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.

A316370 E.g.f.: Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*x.

Original entry on oeis.org

1, 1, 4, 21, 152, 1410, 15774, 207984, 3153632, 54074952, 1034749080, 21858562440, 505274905992, 12686390177136, 343815306388176, 10003360314147480, 311003061260534400, 10289575224413883840, 360967225620921712704, 13383588039651073512576, 522943874535097662998400, 21477474848621411837159040, 924978962293503284606947200
Offset: 0

Views

Author

Paul D. Hanna, Jul 12 2018

Keywords

Comments

More generally, we have the following identity. Given the biexponential series
W(x,y) = Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*y,
then for fixed p and q,
Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k + p)*x + (k + q)*y = W(x,y)^(p+q+1) / ( (1 + x*W(x,y))^q * (1 + y*W(x,y))^p ).
Further, W(x,y) satisfies the biexponential functional equation
( W(x,y)/(1 + x*W(x,y)) )^x = ( W(x,y)/(1 + y*W(x,y)) )^y.

Examples

			E.g.f.: A(x) = 1 + x + 4*x^2/2! + 21*x^3/3! + 152*x^4/4! + 1410*x^5/5! + 15774*x^6/6! + 207984*x^7/7! + 3153632*x^8/8! + 54074952*x^9/9! + 1034749080*x^10/10! + ...
such that
A(x) = 1 + (1+x)*x + (2 + x)*(1 + 2*x)*x^2/2! + (3 + x)*(2 + 2*x)*(1 + 3*x)*x^3/3! + (4 + x)*(3 + 2*x)*(2 + 3*x)*(1 + 4*x)*x^4/4! + (5 + x)*(4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*x^5/5! + ...
Also,
A(x)^2/(1 + x*A(x)) = 1 + (1 + 2*x)*x + (2 + 2*x)*(1 + 3*x)*x^2/2! + (3 + 2*x)*(2 + 3*x)*(1 + 4*x)*x^3/3! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*x^4/4! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*(1 + 6*x)*x^5/5! + ...
And,
A(x)^3/((1 + x*A(x))*(1 + x^2*A(x))) = 1 + (2 + 2*x)*x + (3 + 2*x)*(2 + 3*x)*x^2/2! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*x^3/3! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*x^4/4! + (6 + 2*x)*(5 + 3*x)*(4 + 4*x)*(3 + 5*x)*(2 + 6*x)*x^5/5! + ...
RELATED SERIES.
A(x)/(1 + x*A(x)) = 1 + 2*x^2/2! + 3*x^3/3! + 32*x^4/4! + 190*x^5/5! + 1974*x^6/6! + 21588*x^7/7! + 289232*x^8/8! + 4387752*x^9/9! + ...
A(x)/(1 + x^2*A(x)) = 1 + x + 2*x^2/2! + 9*x^3/3! + 56*x^4/4! + 450*x^5/5! + 4494*x^6/6! + 53424*x^7/7! + 738464*x^8/8! + 11642184*x^9/9! + ...
where ( A(x)/(1 + x^2*A(x)) )^x = A(x)/(1 + x*A(x)).

Links

Crossrefs

Cf. A316700, A316701, A316702.

Programs

  • PARI
    /* From Biexponential Series: */
{a(n) = my(A); A = sum(m=0,n, x^m/m! * prod(k=1,m, m+1-k + k*x +x*O(x^n))); n!*polcoeff(A,n)}
for(n=0,30, print1(a(n),", "))
  • PARI
    /* From Biexponential Functional Equation: */
{a(n) = my(A=1); for(i=0,n, A = (1 + x*A)*( A/(1 + x^2*A +x*O(x^n) ) )^x ); n!*polcoeff(A,n)}
for(n=0,30, print1(a(n),", "))

Formula

E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
(1) A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*x.
(2) Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)*x = A(x)^(p+q+1) / ( (1 + x*A(x))^q * (1 + x^2*A(x))^p ), for fixed p and q.
(3) A(x)/(1 + x*A(x)) = ( A(x)/(1 + x^2*A(x)) )^x.
a(n) ~ 2^(n+1) * n^n / (sqrt(log(2)) * exp(n)). - Vaclav Kotesovec, Jul 13 2018

A316700 E.g.f. A(x) satisfies: A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k/A(x).

