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.

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A188193 G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x))^n/n!.

Original entry on oeis.org

1, 2, 10, 100, 2500, 224728, 77611032, 95603336016, 411188458873152, 6215509773143124736, 334390128406134844422816, 64839530694681966290325813952, 45813418110052719651124682371286592, 119029898667160345925612801976119712375168, 1145669207542037312420485502021473483147684627584
Offset: 0

Views

Author

Paul D. Hanna, Mar 23 2011

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 10*x^2 + 100*x^3 + 2500*x^4 + 224728*x^5 +...
which equals the series:
A(x) = 1 + log(1+2*x*A(x)) + log(1+4*x*A(x))^2/2! + log(1+8*x*A(x))^3/3! +...
Let B(x) equal the g.f. of A014070, which begins:
B(x) = 1 + 2*x + 6*x^2 + 56*x^3 + 1820*x^4 +...+ C(2^n,n)*x^n +...
then B(x) = A(x/B(x)) and A(x) = B(x*A(x)), so that:
A(x) = 1 + 2*x*A(x) + 6*x^2*A(x)^2 + 56*x^3*A(x)^3 + 1820*x^4*A(x)^4 +...+ C(2^n,n)*x^n*A(x)^n +...

Crossrefs

Cf. A014070, A188194.

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, log(1+2^m*x*A+x*O(x^n))^m/m!)); polcoeff(A, n)}

Formula

G.f. A(x) satisfies:
(1) A(x) = Sum_{n>=0} C(2^n,n)*x^n*A(x)^n,
(2) A(x) = (1/x)*Series_Reversion(x/B(x)),
(3) A(x) = B(x*A(x)) and B(x) = A(x/B(x)),
where B(x) = Sum_{n>=0} C(2^n,n)*x^n is the g.f. of A014070.
(4) A(x) = G(x/A(x)) and G(x) = A(x*G(x)), where G(x) is the g.f. of A188194.

Extensions

More terms from Michel Marcus, Apr 21 2025

A188194 G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x)^2)^n/n!.

Original entry on oeis.org

1, 2, 14, 168, 3756, 261560, 80733232, 96730287424, 412733638204832, 6222933783425122080, 334514554099356252794912, 64846889330532757107162199040, 45814974387230048629026769270192768
Offset: 0

Views

Author

Paul D. Hanna, Mar 23 2011

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 14*x^2 + 168*x^3 + 3756*x^4 + 261560*x^5 +...
which equals the series:
A(x) = 1 + log(1+2*x*A(x)^2) + log(1+4*x*A(x)^2)^2/2! + log(1+8*x*A(x)^2)^3/3! +...
Let B(x) equal the g.f. of A014070, which begins:
B(x) = 1 + 2*x + 6*x^2 + 56*x^3 + 1820*x^4 +...+ C(2^n,n)*x^n +...
then B(x) = A(x/B(x)^2) and A(x) = B(x*A(x)^2), so that:
A(x) = 1 + 2*x*A(x)^2 + 6*x^2*A(x)^4 + 56*x^3*A(x)^6 + 1820*x^4*A(x)^8 +...+ C(2^n,n)*x^n*A(x)^(2n) +...

Crossrefs

Cf. A014070, A188193.

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, log(1+2^m*x*A^2+x*O(x^n))^m/m!)); polcoeff(A, n)}

Formula

G.f. A(x) satisfies:
(1) A(x) = Sum_{n>=0} C(2^n,n)*x^n*A(x)^(2n),
(2) A(x) = sqrt((1/x)*Series_Reversion(x/B(x)^2)),
(3) A(x) = B(x*A(x)^2) and B(x) = A(x/B(x)^2),
where B(x) = Sum_{n>=0} C(2^n,n)*x^n is the g.f. of A014070.
(4) A(x) = F(x*A(x)) and F(x) = A(x/F(x)), where F(x) is the g.f. of A188193.

A333125 a(n) = binomial(Fibonacci(n),n).

Original entry on oeis.org

1, 1, 0, 0, 0, 1, 28, 1716, 203490, 52451256, 29248649430, 36519676207704, 103619293824707388, 681222021538453426360, 10526080837282875691177000, 387340445158332035685509830240, 34306348668342682111244774082795555, 7379087300345635546662027722168990277849
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 08 2020, following a suggestion of Andrew Nelson

Keywords

Links

Crossrefs

Cf. A000045, A014070.

Programs

  • Mathematica
    Table[Binomial[Fibonacci[n], n], {n, 0, 20}]
  • PARI
    a(n) = binomial(fibonacci(n), n); \\ Michel Marcus, Mar 10 2020

Formula

a(n) ~ phi^(n^2) * exp(n) / (sqrt(2*Pi) * 5^(n/2) * n^(n + 1/2)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio.

