1111*x^6 + ...
log(A(x)) = [A(x)+1/A(-x)]*x + [A(x)^2+1/A(-x)^2]*x^2/2 + [A(x)^3+1/A(-x)^3]*x^3/3 + ...

Crossrefs

Cf. A171190, A171199.

Programs

  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=exp(sum(m=1,n,(A^m+subst(A^-m,x,-x)+x*O(x^n))*x^m/m)));polcoeff(A,n)}
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=(1-x*A+x*O(x^n))^-1*(1-x/subst(A,x,-x)+x*O(x^n))^-1);polcoeff(A,n)} \\ Paul D. Hanna, Dec 06 2009

Formula

G.f. satisfies: A(x) = 1/[(1 - x*A(x))*(1 - x/A(-x))]. - Paul D. Hanna, Dec 06 2009

A259845 a(0)=1, a(1)=3, and the INVERT transform of the sequence deletes the 3.

Original entry on oeis.org

1, 3, 4, 11, 38, 136, 512, 1993, 7958, 32420, 134216, 563030, 2388092, 10224320, 44127328, 191783029, 838623654, 3686965308, 16287624440, 72262899994, 321852273332, 1438540956048, 6450223722816, 29006443606746, 130790584554748, 591191800834696
Offset: 0

Views

Author

Gary W. Adamson, Jul 06 2015

Keywords

Comments

The sequence is N = 3 in an infinite set, with the first few being:
A086581, N = 0: (1, 0, 1, 2, 5, 13, 35, 97, ...)
A000108, N = 1: (1, 1, 2, 5, 14, 42, 132, ...)
A171199, N = 2: (1, 2, 3, 8, 25, 83, 289, ...)
... The INVERT transforms of the sequences delete the second terms in the sequences.
The g.f. was contributed by Paul D. Hanna: From the definition of the INVERT transform, 1/(1 - x*A) = A - (N-1)*x. Thus, (1 + (N-1)*x - (1 + (N-1)*x^2)*A) + x*A^2 = 0. The g.f. follows, below.

Examples

			The INVERT transform of (1, 3, 4, 11, 38, 136, ...) is (1, 4, 11, 38, 136, ...).

Links

Crossrefs

Cf. A086581, A000108, A171199.

Programs

  • Mathematica
    CoefficientList[Series[1/(2*x) + x - Sqrt[1 - 4*x - 4*x^2 + 4*x^4]/(2*x), {x, 0, 25}], x] (* Michael De Vlieger, Jun 12 2024 *)

Formula

G.f.: A(x) = 1/(2*x) + x - sqrt(1 - 4*x - 4*x^2 + 4*x^4)/(2*x).

Extensions

More terms from Alois P. Heinz, Jul 07 2015

A346506 G.f. A(x) satisfies: A(x) = (1 + x * A(x)^2) / (1 - x + x^2).

Original entry on oeis.org

1, 2, 5, 17, 66, 274, 1190, 5341, 24577, 115326, 549747, 2654739, 12959468, 63848307, 317064921, 1585380283, 7975134892, 40332823042, 204947059412, 1045859173864, 5357606584326, 27540884494209, 142023060613755, 734506610474205, 3808771672620618, 19798640525731461, 103149287155802941
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 21 2021

Keywords

Crossrefs

Cf. A004148, A006318, A025244, A025258, A078482, A171199, A175934, A307733, A346505.

Programs

  • Mathematica
    nmax = 26; A[] = 0; Do[A[x] = (1 + x A[x]^2)/(1 - x + x^2) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[1] = 2; a[n_] := a[n] = 2 a[n - 1] + a[n - 2] + Sum[a[k] a[n - k - 1], {k, 2, n - 1}]; Table[a[n], {n, 0, 26}]

Formula

a(0) = 1, a(1) = 2; a(n) = 2 * a(n-1) + a(n-2) + Sum_{k=2..n-1} a(k) * a(n-k-1).
From Nikolaos Pantelidis, Jan 08 2023 (Start)
G.f.: 1/G(0), where G(k) = 1-(2*x-x^2)/(1-x/G(k+1)) (continued fraction).
G.f.: (1-x+x^2-sqrt(x^4-2*x^3+3*x^2-6*x+1))/(2*x).
(End)
Showing 1-5 of 5 results.

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