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 1-6 of 6 results.

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).

A316701 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, 13, 157, 2819, 67621, 2036230, 73907639, 3142556933, 153268340377, 8436526507286, 517427997295353, 34994424316034815, 2587503674068863681, 207665084850599068022, 17979537469340405579571, 1670426465731302891946025, 165771247503060676475253809, 17501167047878021578046031334, 1958599892703021903310163005669
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 + 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! + 153268340377*x^9/9! + ...
such that A = A(x) satisfies
A(x) = 1 + (1 + A)*x + (2 + A)*(1 + 2*A)*x^2/2! + (3 + A)*(2 + 2*A)*(1 + 3*A)*x^3/3! + (4 + A)*(3 + 2*A)*(2 + 3*A)*(1 + 4*A)*x^4/4! + (5 + 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*A(x)^2)) = 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*A(x)) = 1 + x + 7*x^2/2! + 85*x^3/3! + 1527*x^4/4! + 36621*x^5/5! + 1102348*x^6/6! + 39996727*x^7/7! + 1700108469*x^8/8! + ...
A(x)/(1 + x*A(x)^2) = 1 + x + 3*x^2/3! + 22*x^3/3! + 299*x^4/4! + 6086*x^5/5! + 164782*x^6/6! + 5553185*x^7/7! + 223540669*x^8/8! + ...
where ( A(x)/(1 + x*A(x)^2) )^A(x) = A(x)/(1 + x*A(x)).
Let G(x) = A(x/G(x)) and A(x) = G(x*A(x)), where G(x) begins
G(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! + ... + A316700(n)*x^n/n! + ...
then G(x)/(1 + x) = ( G(x)/(1 + x*G(x)) )^G(x)
and G(x) = x/Series_Reversion( x*A(x) ).

Links

Crossrefs

Cf. A316370, A316700, 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)*(y/(1 + x*y^2))^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*A(x)^2)^p ), for fixed p and q.
(3) A(x)/(1 + x*A(x)) = ( A(x)/(1 + x*A(x)^2) )^A(x).
a(n) ~ sqrt(((-1 + s) * s^3 * (1 + r*s) * (1 + r*s^2) * (1 + s + r*s^2)) / (-1 - (1 + 2*r)*s - 4*r*s^2 + (4 - 5*r)*r*s^3 + 7*r^2*s^4 + r^2*(1 + 2*r)*s^5 + 2*r^3*s^6 + r^4*s^7)) * n^(n-1) / (exp(n) * r^(n - 1/2)), where r = 0.15709770426545411244999697565973251074761653302546043128667... and s = 2.3042217766396380492962218073654169636152676607799337588129... are roots of the system of equations (s/(1 + r*s^2))^s = s/(1 + r*s), -1 + s - r*s^3 - r^2*s^4 + s*(1 + r*s)*(1 + r*s^2) * log(s/(1 + r*s^2)) = 0. - Vaclav Kotesovec, Oct 13 2020

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

A316704 E.g.f.: Sum_{n>=0} (2*x)^n/n! * Product_{k=1..n} (n+1-k) + k*x.

Original entry on oeis.org

1, 2, 12, 108, 1312, 20320, 381408, 8420160, 213813248, 6139270656, 196691281920, 6956268042240, 269187901974528, 11313963679948800, 513251989767487488, 24995547184468008960, 1300702468667721646080, 72026879559935471124480, 4229000873160355032072192, 262425376836886982897958912, 17161024720479004010930503680, 1179556709319250468710226329600
Offset: 0

