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 1-5 of 5 results.

A317349 G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - (1-x)^n )^n = 1.

Original entry on oeis.org

1, 1, 2, 7, 42, 372, 4269, 59047, 946557, 17175289, 347208299, 7730688884, 187911183701, 4951155672353, 140575561645293, 4279249948000903, 139050095246322895, 4804391579357016747, 175902340755219278039, 6803436418471129704925, 277202774381386656583959, 11868116969794805874111831
Offset: 0

Views

Author

Paul D. Hanna, Aug 02 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 42*x^4 + 372*x^5 + 4269*x^6 + 59047*x^7 + 946557*x^8 + 17175289*x^9 + 347208299*x^10 + ...
such that
1 = 1  +  (1/A(x) - (1-x))  +  (1/A(x) - (1-x)^2)^2  +  (1/A(x) - (1-x)^3)^3  +  (1/A(x) - (1-x)^4)^4  +  (1/A(x) - (1-x)^5)^5  +  (1/A(x) - (1-x)^6)^6  +  (1/A(x) - (1-x)^7)^7  +  (1/A(x) - (1-x)^8)^8  + ...
Also,
A(x) = 1  +  (1/A(x) - (1-x)^2)  +  (1/A(x) - (1-x)^3)^2  +  (1/A(x) - (1-x)^4)^3  +  (1/A(x) - (1-x)^5)^4  +  (1/A(x) - (1-x)^6)^5  +  (1/A(x) - (1-x)^7)^6  +  (1/A(x) - (1-x)^8)^7  +  (1/A(x) - (1-x)^9)^8  + ...

Links

Crossrefs

Cf. A317339, A317348, A317340, A317666, A317667 A317668.

Programs

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

Formula

G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} ( 1/A(x) - (1-x)^n )^n.
(2) A(x) = Sum_{n>=0} ( 1/A(x) - (1-x)^(n+1) )^n.
(3) 1 = Sum_{n>=0} ( 1/A(x) - (1-x)^(n+1) )^n * (1-x)^(n+1).
(4) A(x)^2 = 2*A(x) * [ Sum_{n>=0} (n+1) * ( 1/A(x) - (1-x)^(n+1) )^n ] - [ Sum_{n>=0} (n+1) * ( 1/A(x) - (1-x)^(n+2) )^n ].
(5) A(x) = [ Sum_{n>=1} n*(n+1)/2 * (1-x)^(n+1) * ( 1/Ser(A) - (1-x)^(n+1) )^(n-1) ] / [ Sum_{n>=1} n^2 * (1-x)^n * ( 1/Ser(A) - (1-x)^n )^(n-1) ].
a(n) ~ 2^(log(2)/2 - 5/2) * n^n / (sqrt(1-log(2)) * exp(n) * (log(2))^(2*n+1)). - Vaclav Kotesovec, Aug 06 2018

A317801 G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n) )^n = 1.

Original entry on oeis.org

1, 2, 5, 36, 465, 8504, 196900, 5448960, 174282930, 6304990774, 254130115519, 11285072891720, 547300883009420, 28782595694322682, 1631647871872693931, 99196878522983084624, 6438935162966517263673, 444501575692314862825620, 32520919561355194120182078, 2513781290973908970634293260, 204713526722520414595009119193
Offset: 0