Original entry on oeis.org

1, 2, 5, 19, 87, 481, 3058, 22317, 183501, 1695937, 17383266, 196331895, 2413283755, 32071547509, 457005861978, 6958913121081, 112742453743929, 1940037369861185, 35336786759749378, 679714283742254627, 13755601059097927791, 292116789342048656525, 6489891770655364327818, 150589804371710317610221, 3642747130658567662759333, 91770842180615381158770081
Offset: 0

Views

Author

Paul D. Hanna, Jul 13 2018

Keywords

Comments

More generally, we have the following identity. Given the biexponential series
W(x,y) = Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*y,
then for fixed p and q,
Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k + p)*x + (k + q)*y = W(x,y)^(p+q+1) / ( (1 + x*W(x,y))^q * (1 + y*W(x,y))^p ).
Further, W(x,y) satisfies the biexponential functional equation
( W(x,y)/(1 + x*W(x,y)) )^x = ( W(x,y)/(1 + y*W(x,y)) )^y.

Examples

			E.g.f.: A(x) = 1 + 2*x + 5*x^2/2! + 19*x^3/3! + 87*x^4/4! + 481*x^5/5! + 3058*x^6/6! + 22317*x^7/7! + 183501*x^8/8! + 1695937*x^9/9! + ...
such that A = A(x) satisfies
A(x) = 1 + (1 + 1/A)*x + (2 + 1/A)*(1 + 2/A)*x^2/2! + (3 + 1/A)*(2 + 2/A)*(1 + 3/A)*x^3/3! + (4 + 1/A)*(3 + 2/A)*(2 + 3/A)*(1 + 4/A)*x^4/4! + (5 + 1/A)*(4 + 2/A)*(3 + 3/A)*(2 + 4/A)*(1 + 5/A)*x^5/5! + ...
Also,
A(x)^2/(1 + x*A(x)) = 1 + (1 + 2/A)*x + (2 + 2/A)*(1 + 3/A)*x^2/2! + (3 + 2/A)*(2 + 3/A)*(1 + 4/A)*x^3/3! + (4 + 2/A)*(3 + 3/A)*(2 + 4/A)*(1 + 5/A)*x^4/4! + (5 + 2/A)*(4 + 3/A)*(3 + 4/A)*(2 + 5/A)*(1 + 6/A)*x^5/5! + ...
And,
A(x)^3/((1 + x*A(x))*(1 + x)) = 1 + (2 + 2/A)*x + (3 + 2/A)*(2 + 3/A)*x^2/2! + (4 + 2/A)*(3 + 3/A)*(2 + 4/A)*x^3/3! + (5 + 2/A)*(4 + 3/A)*(3 + 4/A)*(2 + 5/A)*x^4/4! + (6 + 2/A)*(5 + 3/A)*(4 + 4/A)*(3 + 5/A)*(2 + 6/A)*x^5/5! + ...
RELATED SERIES.
A(x)/(1+x) = 1 + x + 3*x^2/2! + 10*x^3/3! + 47*x^4/4! + 246*x^5/5! + 1582*x^6/6! + 11243*x^7/7! + 93557*x^8/8! + 853924*x^9/9! + ...
A(x)/(1 + x*A(x)) = 1 + x - x^2/2! - 5*x^3/3! - 5*x^4/4! + 41*x^5/5! + 256*x^6/6! + 533*x^7/7! - 4451*x^8/8! - 57479*x^9/9! + ...
where ( A(x)/(1 + x*A(x)) )^A(x) = A(x)/(1 + x).
Let G(x) = A(x*G(x)) and A(x) = G(x/A(x)), where G(x) begins
G(x) = 1 + 2*x + 13*x^2/2! + 157*x^3/3! + 2819*x^4/4! + 67621*x^5/5! + 2036230*x^6/6! + 73907639*x^7/7! + 3142556933*x^8/8! + ... + A316701(n)*x^n/n! + ...
then G(x)/(1 + x*G(x)) = ( G(x)/(1 + x*G(x)^2) )^G(x)
and G(x) = (1/x)*Series_Reversion( x/A(x) ).

Links

Crossrefs

Cf. A316370, A316701, A316702.

Programs

  • Mathematica
    nmax = 25; aa = ConstantArray[0, nmax]; aa[[1]] = 2; Do[y = 1 + 2*x + Sum[aa[[k]]*x^k, {k, 2, j - 1}] + koef*x^j; sol = Solve[SeriesCoefficient[(1 + x)*(y/(1 + x*y))^y - y, {x, 0, j + 1}] == 0, koef][[1]]; aa[[j]] = koef /. sol[[1]], {j, 2, nmax}]; Flatten[{1, aa}] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 16 2020 *)
  • PARI
    /* From Biexponential Series: */
{a(n) = my(A=1); for(i=1,n, A = sum(m=0, n, x^m/m! * prod(k=1, m, m+1-k + k/A +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))

Formula

E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
(1) A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k/A(x).
(2) Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)/A(x) = A(x)^(p+q+1) / ( (1 + x*A(x))^q * (1 + x)^p ), for fixed p and q.
(3) A(x)/(1 + x) = ( A(x)/(1 + x*A(x)) )^A(x).