Views

Author

Paul D. Hanna, Jul 16 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 + 12*x^2/2! + 108*x^3/3! + 1312*x^4/4! + 20320*x^5/5! + 381408*x^6/6! + 8420160*x^7/7! + 213813248*x^8/8! + 6139270656*x^9/9! + 196691281920*x^10/10! + ...
such that
A(x) = 1 + (1+x)*(2*x) + (2 + x)*(1 + 2*x)*(2*x)^2/2! + (3 + x)*(2 + 2*x)*(1 + 3*x)*(2*x)^3/3! + (4 + x)*(3 + 2*x)*(2 + 3*x)*(1 + 4*x)*(2*x)^4/4! + (5 + x)*(4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*(2*x)^5/5! + ...
Also,
A(x)^2/(1 + 2*x*A(x)) = 1 + (1 + 2*x)*(2*x) + (2 + 2*x)*(1 + 3*x)*(2*x)^2/2! + (3 + 2*x)*(2 + 3*x)*(1 + 4*x)*(2*x)^3/3! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*(2*x)^4/4! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*(1 + 6*x)*(2*x)^5/5! + ...
And,
A(x)^3/((1 + 2*x*A(x))*(1 + 2*x^2*A(x))) = 1 + (2 + 2*x)*(2*x) + (3 + 2*x)*(2 + 3*x)*(2*x)^2/2! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(2*x)^3/3! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*(2*x)^4/4! + (6 + 2*x)*(5 + 3*x)*(4 + 4*x)*(3 + 5*x)*(2 + 6*x)*(2*x)^5/5! + ...
RELATED SERIES.
B(x) = sqrt( (1 + 2*x*A(x)) * (1 + 2*x^2*A(x)) ) = 1 + x + 5*x^2/2! + 45*x^3/3! + 513*x^4/4! + 7745*x^5/5! + 142485*x^6/6! + 3095421*x^7/7! + 77642145*x^8/8! + 2207145825*x^9/9! + ... + A316705(n)*x^n/n! + ...
where
B(x) = 1 + (1 + x)*x + (3 + x)*(1 + 3*x)*x^2/2! + (5 + x)*(3 + 3*x)*(1 + 5*x)*x^3/3! + (7 + x)*(5 + 3*x)*(3 + 5*x)*(1 + 7*x)*x^4/4! + (9 + x)*(7 + 3*x)*(5 + 5*x)*(3 + 7*x)*(1 + 9*x)*x^5/5! + ...

Links

Crossrefs

Cf. A316370, A316705.

Programs

  • PARI
    /* Using the biexponential series */
{a(n) = my(A); A = sum(m=0,n, (2*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
    /* Using Functional Equation: */
{a(n) = my(A=1); for(i=1,n, A = ( (1 + 2*x*A)/(1 + 2*x^2*A +x*O(x^n))^x )^(1/(1-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} (2*x)^n/n! * Product_{k=1..n} (n+1-k) + k*x.
(2) Sum_{n>=0} (2*x)^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)*x = A(x)^(p+q+1) / ( (1 + 2*x*A(x))^q * (1 + 2*x^2*A(x))^p ), for fixed p and q.
(3) A(x)/(1 + 2*x*A(x)) = ( A(x)/(1 + 2*x^2*A(x)) )^x.
a(n)/n! ~ c * d^n / sqrt(n), where d = 3.346513389529679772056152566067040813392... and c = 1.06774499146514892068040233... - Vaclav Kotesovec, Jul 18 2018
The constant d given above is the root of the equation d^(d-2) = 2^(d-1). - Vaclav Kotesovec, Jan 17 2024

A316705 E.g.f.: Sum_{n>=0} x^n/n! * Product_{k=1..n} (2*n+1-2*k) + (2*k-1)*x.

Original entry on oeis.org

1, 1, 5, 45, 513, 7745, 142485, 3095421, 77642145, 2207145825, 70130493765, 2463100122285, 94752421655265, 3962161404127329, 178943401595685909, 8680576995359894205, 450150904632193002945, 24850264116962803786305, 1455015398837011003805445, 90062955077484708745769133, 5876178416626462668682616385, 403059737428052979318873127425
Offset: 0

Views

Author

Paul D. Hanna, Jul 16 2018

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 5*x^2/2! + 45*x^3/3! + 513*x^4/4! + 7745*x^5/5! + 142485*x^6/6! + 3095421*x^7/7! + 77642145*x^8/8! + 2207145825*x^9/9! + 70130493765*x^10/10! + ...
such that
A(x) = 1 + (1 + x)*x + (3 + x)*(1 + 3*x)*x^2/2! + (5 + x)*(3 + 3*x)*(1 + 5*x)*x^3/3! + (7 + x)*(5 + 3*x)*(3 + 5*x)*(1 + 7*x)*x^4/4! + (9 + x)*(7 + 3*x)*(5 + 5*x)*(3 + 7*x)*(1 + 9*x)*x^5/5! + ...
Also,
A(x) = sqrt( (1 + 2*x*W(x))*(1 + 2*x^2*W(x)) )
where
W(x) = 1 + (1 + x)*(2*x) + (2 + x)*(1 + 2*x)*(2*x)^2/2! + (3 + x)*(2 + 2*x)*(1 + 3*x)*(2*x)^3/3! + (4 + x)*(3 + 2*x)*(2 + 3*x)*(1 + 4*x)*(2*x)^4/4! + (5 + x)*(4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*(2*x)^5/5! + ...
Explicitly,
W(x) = 1 + 2*x + 12*x^2/2! + 108*x^3/3! + 1312*x^4/4! + 20320*x^5/5! + 381408*x^6/6! + 8420160*x^7/7! + 213813248*x^8/8! + ... + A316704(n)*x^n/n! + ...
where W(x) satisfies
W(x)/(1 + 2*x*W(x)) = ( W(x)/(1 + 2*x^2*W(x)) )^x.

