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.

A264228 G.f. A(x) satisfies: A(x)^3 = A( x^3/(1-3*x) ), with A(0) = 0.

Original entry on oeis.org

1, 1, 2, 5, 13, 35, 97, 274, 785, 2275, 6656, 19630, 58295, 174175, 523238, 1579584, 4789919, 14584723, 44577799, 136732988, 420784888, 1298937282, 4021383654, 12483820395, 38853994422, 121220646116, 379062880051, 1187912517953, 3730305167438, 11736596024002, 36994041916973, 116807229667919, 369415244627269, 1170113816365089
Offset: 1

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Author

Paul D. Hanna, Nov 08 2015

Keywords

Comments

Radius of convergence is r = (sqrt(13) - 3)/2, where r = r^3/(1-3*r), with A(r) = 1.
Compare to a g.f. M(x) of Motzkin numbers: M(x)^2 = M(x^2/(1-2*x)) where M(x) = (1-x - sqrt(1-2*x-3*x^2))/(2*x).

Examples

			G.f.: A(x) = x + x^2 + 2*x^3 + 5*x^4 + 13*x^5 + 35*x^6 + 97*x^7 + 274*x^8 + 785*x^9 + 2275*x^10 + 6656*x^11 + 19630*x^12 + 58295*x^13 + 174175*x^14 + ...
where A(x)^3 = A( x^3/(1-3*x) ).
RELATED SERIES.
A(x)^3 = x^3 + 3*x^4 + 9*x^5 + 28*x^6 + 87*x^7 + 270*x^8 + 839*x^9 + 2610*x^10 + 8127*x^11 + 25331*x^12 + 79035*x^13 + 246852*x^14 + 771808*x^15 + ...
A( x/(1+x+x^2) ) = x + x^4 + 2*x^7 + 6*x^10 + 22*x^13 + 88*x^16 + 367*x^19 + 1570*x^22 + 6843*x^25 + 30271*x^28 + 135530*x^31 + 612852*x^34 + 2794412*x^37 + 12832472*x^40 + ...
Let B(x) = x/Series_Reversion(A(x)), then A(x) = x*B(A(x)), where
B(x) = 1 + x + x^2 + x^3 - x^5 - x^6 + 2*x^8 + 3*x^9 - 6*x^11 - 9*x^12 + 20*x^14 + 30*x^15 - 71*x^17 - 110*x^18 + 267*x^20 + 419*x^21 - 1041*x^23 + ...
Let C0(x) and C2(x) be series trisections of B(x), B(x) = C0(x) + x + C2(x):
C0(x) = 1 + x^3 - x^6 + 3*x^9 - 9*x^12 + 30*x^15 - 110*x^18 + 419*x^21 - 1648*x^24 + 6652*x^27 - 27369*x^30 + 114384*x^33 - 484276*x^36 + ...
C2(x) = x^2 - x^5 + 2*x^8 - 6*x^11 + 20*x^14 - 71*x^17 + 267*x^20 - 1041*x^23 + 4168*x^26 - 17047*x^29 + 70902*x^32 + ... + (-1)^(n-1)*A370446(n)*x^(3*n-1) + ...
then C0(x) = x^2/C2(x).
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=x); for(i=1, n, A = ( subst(A, x, x^3/(1-3*x +x*O(x^n))) )^(1/3) ); polcoeff(A, n)}
    for(n=1, 40, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following.
(1) A(x)^3 = A( x^3/(1-3*x) ).
(2) A( x/(1+3*x) )^3 = A( x^3/(1+3*x)^2 ). - Paul D. Hanna, Mar 25 2023
(3) A( x/(1+x+x^2) )^3 = A( x^3/(1-x^3)^2 ). - Paul D. Hanna, Mar 11 2024

A370446 Expansion of g.f. A(x) satisfying A(x)^3 + x^4/A(x)^3 = A(x^3) + x^4/A(x^3) - 3*x^2.

