A317802 G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n) )^n = 1.

1, 3, 12, 127, 2445, 66939, 2324026, 96491718, 4631150520, 251413638241, 15206137508067, 1013223645173301, 73729926406815893, 5817609547850902791, 494790115210979151063, 45129281235546080750387, 4394695321061357601501585, 455127430187799524613334185, 49952816657399856543050669882, 5792366218971732073257841216098, 707622192835283858272032714820854

Offset

0,2

Links

Paul D. Hanna, Table of n, a(n) for n = 0..200

Formula

G.f. A(x) satisfies:

(1) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n) )^n.

(2) A(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+3) )^n.

(3) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+3) )^n / (1+x)^(3*n+3).

(4) Let B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+1) )^n , then

B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+3) )^n / (1+x)^(2*n+2).

(5) Let C(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+2) )^n , then

C(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+3) )^n / (1+x)^(n+1).

a(n) ~ 2^(-log(2)/6 - 5/2) * 3^n * n^n / (sqrt(1-log(2)) * exp(n) * (log(2))^(2*n+1)). - Vaclav Kotesovec, Aug 13 2018

Example

G.f.: A(x) = 1 + 3*x + 12*x^2 + 127*x^3 + 2445*x^4 + 66939*x^5 + 2324026*x^6 + 96491718*x^7 + 4631150520*x^8 + 251413638241*x^9 + 15206137508067*x^10 + ...
such that
1 = 1  +  (1/A(x) - 1/(1+x)^3)  +  (1/A(x) - 1/(1+x)^6)^2  +  (1/A(x) - 1/(1+x)^9)^3  +  (1/A(x) - 1/(1+x)^12)^4  +  (1/A(x) - 1/(1+x)^15)^5  +  (1/A(x) - 1/(1+x)^18)^6  +  (1/A(x) - 1/(1+x)^21)^7  +  (1/A(x) - 1/(1+x)^24)^8  + ...
Also,
A(x) = 1  +  (1/A(x) - 1/(1+x)^6)  +  (1/A(x) - 1/(1+x)^9)^2  +  (1/A(x) - 1/(1+x)^12)^3  +  (1/A(x) - 1/(1+x)^15)^4  +  (1/A(x) - 1/(1+x)^18)^5  +  (1/A(x) - 1/(1+x)^21)^6  +  (1/A(x) - 1/(1+x)^24)^7  +  (1/A(x) - 1/(1+x)^27)^8  + ...
RELATED SERIES.
(1) The series B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+1) )^n begins
B(x) = 1 + x + 3*x^2 + 33*x^3 + 634*x^4 + 17326*x^5 + 601161*x^6 + 24961740*x^7 + 1198455358*x^8 + 65087157334*x^9 + 3938132342935*x^10 + ...
restated,
B(x) = 1  +  (1/A(x) - 1/(1+x)^4)  +  (1/A(x) - 1/(1+x)^7)^2  +  (1/A(x) - 1/(1+x)^10)^3  +  (1/A(x) - 1/(1+x)^13)^4  +  (1/A(x) - 1/(1+x)^16)^5  +  (1/A(x) - 1/(1+x)^19)^6  +  (1/A(x) - 1/(1+x)^22)^7  +  (1/A(x) - 1/(1+x)^25)^8  + ...
which can also be written
B(x) = 1/(1+x)^2  +  (1/A(x) - 1/(1+x)^6)/(1+x)^4  +  (1/A(x) - 1/(1+x)^9)^2/(1+x)^6  +  (1/A(x) - 1/(1+x)^12)^3/(1+x)^8  +  (1/A(x) - 1/(1+x)^15)^4/(1+x)^10  +  (1/A(x) - 1/(1+x)^18)^5/(1+x)^12  +  (1/A(x) - 1/(1+x)^21)^6/(1+x)^14  +  (1/A(x) - 1/(1+x)^24)^7/(1+x)^16  +  (1/A(x) - 1/(1+x)^27)^8/(1+x)^18  + ...
...
(2) The series C(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(3*n+2) )^n begins
C(x) = 1 + 2*x + 7*x^2 + 75*x^3 + 1442*x^4 + 39413*x^5 + 1367095*x^6 + 56736076*x^7 + 2722528369*x^8 + 147785496105*x^9 + 8937999326808*x^10 + ...
restated,
C(x) = 1  +  (1/A(x) - 1/(1+x)^5)  +  (1/A(x) - 1/(1+x)^8)^2  +  (1/A(x) - 1/(1+x)^11)^3  +  (1/A(x) - 1/(1+x)^14)^4  +  (1/A(x) - 1/(1+x)^17)^5  +  (1/A(x) - 1/(1+x)^20)^6  +  (1/A(x) - 1/(1+x)^23)^7  +  (1/A(x) - 1/(1+x)^26)^8  + ...
which can also be written
C(x) = 1/(1+x)  +  (1/A(x) - 1/(1+x)^6)/(1+x)^2  +  (1/A(x) - 1/(1+x)^9)^2/(1+x)^3  +  (1/A(x) - 1/(1+x)^12)^3/(1+x)^4  +  (1/A(x) - 1/(1+x)^15)^4/(1+x)^5  +  (1/A(x) - 1/(1+x)^18)^5/(1+x)^6  +  (1/A(x) - 1/(1+x)^21)^6/(1+x)^7  +  (1/A(x) - 1/(1+x)^24)^7/(1+x)^8  +  (1/A(x) - 1/(1+x)^27)^8/(1+x)^9  + ...
...
Compare the above series to
1 = 1/(1+x)^3  +  (1/A(x) - 1/(1+x)^6)/(1+x)^6  +  (1/A(x) - 1/(1+x)^9)^2/(1+x)^9  +  (1/A(x) - 1/(1+x)^12)^3/(1+x)^12  +  (1/A(x) - 1/(1+x)^15)^4/(1+x)^15  +  (1/A(x) - 1/(1+x)^18)^5/(1+x)^18  +  (1/A(x) - 1/(1+x)^21)^6/(1+x)^21  +  (1/A(x) - 1/(1+x)^24)^7/(1+x)^24  +  (1/A(x) - 1/(1+x)^27)^8/(1+x)^27  + ...

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))^(3*m+3) )^m ) )[#A]/2 ); A[n+1]}
for(n=0, 25, print1(a(n), ", "))

Crossrefs

Cf. A317339, A317801, A317803, A317667.

Sequence in context: A124153 A162127 A073987 * A138392 A152544 A280115

Adjacent sequences: A317799 A317800 A317801 * A317803 A317804 A317805

Keyword

nonn

Author

Paul D. Hanna, Aug 12 2018

Status

approved