Views

Author

Paul D. Hanna, Aug 12 2018

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 5*x^2 + 36*x^3 + 465*x^4 + 8504*x^5 + 196900*x^6 + 5448960*x^7 + 174282930*x^8 + 6304990774*x^9 + 254130115519*x^10 + ...
such that
1 = 1  +  (1/A(x) - 1/(1+x)^2)  +  (1/A(x) - 1/(1+x)^4)^2  +  (1/A(x) - 1/(1+x)^6)^3  +  (1/A(x) - 1/(1+x)^8)^4  +  (1/A(x) - 1/(1+x)^10)^5  +  (1/A(x) - 1/(1+x)^12)^6  +  (1/A(x) - 1/(1+x)^14)^7  +  (1/A(x) - 1/(1+x)^16)^8  + ...
Also,
A(x) = 1  +  (1/A(x) - 1/(1+x)^4)  +  (1/A(x) - 1/(1+x)^6)^2  +  (1/A(x) - 1/(1+x)^8)^3  +  (1/A(x) - 1/(1+x)^10)^4  +  (1/A(x) - 1/(1+x)^12)^5  +  (1/A(x) - 1/(1+x)^14)^6  +  (1/A(x) - 1/(1+x)^16)^7  +  (1/A(x) - 1/(1+x)^18)^8  + ...
RELATED SERIES.
The series B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+1) )^n begins
B(x) = 1 + x + 2*x^2 + 15*x^3 + 193*x^4 + 3523*x^5 + 81497*x^6 + 2254695*x^7 + 72114516*x^8 + 2609080565*x^9 + 105173092838*x^10 + ...
restated,
B(x) = 1  +  (1/A(x) - 1/(1+x)^3)  +  (1/A(x) - 1/(1+x)^5)^2  +  (1/A(x) - 1/(1+x)^7)^3  +  (1/A(x) - 1/(1+x)^9)^4  +  (1/A(x) - 1/(1+x)^11)^5  +  (1/A(x) - 1/(1+x)^13)^6  +  (1/A(x) - 1/(1+x)^15)^7  +  (1/A(x) - 1/(1+x)^17)^8  + ...
which can also be written
B(x) = 1/(1+x)  +  (1/A(x) - 1/(1+x)^4)/(1+x)^2  +  (1/A(x) - 1/(1+x)^6)^2/(1+x)^3  +  (1/A(x) - 1/(1+x)^8)^3/(1+x)^4  +  (1/A(x) - 1/(1+x)^10)^4/(1+x)^5  +  (1/A(x) - 1/(1+x)^12)^5/(1+x)^6  +  (1/A(x) - 1/(1+x)^14)^6/(1+x)^7  +  (1/A(x) - 1/(1+x)^16)^7/(1+x)^8  +  (1/A(x) - 1/(1+x)^18)^8/(1+x)^9  + ...
Compare the above to
1 = 1/(1+x)^2  +  (1/A(x) - 1/(1+x)^4)/(1+x)^4  +  (1/A(x) - 1/(1+x)^6)^2/(1+x)^6  +  (1/A(x) - 1/(1+x)^8)^3/(1+x)^8  +  (1/A(x) - 1/(1+x)^10)^4/(1+x)^10  +  (1/A(x) - 1/(1+x)^12)^5/(1+x)^12  +  (1/A(x) - 1/(1+x)^14)^6/(1+x)^14  +  (1/A(x) - 1/(1+x)^16)^7/(1+x)^16  +  (1/A(x) - 1/(1+x)^18)^8/(1+x)^18  + ...

Links

Crossrefs

Cf. A317339, A317802, A317803, A317666.

Programs

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

Formula

G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n) )^n.
(2) A(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n.
(3) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n / (1+x)^(2*n+2).
(4) Let B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+1) )^n ,
then B(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(2*n+2) )^n / (1+x)^(n+1).
a(n) ~ 2^(n - log(2)/4 - 5/2) * n^n / (sqrt(1-log(2)) * exp(n) * (log(2))^(2*n+1)). - Vaclav Kotesovec, Aug 13 2018

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

Original entry on oeis.org

1, 3, 12, 127, 2445, 66939, 2324026, 96491718, 4631150520, 251413638241, 15206137508067, 1013223645173301, 73729926406815893, 5817609547850902791, 494790115210979151063, 45129281235546080750387, 4394695321061357601501585, 455127430187799524613334185, 49952816657399856543050669882, 5792366218971732073257841216098, 707622192835283858272032714820854
Offset: 0

Views

Author

Paul D. Hanna, Aug 12 2018

Keywords

Examples

			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  + ...

Links

Crossrefs

Cf. A317339, A317801, A317803, A317667.

Programs

  • 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), ", "))

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

A317803 G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n) )^n = 1.

