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.

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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...

Links

Crossrefs

Cf. A371716, A370440, A370441, A370446, A264228, A198951.
Cf. A005704, A038464.

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

A369443 Expansion of (1/x) * Series_Reversion( x / ((1+x)^2 * (1+x^3)) ).

Original entry on oeis.org

1, 2, 5, 15, 52, 198, 796, 3306, 14042, 60698, 266235, 1182315, 5306085, 24028162, 109654887, 503797703, 2328343326, 10816971516, 50487762906, 236635814984, 1113297830297, 5255647026534, 24888156618738, 118194065746758, 562773777767295, 2686074452484012
Offset: 0

Views

Author

Seiichi Manyama, Jan 23 2024

Keywords

Links

Crossrefs

Cf. A198951.

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)^2*(1+x^3)))/x)
  • PARI
    a(n, s=3, t=1, u=2) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial(u*(n+1), n-s*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(n+1,k) * binomial(2*n+2,n-3*k).

A211248 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^4).

Original entry on oeis.org

1, 1, 4, 20, 114, 703, 4565, 30752, 212921, 1505916, 10833164, 79018804, 583062388, 4344431508, 32641910199, 247033970128, 1881402836376, 14408753414558, 110897147057354, 857307054338476, 6653979156676983, 51831065993122915, 405060413133136902, 3175019470333290488
Offset: 0

Views

Author

Paul D. Hanna, Apr 05 2012

Keywords

Comments

More generally, for fixed parameters p and q, if F(x) satisfies:
F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ),
then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)); here p=2 and q=1.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 20*x^3 + 114*x^4 + 703*x^5 + 4565*x^6 +...
where A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 85*x^3 + 522*x^4 + 3381*x^5 + 22735*x^6 +...
A(x)^4 = 1 + 4*x + 22*x^2 + 132*x^3 + 841*x^4 + 5588*x^5 + 38288*x^6 +...
A(x)^7 = 1 + 7*x + 49*x^2 + 343*x^3 + 2429*x^4 + 17430*x^5 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^4 + x^3*A(x)^7.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x*A(x))*x*A(x)^2 + (1 + 2^2*x*A(x) + x^2*A(x)^2)*x^2*A(x)^4/2 +
(1 + 3^2*x*A(x) + 3^2*x^2*A(x)^2 + x^3*A(x)^3)*x^3*A(x)^6/3 +
(1 + 4^2*x*A(x) + 6^2*x^2*A(x)^2 + 4^2*x^3*A(x)^3 + x^4*A(x)^4)*x^4*A(x)^8/4 +
(1 + 5^2*x*A(x) + 10^2*x^2*A(x)^2 + 10^2*x^3*A(x)^3 + 5^2*x^4*A(x)^4 + x^5*A(x)^5)*x^5*A(x)^10/5 +
(1 + 6^2*x*A(x) + 15^2*x^2*A(x)^2 + 20^2*x^3*A(x)^3 + 15^2*x^4*A(x)^4 + 6^2*x^5*A(x)^5 + x^6*A(x)^6)*x^6*A(x)^12/6 +...
more explicitly,
log(A(x)) = x + 7*x^2/2 + 49*x^3/3 + 359*x^4/4 + 2706*x^5/5 + 20767*x^6/6 +...

Crossrefs

Cf. A200716, A200717, A199876, A211249, A200731, A199874, A199877.
Cf. A198951, A198953, A198957, A192415, A198888, A036765.

Programs

  • Mathematica
    CoefficientList[Sqrt[1/x * InverseSeries[Series[x*(1 - x - x^3)^2/(1 + x^2)^2, {x, 0, 20}], x]], x] (* Vaclav Kotesovec, Nov 22 2017 *)
  • PARI
    {a(n)=polcoeff(sqrt( (1/x)*serreverse( x*(1-x-x^3)^2/(1+x^2+x*O(x^n))^2 ) ), n)}
for(n=0,30,print1(a(n),", "))
  • PARI
    {a(n)=local(p=2, q=1, A=1+x); for(i=1, n, A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n)); polcoeff(A, n)}
  • PARI
    {a(n)=local(p=2, q=1, A=1+x); for(i=1, n, A=exp(sum(m=1, n, x^m*(A+x*O(x^n))^(p*m)/m*sum(j=0, m, binomial(m, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}
  • PARI
    {a(n)=local(p=2, q=1, A=1+x); for(i=1, n, A=exp(sum(m=1, n, x^m*(A+x*O(x^n))^(p*m)/m*(1-x*A^q)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}

