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 1-4 of 4 results.

A361763 Expansion of g.f. A(x) satisfying A(x)^3 = A( x^3/(1 - 3*x)^3 ).

Original entry on oeis.org

1, 3, 9, 28, 93, 333, 1271, 5064, 20673, 85460, 355659, 1486719, 6238608, 26278281, 111114558, 471608944, 2008906581, 8586410085, 36816550550, 158332335279, 682843960665, 2952865525730, 12802463157570, 55646477022330, 242465061290160, 1059022767175173, 4636452916770489
Offset: 1

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Author

Paul D. Hanna, Mar 23 2023

Keywords

Comments

Related Catalan identity: F(x)^2 = F( x^2/(1 - 2*x)^2 ), where F(x) = x*C(x)^2 and C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
Radius of convergence of g.f. A(x) is r where r is the real root of r = (1 - 3*r)^(3/2) with A(r) = 1 and r = (52 - (324*sqrt(717) + 8108)^(1/3) + (324*sqrt(717) - 8108)^(1/3))/162 = 0.214054846272632706742...

Examples

			G.f.: A(x) = x + 3*x^2 + 9*x^3 + 28*x^4 + 93*x^5 + 333*x^6 + 1271*x^7 + 5064*x^8 + 20673*x^9 + 85460*x^10 + 355659*x^11 + 1486719*x^12 + ...
where
A( x^3/(1 - 3*x)^3 ) = x^3 + 9*x^4 + 54*x^5 + 273*x^6 + 1269*x^7 + 5670*x^8 + 24957*x^9 + 109593*x^10 + 482598*x^11 + 2133082*x^12 + ...
which equals A(x)^3.
RELATED SERIES.
Notice that the following cube root is an integer series
( A(x)/x )^(1/3) = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 197*x^6 + 779*x^7 + 3135*x^8 + 12709*x^9 + 51757*x^10 + ... + A361762(n)*x^n + ...
Also, let B(x) satisfy A(x/B(x)) = x and B(A(x)) = A(x)/x,
then B(x) = x/Series_Reversion(A(x)) is the g.f. of A107092,
B(x) = 1 + 3*x + x^3 - x^6 + 2*x^9 - 4*x^12 + 9*x^15 - 22*x^18 + 55*x^21 - 142*x^24 + 376*x^27 - 1011*x^30 + ...
such that B(x)^3 = B(x^3) + 3*x,
as shown by the series
B(x)^(1/3) = 1 + x - x^2 + 2*x^3 - 4*x^4 + 9*x^5 - 22*x^6 + 55*x^7 - 142*x^8 + 376*x^9 - 1011*x^10 + ...
SPECIFIC VALUES.
A(1/5) = A(1/8)^(1/3) = 0.586384210523490911367880492498...
A(1/5) = (1/5) * (1 - 3/5)^(-1) * (1 - 3/8)^(-1/3) * (1 - 3/125)^(-1/9) * (1 - 3/1815848)^(-1/27) * ...
A(1/6) = A(1/27)^(1/3) = 0.346688997573685318336777346240...
A(1/6) = (1/6) * (1 - 3/6)^(-1) * (1 - 3/27)^(-1/3) * (1 - 3/13824)^(-1/9) * (1 - 3/2640087986661)^(-1/27) * ...
A(1/9) = A(1/216)^(1/3) = 0.16744549995321182031691216552466...
A(1/12) = A(1/729)^(1/3) = 0.11126394649161862248626102306202...

Links

Crossrefs

Cf. A361762 ((A(x)/x)^(1/3)), A264230, A107092, A091190, A361765.
Cf. A264228, A264229, A264230.

Programs

  • PARI
    {a(n) = my(A=x); for(i=1, #binary(n+1), A = ( subst(A, x, x^3/(1 - 3*x +x*O(x^n))^3 ) )^(1/3) ); polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies:
(1) A(x)^3 = A( x^3/(1 - 3*x)^3 ).
(2) A(x^3) = A( x/(1 + 3*x) )^3.
(3) A(x) = x * Product_{n>=0} 1/(1 - 3/F(n,x))^(1/3^n), where F(0,x) = 1/x, F(m,x) = (F(m-1,x) - 3)^3 for m > 0.
(4) x/Series_Reversion(A(x)) = B(x) such that B(x)^3 = B(x^3) + 3*x (cf. A107092).

A091200 G.f. A(x) satisfies xA(x)^5 = B(xA(x^5)) where B(x) = x/(1-5x).

Original entry on oeis.org

1, 1, 3, 11, 44, 185, 802, 3553, 15994, 72886, 335387, 1555487, 7261310, 34083382, 160730900, 761039051, 3616102911, 17235223345, 82372594183, 394648349447, 1894921311499, 9116598414141, 43939539520427, 212124129983285
Offset: 0

Views

Author

Paul D. Hanna, Feb 23 2004

Keywords

Comments

More generally, given A(x) satisfies xA(x)^p = B(xA(x^p)) where B(x) = x/(1-p*x), then it appears that A(x) is an integer series only when p is prime. This is a special case of sequences with g.f.s that satisfy the more general functional equation xA(x)^m = B(xA(x^m)) originated by Michael Somos; some other examples are A085748, A091188 and A091190.

Crossrefs

Cf. A085748, A091188, A091190.

