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.

A182954 G.f. satisfies: A(x) = 1 + x*A(x) * A( x*A(x) )^4.

Original entry on oeis.org

1, 1, 5, 39, 381, 4284, 53163, 710810, 10085621, 150326044, 2336828792, 37687170215, 628069684439, 10782885724300, 190248852445782, 3442896376032300, 63804661588968521, 1209314277690837796
Offset: 0

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Author

Paul D. Hanna, Dec 15 2010

Keywords

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 39*x^3 + 381*x^4 + 4284*x^5 + ...
Related expansions:
A(x*A(x)) = 1 + x + 6*x^2 + 54*x^3 + 592*x^4 + 7331*x^5 + 98870*x^6 + ...
A(x*A(x))^4 = 1 + 4*x + 30*x^2 + 292*x^3 + 3305*x^4 + 41420*x^5 + ...
The g.f. satisfies:
log(A(x)) = A(x)^4*x + {d/dx x*A(x)^8}*x^2/2! + {d^2/dx^2 x^2*A(x)^12}*x^3/3! + {d^3/dx^3 x^3*A(x)^16}*x^4/4! + ...

Links

Crossrefs

Cf. A384265, A030266, A121687, A182953, A182955.

Programs

  • PARI
    {a(n) = my(A = 1 + sum(i=1,n-1,a(i)*x^i+x*O(x^n)));
for(i=1,n, A = exp( sum(m=1,n, sum(k=0,n-m, binomial(m+k-1,k)*polcoef(A^(4*m),k)*x^k) * x^m/m ) + x*O(x^n))); polcoef(A,n)}
  • PARI
    {a(n, m=1) = if(n==0, 1, if(m==0, 0^n, sum(k=0, n, m*binomial(n+m, k)/(n+m)*a(n-k, 4*k))))}

Formula

G.f. A(x) satisfies:
* A(x) = exp( Sum_{m>=0} {d^m/dx^m x^m*A(x)^(4m+4)} * x^(m+1)/(m+1)! );
* A(x) = exp( Sum_{m>=1} [Sum_{k>=0} C(m+k-1,k)*{[y^k] A(y)^(4m)}*x^k]*x^m/m);
which are equivalent.
Recurrence:
Let A(x)^m = Sum_{n>=0} a(n,m)*x^n with a(0,m)=1, then
a(n,m) = Sum_{k=0..n} m*C(n+m,k)/(n+m) * a(n-k,4k).

A145347 G.f. satisfies: A(x/A(x)) = 1 + x*A(x)^3.

Original entry on oeis.org

1, 1, 4, 26, 220, 2203, 24836, 306104, 4047988, 56713521, 834286612, 12801754120, 203889888832, 3357619794321, 56999146850380, 995081586539016, 17830012791062632, 327376145842252333, 6151225530281186372, 118142009771446643592, 2317165307900630229384
Offset: 0

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Author

Paul D. Hanna, Nov 09 2008

Keywords

Comments

More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 26*x^3 + 220*x^4 + 2203*x^5 + 24836*x^6 +...
A(x)^3 = 1 + 3*x + 15*x^2 + 103*x^3 + 876*x^4 + 8679*x^5 + 96382*x^6 +...
A(x/A(x)) = 1 + x + 3*x^2 + 15*x^3 + 103*x^4 + 876*x^5 + 8679*x^6 +...
A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)):
G(x) = 1 + x + 5*x^2 + 39*x^3 + 381*x^4 + 4284*x^5 + 53163*x^6 +...
To illustrate the formula a(n) = [x^(n-1)] 4*A(x)^(n+3)/(n+3),
form a table of coefficients in A(x)^(n+3) as follows:
A^4: [(1), 4, 22, 156, 1337, 13220, 145988, 1759876, ...];
A^5: [1, (5), 30, 220, 1905, 18836, 207100, 2481740, ...];
A^6: [1, 6, (39), 296, 2595, 25704, 281727, 3358488, ...];
A^7: [1, 7, 49, (385), 3423, 34020, 372141, 4416658, ...];
A^8: [1, 8, 60, 488, (4406), 44000, 480900, 5686480, ...];
A^9: [1, 9, 72, 606, 5562, (55881), 610872, 7202268, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[4/4*(1), 4/5*(5), 4/6*(39), 4/7*(385), 4/8*(4406), 4/9*(55881), ...].

Links

Crossrefs

Cf. A384265, A145348, A145350, A147664, A120972.