Links

Crossrefs

Cf. A316704, A316370.

Programs

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

Formula

E.g.f.: A(x) = sqrt( (1 + 2*x*W(x))*(1 + 2*x^2*W(x)) ) such that W(x) satisfies: W(x)/(1 + 2*x*W(x)) = ( W(x)/(1 + 2*x^2*W(x)) )^x.
a(n)/n! ~ c * d^n / sqrt(n), where d = 3.346513389529679772056152566067040813392... and c = 0.34882587166136471331152567... - Vaclav Kotesovec, Jul 18 2018

A363110 G.f.: Sum_{n>=0} x^n * Product_{k=1..n} (k + (n-k+1)*x) / (1 + k*x + (n-k+1)*x^2).

Original entry on oeis.org

1, 1, 2, 4, 10, 28, 88, 306, 1158, 4730, 20722, 96776, 479340, 2507510, 13804014, 79718782, 481614806, 3036358968, 19932689952, 135981543762, 962319171782, 7053068549250, 53458038451082, 418440466421960, 3378290373259300, 28099682071640734, 240537280709926718
Offset: 0

Views

Author

Paul D. Hanna, Jun 02 2023

Keywords

Comments

Compare to the following identities, which hold for any fixed b and c:
(1) Sum_{n>=0} x^n * Product_{k=1..n} (b + k*x)/(1 + b*x + k*x^2) = (1 + b*x)/(1 - x^2).
(2) Sum_{n>=0} x^n * Product_{k=1..n} (k + c*x)/(1 + k*x + c*x^2) = (1 + c*x^2)/(1 - x).
(3) Sum_{n>=0} x^n * Product_{k=1..n} (b*k + c*k*x)/(1 + b*k*x + c*k*x^2) = 1/(1 - b*x - c*x^2).
Conjectures:
(1) a(6*n + k) == 0 (mod 4) for n > 0 when k = {0,5},
(2) a(6*n + k) == 2 (mod 4) for n > 0 when k = {1,2,3,4}.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 4*x^3 + 10*x^4 + 28*x^5 + 88*x^6 + 306*x^7 + 1158*x^8 + 4730*x^9 + 20722*x^10 + 96776*x^11 + 479340*x^12 + ...
where
A(x) = 1 + x*(1+x)/(1+x+x^2) + x^2*(1 + 2*x)*(2 + x)/((1 + x + 2*x^2)*(1 + 2*x + x^2)) + x^3*(1 + 3*x)*(2 + 2*x)*(3 + x)/((1 + x + 3*x^2)*(1 + 2*x + 2*x^2)*(1 + 3*x + x^2)) + x^4*(1 + 4*x)*(2 + 3*x)*(3 + 2*x)*(4 + x)/((1 + x + 4*x^2)*(1 + 2*x + 3*x^2)*(1 + 3*x + 2*x^2)*(1 + 4*x + x^2)) + ...

Links

Crossrefs

Cf. A204064, A204066, A316370, A067948.

Programs

  • PARI
    {a(n) = polcoeff( A = sum(m=0, n, x^m*prod(k=1, m, (k + (m-k+1)*x)/(1 + k*x + (m-k+1)*x^2 +x*O(x^n))) ), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n may be described by the following.
(1) Sum_{n>=0} x^n * Product_{k=1..n} (k + (n-k+1)*x) / (1 + k*x + (n-k+1)*x^2).
(2) Sum_{n>=0} x^n * (Sum_{k=0..n} A067948(n,k) * x^k) / Product_{k=1..n} (1 + k*x + (n-k+1)*x^2).
Showing 1-6 of 6 results.

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