|
| |
|
|
A266328
|
|
E.g.f. A(x) satisfies: A(x) = exp( Integral B(x) dx ) such that B(x) = exp(-x) * exp( Integral A(x) dx ), where the constant of integration is zero.
|
|
4
|
|
|
|
1, 1, 1, 2, 6, 21, 92, 469, 2731, 17985, 131528, 1059616, 9319363, 88833422, 912393381, 10043727089, 117969438513, 1472593659884, 19467505081458, 271704942613323, 3992343851680466, 61603531051030691, 995949139457447931, 16835191741257445589, 296976010796327785530, 5457427389713208932740, 104308245862443706265341, 2070461793105333579698992, 42622090166454492404075635
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,4
|
|
|
COMMENTS
|
Compare to: G(x) = exp( Integral G(x) dx ) when G(x) = 1/(1-x).
What is Limit (a(n)/n!)^(1/n) ? Example: (a(500)/500!)^(1/500) = 0.7353325805...
Limit (a(n)/n!)^(1/n) = 1/Integral_{x=0..oo} 1/(exp(x) - x) dx = 0.73578196429164719984313538... - Vaclav Kotesovec, Aug 21 2017
|
|
|
LINKS
|
|
|
|
FORMULA
|
E.g.f. A(x) satisfies:
(1) A(x) = exp( Integral A(x) - log(A(x)) dx ).
(2) A(x) = log(A(x)) + A'(x)/A(x).
(3) log(A(x)) = exp(-x) * Integral exp(x)*A(x) dx.
(4) A(x) = exp( Series_Reversion( Integral 1/(exp(x) - x) dx ) ).
a(n) ~ c^(n+1) * n!, where c = 1/Integral_{x=0..oo} 1/(exp(x) - x) dx = 0.7357819642916471998431353808137704665788888148929882090175... - Vaclav Kotesovec, Aug 21 2017
a(n) = R_1(n-1, 0) for n > 0 with a(0) = 1 where R_1(n, q) = R_1(n-1, q+1) + Sum_{j=0..q-1} binomial(q+1, j)*R_1(n-1, j) for n > 0, q >= 0 with R_1(0, q) = 1 for q >= 0. Compare to: R_2(n-1, 0) = A000142(n) for n > 0 where R_2(n, q) = Sum_{j=0..q+1} binomial(q+1, j)*R_2(n-1, j) for n > 0, q >= 0 with R_2(0, q) = 1 for q >= 0. - Mikhail Kurkov, Dec 26 2023 [verification needed]
|
|
|
EXAMPLE
|
E.g.f.: A(x) = 1 + x + x^2/2! + 2*x^3/3! + 6*x^4/4! + 21*x^5/5! + 92*x^6/6! + 469*x^7/7! + 2731*x^8/8! + 17985*x^9/9! + 131528*x^10/10! +...
such that log(A(x)) = Integral B(x) dx
where
B(x) = 1 + x^2/2! + x^3/3! + 5*x^4/4! + 16*x^5/5! + 76*x^6/6! + 393*x^7/7! + 2338*x^8/8! + 15647*x^9/9! + 115881*x^10/10! +...
and A(x) and B(x) satisfy:
(1) A(x) = B'(x)/B(x) + 1,
(2) B(x) = A'(x)/A(x),
(3) B(x) = A(x) - log(A(x)),
(4) log(A(x)) = Integral B(x) dx,
(5) log(B(x)) = Integral A(x) dx - x.
The Series Reversion of log(A(x)) equals Integral 1/(exp(x) - x) dx:
Integral 1/(exp(x) - x) dx = x - x^3/3! - x^4/4! + 5*x^5/5! + 19*x^6/6! - 41*x^7/7! - 519*x^8/8! - 183*x^9/9! + 19223*x^10/10! +...+ A089148(n-1)*x^n/n! +...
so that A( Integral 1/(exp(x) - x) dx ) = exp(x).
|
|
|
PROG
|
(PARI) {a(n) = my(A=1+x, B=1+x); for(i=0, n, A = exp( intformal( B + x*O(x^n) ) ); B = exp( intformal( A - 1 ) ) ); n!*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(exp(x +x*O(x^n)) - x) ) )), n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) Vec( serlaplace( exp( serreverse( intformal( 1/(exp(x +x*O(x^25)) - x)))))) \\ Joerg Arndt, Dec 26 2023
(PARI) upto(n) = n--; my(v1, v2, v3, v4); v1 = vector(n + 1, i, 1); v2 = v1; v3 = vector(n + 1, i, 0); v3[1] = 1; v4 = vector(n + 1, i, vector(i, j, (j == 1) || (j == i))); for(i = 3, n + 1, for(j = 2, i - 1, v4[i][j] = v4[i - 1][j] + v4[i - 1][j - 1])); for(i = 1, n, for(q = 0, n - i, v2[q + 1] = v1[q + 2] + sum(j = 0, q - 1, v4[q + 2][j + 1] * v1[j + 1])); v1 = v2; v3[i + 1] = v1[1]); concat(1, v3) \\ Mikhail Kurkov, Dec 26 2023 [verification needed]
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,changed
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