Original entry on oeis.org

1, 1, 2, 6, 20, 71, 267, 1041, 4168, 17047, 70902, 298967, 1275141, 5491504, 23846271, 104295430, 459023543, 2031459236, 9034769573, 40358643042, 180998556943, 814645632727, 3678542796070, 16659932961647, 75657738747396, 344446195875766, 1571786529601990, 7187790264787872
Offset: 1

Views

Author

Paul D. Hanna, Mar 09 2024

Keywords

Examples

			G.f.: A(x) = x + x^2 + 2*x^3 + 6*x^4 + 20*x^5 + 71*x^6 + 267*x^7 + 1041*x^8 + 4168*x^9 + 17047*x^10 + 70902*x^11 + 298967*x^12 + 1275141*x^13 + 5491504*x^14 + 23846271*x^15 + ...
RELATED SERIES.
We can illustrate the formulas with the following related expansions.
(1) A(x)^3 + 2*x^2 + x^4/A(x)^3 = x - x^2 + x^3 - x^4 + x^6 - x^7 + 2*x^9 - 3*x^10 + 6*x^12 - 9*x^13 + 20*x^15 - 30*x^16 + 71*x^18 - 110*x^19 + 267*x^21 + ...
which equals A(x^3) - x^2 + x^4/A(x^3), as can be seen from
x^4/A(x^3) = x - x^4 - x^7 - 3*x^10 - 9*x^13 - 30*x^16 - 110*x^19 + ...
A(x)^3 = x^3 + 3*x^4 + 9*x^5 + 31*x^6 + 114*x^7 + 435*x^8 + 1715*x^9 + ...
x^4/A(x)^3 = x - 3*x^2 - 4*x^4 - 9*x^5 - 30*x^6 - 115*x^7 - 435*x^8 - 1713*x^9 + ...
(2) Let F(x) be the g.f. of A370440, which begins
F(x) = x + x^2 + x^3 + x^4 + 2*x^5 + 6*x^6 + 15*x^7 + 30*x^8 + 55*x^9 + 113*x^10 + 274*x^11 + 683*x^12 + ...
where F(x)^3 = F( x^3 + 3*x^2*F(x)^2 ),
then the series reversion of F(x) begins
A(x^3) - x^2 + x^4/A(x^3) = x - x^2 + x^3 - x^4 + x^6 - x^7 + 2*x^9 - 3*x^10 + 6*x^12 - 9*x^13 + 20*x^15 - 30*x^16 + 71*x^18 - 110*x^19 + 267*x^21 - 419*x^22 + 1041*x^24 + ...
(3) Let G(x) be the g.f. of A264228, which begins
G(x) = x + x^2 + 2*x^3 + 5*x^4 + 13*x^5 + 35*x^6 + 97*x^7 + 274*x^8 + 785*x^9 + 2275*x^10 + 6656*x^11 + 19630*x^12 +...
where G(x)^3 = G( x^3/(1 - 3*x) ),
then the series reversion of G(x) begins
-x^2/(A(-x^3) - x^2 + x^4/A(-x^3)) = x^2/(x + x^2 + x^3 + x^4 - x^6 - x^7 + 2*x^9 + 3*x^10 - 6*x^12 - 9*x^13 + 20*x^15 + 30*x^16 - 71*x^18 - 110*x^19 + 267*x^21 + 419*x^22 - 1041*x^24 +...).
SPECIFIC VALUES.
A(1/4.834464) = 0.349644497578571280258023712232522068793519739...
A(1/5) = 0.29940801195429552263938628184744484915469836164855...
A(1/6) = 0.21539123666426270273178791857213676628593723946879...
A(1/7) = 0.17414937372444126736977770687571455113383911571251...
A(1/8) = 0.14713126344900776621336355426627444003268957268553...
A(1/5^3) = 0.00806504925055020701973761348380106375185943151538...
A(1/6^3) = 0.00465126435780731657600811126033650347236250831668...
A(1/7^3) = 0.00292400175440295890949208907819991271975334925594...
which may be used to verify that the formula
A(x)^3 + x^4/A(x)^3 = A(x^3) + x^4/A(x^3) - 3*x^2
holds for these specific values.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=x); for(m=1,n, A=truncate(A) +x^4*O(x^m); A = ( x^4/(x^4/subst(A,x,x^3) + subst(A,x,x^3) - A^3 - 3*x^2) +x^4*O(x^n))^(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 the following formulas.
(1) A(x)^3 + 2*x^2 + x^4/A(x)^3 = A(x^3) - x^2 + x^4/A(x^3).
(2) F( A(x^3) - x^2 + x^4/A(x^3) ) = x, where F(x) = F( x^3 + 3*x^2*F(x)^2 )^(1/3) is the g.f. of A370440.
(3) G( -x^2/(A(-x^3) - x^2 + x^4/A(-x^3)) ) = x, where G(x) = G( x^3/(1 - 3*x) )^(1/3) is the g.f. of A264228.
a(n) ~ c * d^n / n^(3/2), where d = 4.8344630246454026903035642546835542141482126303313357979263... and c = 0.0713578385738499677445741870058758452888939567284935382... - Vaclav Kotesovec, Mar 13 2024
The radius of convergence r = 0.20684820525095397... = 1/d (where d is given above), and A(r) = 0.3497581458819115559285308998459940399916633464611700768... satisfy A(r) = r^(2/3) and A(r^3) = (5 - sqrt(21))/2 * r^2. - Paul D. Hanna, Mar 13 2024