Programs

  • PARI
    {a(n) = my(F=1+x); for(i=0,n, G=serreverse(x/(F+x*O(x^n))^1)/x; F=1+x*G^4); polcoef(F,n)}
  • PARI
    /* This sequence is generated when k=1, m=3: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=1, m=3) = my(A=sum(i=0, n-1, a(i, k, m)*x^i) +x*O(x^n)); if(n==0, 1, polcoef((m+k)/(m+k*n)*A^(m+k*n), n-1))}
for(n=0,20,print1(a(n),", "))

Formula

G.f.: A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)) and A(x) = G(x/A(x)).
a(n) = [x^(n-1)] 4*A(x)^(n+3)/(n+3) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 4*A(x)^(n+3)/(n+3) for n>=1 (see comment).

A383563 G.f. A(x) satisfies A( x*(1+x)/A(x)^2 ) = 1 + x.

Original entry on oeis.org

1, 1, 1, 3, 13, 72, 465, 3362, 26531, 224856, 2024188, 19202830, 190857879, 1978567663, 21319434418, 238109360460, 2750229390071, 32789591062124, 402891169846242, 5094855923807780, 66229610059651788, 884081025776797026, 12107164229698851942, 169954380180177899277, 2443554376412586234247
Offset: 0

Views

Author

Paul D. Hanna, May 26 2025

Keywords

Examples

			G.f.: A(x) = 1 + x + x^2 + 3*x^3 + 13*x^4 + 72*x^5 + 465*x^6 + 3362*x^7 + 26531*x^8 + 224856*x^9 + 2024188*x^10 + ...
 where A( x*(1+x)/A(x)^2 ) = 1 + x.
RELATED SERIES.
The g.f. of A145345 begins
B(x) = 1 + x + 2*x^2 + 7*x^3 + 34*x^4 + 203*x^5 + 1398*x^6 + 10706*x^7 + 89120*x^8 + 794347*x^9 + ...
 where B(x/B(x)) = 1 + x*B(x)
 also, B( x/A(x)^2 ) = 1 + x.
The g.f. of A121687 begins
C(x) = 1 + x + 3*x^2 + 14*x^3 + 83*x^4 + 574*x^5 + 4432*x^6 + 37244*x^7 + 335153*x^8 + 3194510*x^9 + ...
 where C(x) = 1/(1 - x*C(x*C(x))^2)
 also, C( (x/(1+x))/A(x)^2 ) = 1 + x.
C(x)^2 = 1 + 2*x + 7*x^2 + 34*x^3 + 203*x^4 + 1398*x^5 + 10706*x^6 + 89120*x^7 + 794347*x^8 + ...
 where B(x) = 1 + x*C(x)^2.

Links

Crossrefs

Cf. A384265, A145345 (B(x)), A121687 (C(x)).

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
A[#A] = polcoef(x - serreverse(x/Ser(A)) + x*serreverse(x/Ser(A)^2),#A) ); A[n+1]}
for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n along with B(x) = g.f. of A145345 and C(x) = g.f. of A121687 satisfies the following formulas.
(1) A( x*(1+x)/A(x)^2 ) = 1 + x.
(2.a) Series_Reversion( x/A(x) ) = x + x*Series_Reversion( x/A(x)^2 ).
(2.b) [x^n] A(x)^(n+1)/(n+1) = [x^(n-1)] A(x)^(2*n)/n for n >= 1.
(2.c) B(x) = 1 + x*C(x)^2.
(3.a) A(x) = B(x/A(x)) where B(x) = A(x*B(x)) = C(x/B(x)).
(3.b) A(x) = C(x/A(x)^2) where C(x) = A(x*C(x)^2) = B(x*C(x)).
(4.a) A(x) = A(x)^2 - x*C(x/A(x))^2.
(4.b) B( x/A(x)^2 ) = 1 + x.
(4.c) C( (x/(1+x))/A(x)^2 ) = 1 + x.
(4.d) B( (x/(1+x))/A(x)^2 ) = 1 + x*(1+x)/A(x)^2.
(4.e) A( (x/(1+x))/A(x)^2 + x^2/A(x)^4 ) = 1 + x*(1+x)/A(x)^2.
(5.a) A145345(n) = [x^n] B(x) = [x^n] A(x)^(n+1)/(n+1) for n >= 0.
(5.b) A121687(n) = [x^n] C(x) = [x^n] A(x)^(2*n+1)/(2*n+1) for n >= 0.
(5.c) A145345(n) = [x^(n-1)] C(x)^2 = [x^(n-1)] A(x)^(2*n)/n for n >= 1.
Showing 1-3 of 3 results.

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