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 1-3 of 3 results.

A303056 G.f. A(x) satisfies: 1 = Sum_{n>=0} ((1+x)^n - A(x))^n.

Original entry on oeis.org

1, 1, 1, 8, 89, 1326, 24247, 521764, 12867985, 357229785, 11017306489, 373675921093, 13825260663882, 554216064798423, 23934356706763264, 1108017262467214486, 54747529760516714323, 2876096694574711401525, 160092696678371426933342, 9413031424290635395882462, 583000844360279565483710624
Offset: 0

Views

Author

Paul D. Hanna, Apr 19 2018

Keywords

Comments

More generally, the following sums are equal:
(1) Sum_{n>=0} binomial(n+k-1, n) * r^n * (p + q^n)^n,
(2) Sum_{n>=0} binomial(n+k-1, n) * r^n * q^(n^2) / (1 - r*p*q^n)^(n+k),
for any fixed integer k; here, k = 1 with r = 1, p = -A(x), q = (1+x). - Paul D. Hanna, Jun 22 2019

Examples

			G.f.: A(x) = 1 + x + x^2 + 8*x^3 + 89*x^4 + 1326*x^5 + 24247*x^6 + 521764*x^7 + 12867985*x^8 + 357229785*x^9 + 11017306489*x^10 + ...
such that
1 = 1  +  ((1+x) - A(x))  +  ((1+x)^2 - A(x))^2  +  ((1+x)^3 - A(x))^3  +  ((1+x)^4 - A(x))^4  +  ((1+x)^5 - A(x))^5  +  ((1+x)^6 - A(x))^6  +  ((1+x)^7 - A(x))^7 + ...
Also,
1 = 1/(1 + A(x))  +  (1+x)/(1 + (1+x)*A(x))^2  +  (1+x)^4/(1 + (1+x)^2*A(x))^3  +  (1+x)^9/(1 + (1+x)^3*A(x))^4  +  (1+x)^16/(1 + (1+x)^4*A(x))^5  +  (1+x)^25/(1 + (1+x)^5*A(x))^6  +  (1+x)^36/(1 + (1+x)^6*A(x))^7 + ...
RELATED SERIES.
log(A(x)) = x + x^2/2 + 22*x^3/3 + 325*x^4/4 + 6186*x^5/5 + 137380*x^6/6 + 3478651*x^7/7 + 98674253*x^8/8 + 3096911434*x^9/9 + ...
PARTICULAR VALUES.
Although the power series A(x) diverges at x = -1/2, it may be evaluated formally.
Let t = A(-1/2) = 0.545218973635949431234950245034944106957612798888179456724264...
then t satisfies
(1) 1 = Sum_{n>=0} ( 1/2^n - t )^n.
(2) 1 = Sum_{n>=0} 2^n / ( 2^n + t )^(n+1).
Also,
A(r) = 1/2 at r = -0.54683649902292991492196620520872286547799291909992048564578...
where
(1) 1 = Sum_{n>=0} ( (1+r)^n - 1/2 )^n.
(2) 1 = Sum_{n>=0} (1+r)^(-n) / ( 1/(1+r)^n + 1/2 )^(n+1).
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=0,n, A=concat(A,0); A[#A] = Vec( sum(m=0,#A, ((1+x)^m - Ser(A))^m ) )[#A] );A[n+1]}
    for(n=0,30, print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} ((1+x)^n - A(x))^n.
(2) 1 = Sum_{n>=0} (1+x)^(n^2) / (1 + (1+x)^n*A(x))^(n+1).
a(n) ~ c * d^n * n! / sqrt(n), where d = A317855 = 3.1610886538654... and c = 0.11739505492506... - Vaclav Kotesovec, Sep 26 2020

A337755 G.f. A(x) satisfies: 1 = Sum_{n>=0} (n+1) * 2^n * ((1+x)^n - A(x))^n.

Original entry on oeis.org

1, 1, 3, 52, 1320, 43440, 1722712, 79186272, 4118457732, 238450436416, 15189543467688, 1055122226778720, 79359687454230296, 6425540170275120528, 557306222539540276176, 51558846502494563714080, 5068865533417385007925076, 527798429103621760357553448, 58032815995114574483132149504
Offset: 0

Views

Author

Paul D. Hanna, Sep 18 2020

Keywords

Comments

In general, the following sums are equal:
(1) Sum_{n>=0} binomial(n+k-1, n) * r^n * (p + q^n)^n,
(2) Sum_{n>=0} binomial(n+k-1, n) * r^n * q^(n^2) / (1 - r*p*q^n)^(n+k),
for any fixed integer k; here, k = 2 with r = 2, p = -A(x), q = (1+x).