A371709 Expansion of g.f. A(x) satisfying A( x*A(x)^2 + x*A(x)^3 ) = A(x)^3.

Original entry on oeis.org

1, 1, 1, 2, 6, 16, 39, 99, 271, 764, 2157, 6128, 17658, 51534, 151635, 448962, 1337493, 4008040, 12072594, 36524898, 110943633, 338218626, 1034509917, 3173811240, 9763898994, 30113782641, 93094164244, 288415278638, 895332445053, 2784580242557, 8675408291598, 27072326322939
Offset: 1

Views

Author

Paul D. Hanna, May 02 2024

Keywords

Comments

Compare to the following identities of the Catalan function C(x) = x + C(x)^2 (A000108):
(1) C(x)^2 = C( x*C(x)*(1 + C(x)) ),
(2) C(x)^4 = C( x*C(x)^3*(1 + C(x))*(1 + C(x)^2) ),
(3) C(x)^8 = C( x*C(x)^7*(1 + C(x))*(1 + C(x)^2)*(1 + C(x)^4) ),
(4) C(x)^(2^n) = C( x*C(x)^(2^n-1)*Product_{k=0..n-1} (1 + C(x)^(2^k)) ) for n > 0.
a(3^n) == 1 (mod 3) for n >= 0.
a(2*3^n) == 1 (mod 3) for n >= 0.
a(n) == 2 (mod 3) iff n is the sum of 2 distinct powers of 3 (A038464).

