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

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
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