Formula

G.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion( x*(1-x-x^3)^2/(1+x^2)^2 ) ).
(2) A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
(3) a(n) = [x^n] ((1+x^2)/(1-x-x^3))^(2*n+2) / (n+1).
(4) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^k) * x^n*A(x)^(2*n)/n ).
(5) A(x) = exp( Sum_{n>=1} (1-x*A(x))^(2*n+1) * (Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^k) * x^n*A(x)^(2*n)/n ).
(6) A(x) = (1/x)*Series_Reversion(x/G(x)) where A(x) = G(x*A(x)) and A(x/G(x)) = G(x) = (1 + x*G(x)^2)*(1 + x^2*G(x)^2) is the g.f. of A199874.
a(n) ~ s * sqrt((1 + 2*r*s + 3*r^2*s^4) / (3*Pi*(1 + 2*r*s + 7*r^2*s^4))) / (2*n^(3/2)*r^n), where r = 0.1194948955213353102456218138370139612914667337222... and s = 1.428770161302757679335810379290625953730830139744... are real roots of the system of equations (1 + r*s^3)*(1 + r^2*s^4) = s, r*s^2*(3 + 4*r*s + 7*r^2*s^4) = 1. - Vaclav Kotesovec, Nov 22 2017

A211249 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^5).

Original entry on oeis.org

1, 1, 4, 21, 126, 819, 5611, 39900, 291719, 2179181, 16560175, 127617168, 994951887, 7833555324, 62196300997, 497425570173, 4003607595960, 32404662671330, 263586896132154, 2153631763231319, 17666722629907960, 145449082369322208, 1201414340736684702
Offset: 0

Views

Author

Paul D. Hanna, Apr 05 2012

Keywords

Comments

More generally, for fixed parameters p and q, if F(x) satisfies:
F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ),
then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)); here p=2 and q=2.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 21*x^3 + 126*x^4 + 819*x^5 + 5611*x^6 +...
where A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 88*x^3 + 564*x^4 + 3828*x^5 + 27040*x^6 +...
A(x)^5 = 1 + 5*x + 30*x^2 + 195*x^3 + 1335*x^4 + 9486*x^5 + 69305*x^6 +...
A(x)^8 = 1 + 8*x + 60*x^2 + 448*x^3 + 3374*x^4 + 25704*x^5 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^5 + x^3*A(x)^8.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x*A(x)^2)*x*A(x)^2 +
(1 + 2^2*x*A(x)^2 + x^2*A(x)^4)*x^2*A(x)^4/2 +
(1 + 3^2*x*A(x)^2 + 3^2*x^2*A(x)^4 + x^3*A(x)^6)*x^3*A(x)^6/3 +
(1 + 4^2*x*A(x)^2 + 6^2*x^2*A(x)^4 + 4^2*x^3*A(x)^6 + x^4*A(x)^8)*x^4*A(x)^8/4 +
(1 + 5^2*x*A(x)^2 + 10^2*x^2*A(x)^4 + 10^2*x^3*A(x)^6 + 5^2*x^4*A(x)^8 + x^5*A(x)^10)*x^5*A(x)^10/5 +
(1 + 6^2*x*A(x)^2 + 15^2*x^2*A(x)^4 + 20^2*x^3*A(x)^6 + 15^2*x^4*A(x)^8 + 6^2*x^5*A(x)^10 + x^6*A(x)^12)*x^6*A(x)^12/6 +...
more explicitly,
log(A(x)) = x + 7*x^2/2 + 52*x^3/3 + 403*x^4/4 + 3211*x^5/5 + 26050*x^6/6 +...

Crossrefs

Cf. A200716, A200717, A199876, A211248, A200731, A200075, A199877.
Cf. A199874, A198951, A198953, A198957, A192415, A198888, A036765.

Programs

  • Mathematica
    CoefficientList[Sqrt[1/x * InverseSeries[Series[x*(1-2*x-x^2+x^4 + (1-x-x^2) * Sqrt[(1+x+x^2)*(1-3*x+x^2)])/2, {x, 0, 20}], x]], x] (* Vaclav Kotesovec, Nov 22 2017 *)
  • PARI
    {a(n)=polcoeff(sqrt( (1/x)*serreverse( x*(1-2*x-x^2+x^4 + (1-x-x^2)*sqrt( (1+x+x^2)*(1-3*x+x^2) +x*O(x^n)))/2 ) ), n)}
for(n=0,30,print1(a(n),", "))
  • PARI
    {a(n)=local(p=2, q=2, A=1+x); for(i=1, n, A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n)); polcoeff(A, n)}
  • PARI
    {a(n)=local(p=2, q=2, A=1+x); for(i=1, n, A=exp(sum(m=1, n, x^m*(A+x*O(x^n))^(p*m)/m*sum(j=0, m, binomial(m, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}
  • PARI
    {a(n)=local(p=2, q=2, A=1+x); for(i=1, n, A=exp(sum(m=1, n, x^m*(A+x*O(x^n))^(p*m)/m*(1-x*A^q)^(2*m+1)*sum(j=0, n, binomial(m+j,j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}

