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.

A317801 G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n) )^n = 1.

Original entry on oeis.org

1, 2, 5, 36, 465, 8504, 196900, 5448960, 174282930, 6304990774, 254130115519, 11285072891720, 547300883009420, 28782595694322682, 1631647871872693931, 99196878522983084624, 6438935162966517263673, 444501575692314862825620, 32520919561355194120182078, 2513781290973908970634293260, 204713526722520414595009119193
Offset: 0

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Author

Paul D. Hanna, Aug 12 2018

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 5*x^2 + 36*x^3 + 465*x^4 + 8504*x^5 + 196900*x^6 + 5448960*x^7 + 174282930*x^8 + 6304990774*x^9 + 254130115519*x^10 + ...
such that
1 = 1  +  (1/A(x) - 1/(1+x)^2)  +  (1/A(x) - 1/(1+x)^4)^2  +  (1/A(x) - 1/(1+x)^6)^3  +  (1/A(x) - 1/(1+x)^8)^4  +  (1/A(x) - 1/(1+x)^10)^5  +  (1/A(x) - 1/(1+x)^12)^6  +  (1/A(x) - 1/(1+x)^14)^7  +  (1/A(x) - 1/(1+x)^16)^8  + ...
Also,
A(x) = 1  +  (1/A(x) - 1/(1+x)^4)  +  (1/A(x) - 1/(1+x)^6)^2  +  (1/A(x) - 1/(1+x)^8)^3  +  (1/A(x) - 1/(1+x)^10)^4  +  (1/A(x) - 1/(1+x)^12)^5  +  (1/A(x) - 1/(1+x)^14)^6  +  (1/A(x) - 1/(1+x)^16)^7  +  (1/A(x) - 1/(1+x)^18)^8  + ...
RELATED SERIES.
The series B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+1) )^n begins
B(x) = 1 + x + 2*x^2 + 15*x^3 + 193*x^4 + 3523*x^5 + 81497*x^6 + 2254695*x^7 + 72114516*x^8 + 2609080565*x^9 + 105173092838*x^10 + ...
restated,
B(x) = 1  +  (1/A(x) - 1/(1+x)^3)  +  (1/A(x) - 1/(1+x)^5)^2  +  (1/A(x) - 1/(1+x)^7)^3  +  (1/A(x) - 1/(1+x)^9)^4  +  (1/A(x) - 1/(1+x)^11)^5  +  (1/A(x) - 1/(1+x)^13)^6  +  (1/A(x) - 1/(1+x)^15)^7  +  (1/A(x) - 1/(1+x)^17)^8  + ...
which can also be written
B(x) = 1/(1+x)  +  (1/A(x) - 1/(1+x)^4)/(1+x)^2  +  (1/A(x) - 1/(1+x)^6)^2/(1+x)^3  +  (1/A(x) - 1/(1+x)^8)^3/(1+x)^4  +  (1/A(x) - 1/(1+x)^10)^4/(1+x)^5  +  (1/A(x) - 1/(1+x)^12)^5/(1+x)^6  +  (1/A(x) - 1/(1+x)^14)^6/(1+x)^7  +  (1/A(x) - 1/(1+x)^16)^7/(1+x)^8  +  (1/A(x) - 1/(1+x)^18)^8/(1+x)^9  + ...
Compare the above to
1 = 1/(1+x)^2  +  (1/A(x) - 1/(1+x)^4)/(1+x)^4  +  (1/A(x) - 1/(1+x)^6)^2/(1+x)^6  +  (1/A(x) - 1/(1+x)^8)^3/(1+x)^8  +  (1/A(x) - 1/(1+x)^10)^4/(1+x)^10  +  (1/A(x) - 1/(1+x)^12)^5/(1+x)^12  +  (1/A(x) - 1/(1+x)^14)^6/(1+x)^14  +  (1/A(x) - 1/(1+x)^16)^7/(1+x)^16  +  (1/A(x) - 1/(1+x)^18)^8/(1+x)^18  + ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); A[#A] = Vec( sum(m=0, #A, ( 1/Ser(A) - 1/(1+x +x*O(x^#A))^(2*m+2) )^m ) )[#A]/2 ); A[n+1]}
    for(n=0, 25, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n) )^n.
(2) A(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n.
(3) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n / (1+x)^(2*n+2).
(4) Let B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+1) )^n ,
then B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n / (1+x)^(n+1).
a(n) ~ 2^(n - log(2)/4 - 5/2) * n^n / (sqrt(1-log(2)) * exp(n) * (log(2))^(2*n+1)). - Vaclav Kotesovec, Aug 13 2018