Original entry on oeis.org

1, 4, 22, 308, 7877, 287224, 13293116, 735955720, 47105160785, 3410314286768, 275071315285416, 24442342714268592, 2371821148074889444, 249559207019813962752, 28303003280888905543584, 3442273720243525242224992, 446977352681757476329452018, 61724119095080041604018873868, 9033234491867095630258647812994, 1396682556807057529868101744945708, 227509260041431637641628131782970335
Offset: 0

Views

Author

Paul D. Hanna, Aug 12 2018

Keywords

Examples

			G.f.: A(x) = 1 + 4*x + 22*x^2 + 308*x^3 + 7877*x^4 + 287224*x^5 + 13293116*x^6 + 735955720*x^7 + 47105160785*x^8 + 3410314286768*x^9 + 275071315285416*x^10 + ...
such that
1 = 1  +  (1/A(x) - 1/(1+x)^4)  +  (1/A(x) - 1/(1+x)^8)^2  +  (1/A(x) - 1/(1+x)^12)^3  +  (1/A(x) - 1/(1+x)^16)^4  +  (1/A(x) - 1/(1+x)^20)^5  +  (1/A(x) - 1/(1+x)^24)^6  +  (1/A(x) - 1/(1+x)^28)^7  +  (1/A(x) - 1/(1+x)^32)^8  + ...
Also,
A(x) = 1  +  (1/A(x) - 1/(1+x)^8)  +  (1/A(x) - 1/(1+x)^12)^2  +  (1/A(x) - 1/(1+x)^16)^3  +  (1/A(x) - 1/(1+x)^20)^4  +  (1/A(x) - 1/(1+x)^24)^5  +  (1/A(x) - 1/(1+x)^28)^6  +  (1/A(x) - 1/(1+x)^32)^7  +  (1/A(x) - 1/(1+x)^36)^8  + ...
RELATED SERIES.
(1) The series B(x,1) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+1) )^n begins
B(x,1) = 1 + x + 4*x^2 + 58*x^3 + 1482*x^4 + 53953*x^5 + 2496149*x^6 + 138245508*x^7 + 8853719964*x^8 + 641386920943*x^9 + 51762649442019*x^10 + ...
where B(x,1) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n / (1+x)^(3*n+3).
(2) The series B(x,2) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+2) )^n begins
B(x,2) = 1 + 2*x + 9*x^2 + 128*x^3 + 3270*x^4 + 119002*x^5 + 5502295*x^6 + 304531768*x^7 + 19491119849*x^8 + 1411222743454*x^9 + 113839065423087*x^10 + ...
where B(x,2) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n / (1+x)^(2*n+2).
(3) The series B(x,3) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+3) )^n begins
B(x,3) = 1 + 3*x + 15*x^2 + 211*x^3 + 5392*x^4 + 196341*x^5 + 9079538*x^6 + 502467023*x^7 + 32153605481*x^8 + 2327561975059*x^9 + 187722580703289*x^10 + ...
where B(x,3) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n / (1+x)^(n+1).

Links

Crossrefs

Cf. A317339, A317801, A317802, A317995, A317668.

Programs

  • 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))^(4*m+4) )^m ) )[#A]/2 ); A[n+1]}
for(n=0, 25, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n) )^n.
(2) A(x) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n.
(3) 1 = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n+4) )^n / (1+x)^(4*n+4).
(4) Let B(x,p) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*n + p) )^n ,
then B(x,p) = Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(4*(n+1)) )^n / (1+x)^((4-p)*(n+1)), where B(x,0) = 1 and B(x,4) = A(x).
a(n) ~ 2^(2*n - log(2)/8 - 5/2) * n^n / (sqrt(1-log(2)) * exp(n) * (log(2))^(2*n+1)). - Vaclav Kotesovec, Aug 13 2018

A317995 G.f. A(x) satisfies: Sum_{n>=0} ( 1/A(x) - 1/(1+x)^(5*n) )^n = 1.

Original entry on oeis.org

1, 5, 35, 610, 19455, 886126, 51256460, 3547342545, 283841669495, 25689974114785, 2590438823559751, 287755717118442960, 34906792324639545345, 4591374110875921928770, 650935065832755644508135, 98965182089496736423674254, 16063900800630675693846054095, 2772975952788175401479179760640, 507291948247657812718949908038315
Offset: 0

Views

Author

Paul D. Hanna, Aug 15 2018

Keywords

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

Examples

			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).

Links

Crossrefs

Cf. A317339, A317801, A317802, A317803.

Programs

  • 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), ", "))

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
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