A316702 E.g.f.: Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*x^2.

Original entry on oeis.org

1, 1, 2, 12, 84, 640, 6060, 70728, 941808, 13950144, 230971680, 4242680640, 85192002720, 1854377366400, 43570277097984, 1099505252240640, 29642211339068160, 850166713775554560, 25852506567901839360, 830856828456304128000, 28137892587325700198400, 1001532282143426144133120, 37379628178079964459217920, 1459734364264707546159513600
Offset: 0

Views

Author

Paul D. Hanna, Jul 13 2018

Keywords

Comments

More generally, we have the following identity. Given the biexponential series
W(x,y) = Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*y,
then for fixed p and q,
Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k + p)*x + (k + q)*y = W(x,y)^(p+q+1) / ( (1 + x*W(x,y))^q * (1 + y*W(x,y))^p ).
Further, W(x,y) satisfies the biexponential functional equation
( W(x,y)/(1 + x*W(x,y)) )^x = ( W(x,y)/(1 + y*W(x,y)) )^y.

Examples

			E.g.f.: A(x) = 1 + x + 2*x^2/2! + 12*x^3/3! + 84*x^4/4! + 640*x^5/5! + 6060*x^6/6! + 70728*x^7/7! + 941808*x^8/8! + 13950144*x^9/9! + 230971680*x^10/10! + ...
such that
A(x) = 1 + (1 + x^2)*x + (2 + x^2)*(1 + 2*x^2)*x^2/2! + (3 + x^2)*(2 + 2*x^2)*(1 + 3*x^2)*x^3/3! + (4 + x^2)*(3 + 2*x^2)*(2 + 3*x^2)*(1 + 4*x^2)*x^4/4! + (5 + x^2)*(4 + 2*x^2)*(3 + 3*x^2)*(2 + 4*x^2)*(1 + 5*x^2)*x^5/5! + ...
Also,
A(x)^2/(1 + x*A(x)) = 1 + (1 + 2*x^2)*x + (2 + 2*x^2)*(1 + 3*x^2)*x^2/2! + (3 + 2*x^2)*(2 + 3*x^2)*(1 + 4*x^2)*x^3/3! + (4 + 2*x^2)*(3 + 3*x^2)*(2 + 4*x^2)*(1 + 5*x^2)*x^4/4! + (5 + 2*x^2)*(4 + 3*x^2)*(3 + 4*x^2)*(2 + 5*x^2)*(1 + 6*x^2)*x^5/5! + ...
And,
A(x)^3/((1 + x*A(x))*(1 + x^3*A(x))) = 1 + (2 + 2*x^2)*x + (3 + 2*x^2)*(2 + 3*x^2)*x^2/2! + (4 + 2*x^2)*(3 + 3*x^2)*(2 + 4*x^2)*x^3/3! + (5 + 2*x^2)*(4 + 3*x^2)*(3 + 4*x^2)*(2 + 5*x^2)*x^4/4! + (6 + 2*x^2)*(5 + 3*x^2)*(4 + 4*x^2)*(3 + 5*x^2)*(2 + 6*x^2)*x^5/5! + ...
RELATED SERIES.
A(x)/(1 + x*A(x)) = 1 + 6*x^3/3! + 12*x^4/4! + 40*x^5/5! + 900*x^6/6! + 7728*x^7/7! + 68880*x^8/8! + 1031616*x^9/9! + ...
A(x)/(1 + x^3*A(x)) = 1 + x + 2*x^2/2! + 6*x^3/3! + 36*x^4/4! + 280*x^5/5! + 2460*x^6/6! + 25368*x^7/7! + 310128*x^8/8! + 4333824*x^9/9! + ...
where ( A(x)/(1 + x^3*A(x)) )^(x^2) = A(x)/(1 + x*A(x)).

Links

Crossrefs

Cf. A316370, A316700, A316701.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sum[(x^k*(x^2 - 1)^k * Pochhammer[(k + x^2)/(x^2 - 1), k])/k!, {k, 0, nmax}], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Jul 15 2018 *)
  • PARI
    {a(n) = my(A=1); A = sum(m=0,n, x^m/m! * prod(k=1,m, m+1-k + k*x^2 +x*O(x^n))); n!*polcoeff(A,n)}
for(n=0,30, print1(a(n),", "))

Formula

E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
(1) A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*x^2.
(2) Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)*x^2 = A(x)^(p+q+1) / ( (1 + x*A(x))^q * (1 + x^3*A(x))^p ), for fixed p and q.
(3) A(x)/(1 + x*A(x)) = ( A(x)/(1 + x^3*A(x)) )^(x^2).
a(n) ~ 3^((n+1)/2) * n! / sqrt(Pi*log(3)*n). - Vaclav Kotesovec, Jul 15 2018
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