Examples

			G.f. A(x) = x + x^2 + x^3 + 2*x^4 + 6*x^5 + 16*x^6 + 39*x^7 + 99*x^8 + 271*x^9 + 764*x^10 + 2157*x^11 + 6128*x^12 + 17658*x^13 + 51534*x^14 + 151635*x^15 + 448962*x^16 + ...
where A( x*A(x)^2*(1 + A(x)) ) = A(x)^3.
RELATED SERIES.
A(x)^2 = x^2 + 2*x^3 + 3*x^4 + 6*x^5 + 17*x^6 + 48*x^7 + 126*x^8 + 332*x^9 + 918*x^10 + 2616*x^11 + 7504*x^12 + ...
A(x)^3 = x^3 + 3*x^4 + 6*x^5 + 13*x^6 + 36*x^7 + 105*x^8 + 292*x^9 + 801*x^10 + 2256*x^11 + 6515*x^12 + 18981*x^13 + ...
A(x)^2 + A(x)^3 = x^2 + 3*x^3 + 6*x^4 + 12*x^5 + 30*x^6 + 84*x^7 + 231*x^8 + 624*x^9 + 1719*x^10 + 4872*x^11 + 14019*x^12 + 40599*x^13 + ...
Let B(x) be the series reversion of g.f. A(x), B(A(x)) = x, then
B(x) * (1+x)/(1+x^3) = x - 2*x^4 + 3*x^7 - 5*x^10 + 7*x^13 - 9*x^16 + 12*x^19 - 15*x^22 + 18*x^25 - 23*x^28 + ... + (-1)^n*A005704(n)*x^(3*n+1) + ...
where A005704 is the number of partitions of 3*n into powers of 3.
We can show that g.f. A(x) = A( x*A(x)^2*(1 + A(x)) )^(1/3) satisfies
(4) A(x) = x * Product_{n>=0} (1 + A(x)^(3^n))
by substituting x*A(x)^2*(1 + A(x)) for x in (4) to obtain
A(x)^3 = x * A(x)^2*(1 + A(x)) * Product_{n>=1} (1 + A(x)^(3^n))
which is equivalent to formula (4).
SPECIFIC VALUES.
A(3/10) = 0.526165645044542830201162330432965674027415264612114520...
A(1/4) = 0.353259384374080248921564026412797625837830114153200664...
A(1/5) = 0.255218141344695821239609680309162895225297482063273545...
A(t) = 1/2 and A(t*3/8) = 1/8 at t = (1/2)/Product_{n>=0} (1 + 1/2^(3^n)) = 0.295718718466711580562679377308518930409875701753934072...
A(t) = 1/3 and A(t*4/27) = 1/27 at t = (1/3)/Product_{n>=0} (1 + 1/3^(3^n)) = 0.241059181496179959557718992589733756750585121455883861...
A(t) = 1/4 and A(t*5/64) = 1/64 at t = (1/4)/Product_{n>=0} (1 + 1/4^(3^n)) = 0.196922325724019432212969925740117827612003158137366017...
		

Crossrefs

Programs

  • PARI
    /* Using series reversion of x/Product_{n>=0} (1 + x^(3^n)) */
    {a(n) = my(A); A = serreverse( x/prod(k=0,ceil(log(n)/log(3)), (1 + x^(3^k) +x*O(x^n)) ) ); polcoeff(A,n)}
    for(n=1,35, print1(a(n),", "))
    
  • PARI
    /* Using A(x)^3 = A( x*A(x)^2 + x*A(x)^3 ) */
    {a(n) = my(A=[1],F); for(i=1,n, A = concat(A,0); F = x*Ser(A);
    A[#A] = polcoeff( subst(F,x, x*F^2 + x*F^3 ) - F^3, #A+2) ); A[n]}
    for(n=1,35, print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.
(1) A(x)^3 = A( x*A(x)^2*(1 + A(x)) ).
(2) A(x)^9 = A( x*A(x)^8*(1 + A(x))*(1 + A(x)^3) ).
(3) A(x)^27 = A( x*A(x)^26*(1 + A(x))*(1 + A(x)^3)*(1 + A(x)^9) ).
(4) A(x) = x * Product_{n>=0} (1 + A(x)^(3^n)).
(5) A(x) = Series_Reversion( x / Product_{n>=0} (1 + x^(3^n)) ).
a(n) ~ c * d^n / n^(3/2), where d = 3.2753449994351908157330968510747739... and c = 0.1559869008021616116037651076359... - Vaclav Kotesovec, May 03 2024
The radius of convergence r of g.f. A(x) and A(r) satisfy 1 = Sum_{n>=0} 3^n * A(r)^(3^n) / (1 + A(r)^(3^n)) and r = A(r) / Product_{n>=0} (1 + A(r)^(3^n)), where r = 0.30531134893345362211... = 1/d (d is given above) and A(r) = 0.600582105427215700175254768411726892599... - Paul D. Hanna, May 03 2024

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

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.