Formula

G.f.: sqrt( (1/x)*Series_Reversion( x*(1-2*x-x^2+x^4 + (1-x-x^2)*sqrt( (1+x+x^2)*(1-3*x+x^2) ))/2 ) ).
G.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)) where A(x) = G(x*A(x)) and A(x/G(x)) = G(x) is the g.f. of A200075.
(2) A(x) = sqrt( (1/x)*Series_Reversion( x/G(x)^2 ) ) where A(x) = G(x*A(x)^2) and G(x) = A(x/G(x)^2) and 1+x*G(x) is the g.f. of A004148.
(3) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^(2*k)) * x^n*A(x)^(2*n)/n ).
(4) A(x) = exp( Sum_{n>=1} (1-x*A(x)^2)^(2*n+1) * (Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^(2*k)) * x^n*A(x)^(2*n)/n ).
a(n) ~ s * sqrt((1 + 2*r*s^2 + 3*r^2*s^5) / (Pi*(3 + 10*r*s^2 + 28*r^2*s^5))) / (2*n^(3/2)*r^n), where r = 0.1130413665724951344267888513870607581680912144315... and s = 1.385648922830296011590145919380626723251960276539... are real roots of the system of equations (1 + r*s^3)*(1 + r^2*s^5) = s, r*s^2*(3 + 5*r*s^2 + 8*r^2*s^5) = 1. - Vaclav Kotesovec, Nov 22 2017

A369442 Expansion of (1/x) * Series_Reversion( x / ((1+x) * (1+x^3)^2) ).

Original entry on oeis.org

1, 1, 1, 3, 11, 31, 84, 261, 865, 2815, 9131, 30339, 102681, 349376, 1193993, 4111947, 14263137, 49720513, 174040102, 611770893, 2158954383, 7645030641, 27153898487, 96719683491, 345414958227, 1236555046701, 4436564115556, 15950469680836, 57455730349552
Offset: 0

Views

Author

Seiichi Manyama, Jan 23 2024

Keywords

Links

Crossrefs

Cf. A198951.

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)*(1+x^3)^2))/x)
  • PARI
    a(n, s=3, t=2, u=1) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial(u*(n+1), n-s*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(2*n+2,k) * binomial(n+1,n-3*k).

A378427 Expansion of (1/x) * Series_Reversion( x / (1 + x + x^3 * (1 + x)^3) ).

Original entry on oeis.org

1, 1, 1, 2, 8, 29, 88, 253, 775, 2575, 8797, 29833, 100635, 342408, 1181727, 4120223, 14435969, 50738813, 179038408, 634696939, 2259677734, 8072923814, 28924907573, 103915759961, 374302237154, 1351541722226, 4891132336481, 17736792240766, 64440831300682
Offset: 0

Views

Author

Seiichi Manyama, Nov 25 2024

Keywords

Links

Crossrefs

Cf. A300048, A378425.
Cf. A198951, A378407.

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x/(1+x+x^3*(1+x)^3))/x)
  • PARI
    a(n) = sum(k=0, n\3, binomial(n+1, k)*binomial(n+2*k+1, n-3*k))/(n+1);

Formula

G.f.: exp( Sum_{k>=1} A378407(k) * x^k/k ).
a(n) = (1/(n+1)) * [x^n] (1 + x + x^3 * (1 + x)^3)^(n+1).
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(n+1,k) * binomial(n+2*k+1,n-3*k).

A369444 Expansion of (1/x) * Series_Reversion( x / ((1+x) * (1+x^4)) ).

Original entry on oeis.org

1, 1, 1, 1, 2, 7, 22, 57, 131, 298, 738, 2003, 5600, 15380, 41224, 109769, 296009, 813315, 2261647, 6305930, 17554044, 48851034, 136350556, 382408995, 1077164245, 3042452536, 8606495236, 24377127256, 69159381856, 196600128592, 559990599808, 1597797525833
Offset: 0

Views

Author

Seiichi Manyama, Jan 23 2024

Keywords

Links

Crossrefs

Cf. A036765, A198951.

Programs

  • PARI
    my(N=40, x='x+O('x^N)); Vec(serreverse(x/((1+x)*(1+x^4)))/x)
  • PARI
    a(n, s=4, t=1, u=1) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial(u*(n+1), n-s*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/4)} binomial(n+1,k) * binomial(n+1,n-4*k).
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