A042943 Numbers k such that binomial(2^k, k) is divisible by binomial(2^k, 2).

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 11, 13, 14, 17, 19, 22, 23, 25, 26, 27, 29, 31, 33, 35, 37, 38, 39, 41, 43, 45, 46, 47, 49, 50, 51, 53, 55, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 75, 77, 79, 81, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 98, 99, 101, 102, 103, 106, 107, 109, 111
Offset: 1

Views

Author

Labos Elemer, Apr 11 2001

Keywords

Comments

Does not contain multiples of 4 (A008586).

Crossrefs

Cf. A014070, A006516.

Programs

  • Mathematica
    Select[Range[150],Divisible[Binomial[2^#,#],Binomial[2^#,2]]&]  (* Harvey P. Dale, Mar 24 2011 *)
  • PARI
    isok(k) = (binomial(2^k, k) % binomial(2^k, 2)) == 0; \\ Michel Marcus, May 14 2018
  • Python
    from math import comb
from itertools import count, islice
def A042943_gen(startvalue=1): # generator of terms >= startvalue
    for k in count(max(startvalue,1)):
        if comb(m:=1<A042943_list = list(islice(A042943_gen(),30)) # Chai Wah Wu, Jul 31 2025

Formula

k : A014070(k) mod A006516(k) = binomial(2^k, k) mod binomial(2^n, 2) = 0.

A060567 Number of binomial coefficients C(n,j) with j=0..n that are divisible by C(n,2).

Original entry on oeis.org

3, 2, 1, 2, 2, 2, 4, 2, 2, 6, 5, 4, 10, 4, 2, 8, 10, 6, 12, 4, 6, 18, 10, 4, 14, 14, 6, 10, 23, 20, 22, 10, 18, 22, 10, 18, 34, 26, 15, 18, 24, 14, 32, 14, 10, 42, 28, 12, 28, 18, 12, 32, 34, 14, 25, 26, 42, 54, 30, 14, 58, 42, 22, 24, 28, 36, 60, 38, 20, 34, 44, 26, 70, 42, 20, 42
Offset: 2

Views

Author

Labos Elemer, Apr 12 2001

Keywords

Examples

			The relevant residues C(n,j) mod C(n,2) for n=16 are {1, 16, 0, 80, 20, 48, 88, 40, 30, 40, 88, 48, 20, 80, 0, 16, 1}, so a(16) = 2.
For n=227, there are only 6 nonzero residues; the residues are {1, 227, [111 zeros], 454, 454, [111 zeros], 227, 1}, so a(227) = 222.

Crossrefs

Cf. A060430, A014070.

Formula

a(n) = Cardinality{j | C(n,2) divides C(n,j), j=0..n}.

A136584 G.f. A(x) satisfies: 1+x = Sum_{n>=0} C(2^n,n) * x^n / A(x)^(2^n).

Original entry on oeis.org

1, 1, 3, 31, 1327, 170211, 68333813, 89675072255, 397525147082217, 6103188627225900995, 331088233835064606501621, 64490029272314754165301653295, 45679131008965219349145151231118965
Offset: 0

Views

Author

Paul D. Hanna, Jan 09 2008

Keywords

Examples

			G.f. A(x) = 1 + x + 3*x^2 + 31*x^3 + 1327*x^4 + 170211*x^5 + ...
1 + x = Sum_{n>=0} C(2^n,n) * x^n * A(x)^(-2^n) =
(1 - x - 2x^2 - 26x^3 - 1264x^4 - 167480x^5 - 67988870x^6 -...) +
.2x*(1 - 2x - 3x^2 - 48x^3 - 2472x^4 - 332328x^5 -...) +
...6x^2*(1 - 4x - 2x^2 - 84x^3 - 4743x^4 - 654480x^5 -...) +
.......56x^3*(1 - 8x + 12x^2 - 152x^3 - 8810x^4 -...) +
..........1820x^4*(1 - 16x + 88x^2 - 496x^3 - 15044x^4 -...) +
..............201376x^5*(1 - 32x + 432x^2 - 3808x^3 -...) +
..................74974368x^6*(1 - 64x + 1888x^2 +...) + ...

Crossrefs

Cf. A014070 (C(2^n, n)).

Programs

  • PARI
    {a(n)=local(A=[1,1]);if(n<0,0,if(n==0,1,for(i=0,n-1,A=concat(A,0); A[ #A]=Vec(sum(n=0,#A-1,log((1+2^n*x)/Ser(A))^n/n!))[ #A]);A[n+1]))}

Formula

G.f. A(x) satisfies: 1+x = Sum_{n>=0} log( (1 + 2^n*x)/A(x) )^n / n!.
Previous Showing 41-46 of 46 results.

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