Programs

  • PARI
    {a(n)=local(A,m); p=5;if(n<0,0,m=1; A=1+O(x); while(m<=n,m*=p; A=x*subst(A,x,x^p); A=(A/(1-p*A)/x)^(1/p));polcoeff(A,n))}

A091188 G.f. A(x) satisfies both A(-x)*A(x) = A(x^2) and xA(x)^2 = B(xA(x^2)) where B(x) = x*(1+x)/(1-x).

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 5, 10, 12, 23, 31, 58, 79, 145, 207, 374, 540, 964, 1427, 2522, 3775, 6626, 10050, 17532, 26811, 46561, 71795, 124188, 192661, 332228, 518303, 891340, 1396902, 2396912, 3771822, 6459202, 10199912, 17437727, 27622807, 47152952
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2004

Keywords

Comments

This is a special case of sequences with g.f.s that satisfy the more general functional equation xA(x)^m = B(xA(x^m)) originated by Michael Somos; some other examples are A085748, A091190 and A091200.

Examples

			1 + x + x^2 + 2*x^3 + 2*x^4 + 4*x^5 + 5*x^6 + 10*x^7 + 12*x^8 + 23*x^9 + ...
q + q^3 + q^5 + 2*q^7 + 2*q^9 + 4*q^11 + 5*q^13 + 10*q^15 + 12*q^17 + ...

Crossrefs

Cf. A085748, A091190, A091200, A092869.

Programs

  • PARI
    {a(n) = local(A, m); if( n<0, 0, m=1; A = 1 + O(x); while( m<=n, m*=2; A = x * subst(A, x, x^2); A = (A *(1 + A) /(1 - A) / x)^(1/2)); polcoeff(A, n))}

Formula

Given g.f. A(x), then B(x) = x * A(x^2) satisfies 0 = f(B(x), B(x^2)) were f(u, v) = u^2 * (1 - v) - v * (1 + v). - Michael Somos, Aug 02 2011

A370441 Expansion of g.f. A(x) satisfying A(x) = A( x^3 + 3*A(x)^4 )^(1/3), with A(0)=0, A'(0)=1.

Original entry on oeis.org

1, 1, 3, 12, 54, 261, 1324, 6952, 37461, 205977, 1151034, 6518085, 37321748, 215714904, 1256889150, 7374790400, 43537323406, 258417908640, 1541250594499, 9231988699115, 55514033703450, 334993491267955, 2027954403410504, 12312557796833622, 74955173794196890, 457431093085335708
Offset: 1

Views

Author

Paul D. Hanna, Mar 09 2024

Keywords

Comments

Compare the g.f. to the following identities:
(1) C(x) = C( x^2 + 2*x*C(x)^2 )^(1/2),
(2) C(x) = C( x^3 + 3*x*C(x)^3 )^(1/3),
where C(x) = x + C(x)^2 is a g.f. of the Catalan numbers (A000108).

Examples

			G.f.: A(x) = x + x^2 + 3*x^3 + 12*x^4 + 54*x^5 + 261*x^6 + 1324*x^7 + 6952*x^8 + 37461*x^9 + 205977*x^10 + 1151034*x^11 + 6518085*x^12 + ...
where A(x)^3 = A( x^3 + 3*A(x)^4 ).
RELATED SERIES.
A(x)^3 = x^3 + 3*x^4 + 12*x^5 + 55*x^6 + 270*x^7 + 1386*x^8 + 7347*x^9 + 39897*x^10 + 220779*x^11 + 1240392*x^12 + ...
A(x)^4 = x^4 + 4*x^5 + 18*x^6 + 88*x^7 + 451*x^8 + 2388*x^9 + 12958*x^10 + 71668*x^11 + 402489*x^12 + ...
Let B(x) denote the series reversion of A(x), A(B(x)) = x, where
B(x) = x - x^2 - x^3 - 2*x^4 - 4*x^5 - 9*x^6 - 22*x^7 - 55*x^8 - 142*x^9 - 376*x^10 - 1011*x^11 - 2758*x^12 + ... + (-1)^(n+1)*A107092(n)*x^n + ...
then B(x)^3 = B(x^3) - 3*x^4, where
B(x)^3 = x^3 - 3*x^4 - x^6 - x^9 - 2*x^12 - 4*x^15 - 9*x^18 - 22*x^21 - 55*x^24 - 142*x^27 - 376*x^30 - 1011*x^33 - 2758*x^36 + ...
Also, we have D(x) = x/B(x) is the g.f. of A091190, which begins
D(x) = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 35*x^5 + 97*x^6 + 273*x^7 + 778*x^8 + 2240*x^9 + 6499*x^10 + 18976*x^11 + ... + A091190(n)*x^n + ...
such that D(x)^3 = D(x^3)/(1 - 3*x*D(x^3)).

Crossrefs

Cf. A107092, A091190, A370440.

Programs

  • PARI
    {a(n) = my(A=[1],G); for(i=1,n, G = x*Ser(A); A = Vec((subst(G,x, x^3 + 3*x^2*G^2) + x^4*O(x^#A))^(1/3)); );A[n+1]}
for(n=0,40, print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n) * x^n satisfies:
(1) A(x) = A( x^3 + 3*A(x)^4 )^(1/3).
(2) B(x)^3 = B(x^3) - 3*x^4, where A(B(x)) = x.
(3) A(x) = x*D(A(x)) where D(x) = x/Series_Reversion(A(x)) is the g.f. of A091190.
Showing 1-4 of 4 results.

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