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A317995
G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n) )^n = 1.
2
1, 5, 35, 610, 19455, 886126, 51256460, 3547342545, 283841669495, 25689974114785, 2590438823559751, 287755717118442960, 34906792324639545345, 4591374110875921928770, 650935065832755644508135, 98965182089496736423674254, 16063900800630675693846054095, 2772975952788175401479179760640, 507291948247657812718949908038315
OFFSET
0,2
COMMENTS
In general, if k > 0 and g.f. A(x) satisfies Sum_{n>=0} (1/A(x) - 1/(1+x)^(k*n))^n = 1, then a(n,k) ~ k^n * n^n / (2^(5/2 + log(2)/(2*k)) * sqrt(1 - log(2)) * exp(n) * log(2)^(2*n+1)). - Vaclav Kotesovec, Aug 21 2018
LINKS
FORMULA
G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n) )^n.
(2) A(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+5) )^n.
(3) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+5) )^n / (1+x)^(5*n+5).
(4) Let B(x,p) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n + p) )^n ,
then B(x,p) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*(n+1)) )^n / (1+x)^((5-p)*(n+1)), where B(x,0) = 1 and B(x,5) = A(x).
a(n) ~ 5^n * n^n / (2^(5/2 + log(2)/10) * sqrt(1 - log(2)) * exp(n) * log(2)^(2*n+1)). - Vaclav Kotesovec, Aug 21 2018
EXAMPLE
G.f.: A(x) = 1 + 5*x + 35*x^2 + 610*x^3 + 19455*x^4 + 886126*x^5 + 51256460*x^6 + 3547342545*x^7 + 283841669495*x^8 + 25689974114785*x^9 + 2590438823559751*x^10 + 287755717118442960*x^11 + 34906792324639545345*x^12 + ...
such that
1 = 1 + (1/A(x) - 1/(1+x)^5) + (1/A(x) - 1/(1+x)^10)^2 + (1/A(x) - 1/(1+x)^15)^3 + (1/A(x) - 1/(1+x)^20)^4 + (1/A(x) - 1/(1+x)^25)^5 + (1/A(x) - 1/(1+x)^30)^6 + (1/A(x) - 1/(1+x)^35)^7 + (1/A(x) - 1/(1+x)^40)^8 + ...
Also,
A(x) = 1 + (1/A(x) - 1/(1+x)^10) + (1/A(x) - 1/(1+x)^15)^2 + (1/A(x) - 1/(1+x)^20)^3 + (1/A(x) - 1/(1+x)^25)^4 + (1/A(x) - 1/(1+x)^30)^5 + (1/A(x) - 1/(1+x)^35)^6 + (1/A(x) - 1/(1+x)^40)^7 + (1/A(x) - 1/(1+x)^45)^8 + ...
RELATED SERIES.
(1) The series B(x,1) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+1) )^n begins
B(x,1) = 1 + x + 5*x^2 + 90*x^3 + 2870*x^4 + 130540*x^5 + 7549806*x^6 + 522796431*x^7 + 41863962380*x^8 + 3791942099690*x^9 + ...
where B(x,1) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+5) )^n / (1+x)^(4*n+4).
(2) The series B(x,2) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+2) )^n begins
B(x,2) = 1 + 2*x + 11*x^2 + 195*x^3 + 6215*x^4 + 282530*x^5 + 16329027*x^6 + 1129955520*x^7 + 90428513089*x^8 + 8186559207316*x^9 + ...
where B(x,2) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+5) )^n / (1+x)^(3*n+3).
(3) The series B(x,3) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+3) )^n begins
B(x,3) = 1 + 3*x + 18*x^2 + 316*x^3 + 10070*x^4 + 457825*x^5 + 26455758*x^6 + 1830162112*x^7 + 146417823614*x^8 + 13251391771695*x^9 + ...
where B(x,3) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+5) )^n / (1+x)^(2*n+2).
(4) The series B(x,4) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+4) )^n begins
B(x,4) = 1 + 4*x + 26*x^2 + 454*x^3 + 14471*x^4 + 658355*x^5 + 38054529*x^6 + 2632673917*x^7 + 210610397992*x^8 + 19059538561119*x^9 + ...
where B(x,4) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n+5) )^n / (1+x)^(n+1).
PROG
(PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); A[#A] =Vec( sum(m=0, #A, ( 1/Ser(A) - 1/(1+x +x*O(x^#A))^(5*m+5) )^m ) )[#A]/2 ); A[n+1]}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Aug 15 2018
STATUS
approved