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 52*x^3 + 1320*x^4 + 43440*x^5 + 1722712*x^6 + 79186272*x^7 + 4118457732*x^8 + 238450436416*x^9 + ...
where
1 = 1  +  2*2*((1+x) - A(x))  +  3*2^2*((1+x)^2 - A(x))^2  +  4*2^3*((1+x)^3 - A(x))^3  +  5*2^4*((1+x)^4 - A(x))^4  +  6*2^5*((1+x)^5 - A(x))^5  +  7*2^6*((1+x)^6 - A(x))^6  +  8*2^7*((1+x)^7 - A(x))^7 + ... + (n+1)*2^n*((1+x)^n - A(x))^n + ...
Also,
1 = 1/(1 + 2*A(x))^2  +  2*2*(1+x)/(1 + 2*(1+x)*A(x))^3  +  3*2^2*(1+x)^4/(1 + 2*(1+x)^2*A(x))^4  +  4*2^3*(1+x)^9/(1 + 2*(1+x)^3*A(x))^5  +  5*2^4*(1+x)^16/(1 + 2*(1+x)^4*A(x))^6  +  6*2^5*(1+x)^25/(1 + 2*(1+x)^5*A(x))^7  +  7*2^6*(1+x)^36/(1 + 2*(1+x)^6*A(x))^8 + ... + (n+1)*2^n*(1+x)^(n^2)/(1 + 2*(1+x)^n*A(x))^(n+2) + ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=0, n, A=concat(A, 0); A[#A] = Vec( sum(m=0, #A, (m+1) * 2^m * ((1+x)^m - Ser(A))^m ) )[#A]/4 ); A[n+1]}
    for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} (n+1) * 2^n * ((1+x)^n - A(x))^n.
(2) 1 = Sum_{n>=0} (n+1) * 2^n * (1+x)^(n^2) / (1 + 2*(1+x)^n*A(x))^(n+2).
a(n) ~ c * d^n * n! * sqrt(n), where d = (1 + 2*exp(1/r)) * r^2 = 5.9039765456700218004884947864345974878995107118996038057..., where r = 0.925556278640887084941460444526398190071550948416... is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/2 and c = 0.0300441568138... - Vaclav Kotesovec, Sep 25 2020

A337756 G.f. A(x) satisfies: 1 = Sum_{n>=0} (n+1)*(n+2)/2! * 3^n * ((1+x)^n - A(x))^n.

Original entry on oeis.org

1, 1, 6, 180, 7845, 434448, 28594494, 2157238350, 182404049175, 17026549342770, 1735705779016158, 191667825521201286, 22781050822806698709, 2899308092950790588988, 393385952195184523370994, 56691647586489579559334352, 8649001755741912766806347253, 1392791055204268736953260163092
Offset: 0

Views

Author

Paul D. Hanna, Sep 18 2020

Keywords

Comments

In general, the following sums are equal:
(1) Sum_{n>=0} binomial(n+k-1, n) * r^n * (p + q^n)^n,
(2) Sum_{n>=0} binomial(n+k-1, n) * r^n * q^(n^2) / (1 - r*p*q^n)^(n+k),
for any fixed integer k; here, k = 3 with r = 3, p = -A(x), q = (1+x).

Examples

			G.f.: A(x) = 1 + x + 6*x^2 + 180*x^3 + 7845*x^4 + 434448*x^5 + 28594494*x^6 + 2157238350*x^7 + 182404049175*x^8 + ...
where
1 = 1  +  3*3*((1+x) - A(x))  +  6*3^2*((1+x)^2 - A(x))^2  +  10*3^3*((1+x)^3 - A(x))^3  +  15*3^4*((1+x)^4 - A(x))^4  +  21*3^5*((1+x)^5 - A(x))^5  +  28*3^6*((1+x)^6 - A(x))^6  +  38*3^7*((1+x)^7 - A(x))^7 + ... + C(n+2,2)*3^n*((1+x)^n - A(x))^n + ...
Also,
1 = 1/(1 + 3*A(x))^3  +  3*3*(1+x)/(1 + 3*(1+x)*A(x))^4  +  6*3^2*(1+x)^4/(1 + 3*(1+x)^2*A(x))^5  +  10*3^3*(1+x)^9/(1 + 3*(1+x)^3*A(x))^6  +  15*3^4*(1+x)^16/(1 + 3*(1+x)^4*A(x))^7  +  21*3^5*(1+x)^25/(1 + 3*(1+x)^5*A(x))^8  +  28*3^6*(1+x)^36/(1 + 3*(1+x)^6*A(x))^9 + ... + C(n+2,2)*3^n*(1+x)^(n^2)/(1 + 3*(1+x)^n*A(x))^(n+3) + ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=0, n, A=concat(A, 0); A[#A] = Vec( sum(m=0, #A, (m+1)*(m+2)/2! * 3^m * ((1+x)^m - Ser(A))^m ) )[#A]/9 ); A[n+1]}
    for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} C(n+2,2) * 3^n * ( (1+x)^n - A(x) )^n.
(2) 1 = Sum_{n>=0} C(n+2,2) * 3^n * (1+x)^(n^2) / (1 + 3*(1+x)^n*A(x))^(n+3).
a(n) ~ c * (1 + 3*exp(1/r))^n * r^(2*n) * n! * n^(3/2), where r = 0.947093169766093813913446822751643203941993193936... is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/3 and c = 0.00671991787239... - Vaclav Kotesovec, Oct 13 2020
Showing 1-3 of 3 results.