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.

A190015 Triangle T(n,k) for solving differential equation A'(x)=G(A(x)), G(0)!=0.

Original entry on oeis.org

1, 1, 2, 1, 6, 8, 1, 24, 42, 16, 22, 1, 120, 264, 180, 192, 136, 52, 1, 720, 1920, 1248, 540, 1824, 2304, 272, 732, 720, 114, 1, 5040, 15840, 10080, 8064, 18720, 22752, 9612, 7056, 10224, 17928, 3968, 2538, 3072, 240, 1, 40320, 146160, 92160, 70560, 32256, 207360, 249120, 193536, 73728, 61560, 144720, 246816, 101844, 142704, 7936, 51048, 110448, 34304, 8334, 11616, 494, 1
Offset: 0

Views

Author

Vladimir Kruchinin, May 04 2011

Keywords

Comments

For solving the differential equation A'(x)=G(A(x)), where G(0)!=0,
a(n) = 1/n!*sum(pi(i) in P(2*n-1,n), T(n,i)*prod(j=1..n, g(k_j-1))),
where pi(i) is the partition of 2*n-1 into n parts in lexicographic order P(2*n-1,n).
G(x) = g(0)+g(1)*x+g(2)*x^2+...
Examples
A003422 A'(x)=A(x)+1/(1-x)
A000108 A'(x)=1/(1-2*A(x)),
A001147 A'(x)=1/(1-A(x))
A007489 A'(x)=A(x)+x/(1-x)^2+1.
A006351 B'(x)=(1+B(x))/(1-B(x))
A029768 A'(x)=log(1/(1-A(x)))+1.
A001662 B'(x)=1/(1+B(x))
A180254 A'(x)=(1-sqrt(1-4*A(x)))/2
Compare with A145271. There (j')^k = [(d/dx)^j g(x)]^k evaluated at x=0 gives formulas expressed in terms of the coefficients of the Taylor series g(x). If, instead, we express the formulas in terms of the coefficients of the power series of g(x), we obtain a row reversed array for A190015 since the partitions there are in reverse order to the ones here. Simply exchange (j!)^k * (j")^k for (j')^k, where (j")^k = [(d/dx)^j g(x) / j!]^k, to transform from one array to the other. E.g., R^3 g(x) = 1 (0')^1 (1')^3 + 4 (0')^2 (1')^1 (2')^1 + 1 (0')^3 (3')^1 = 1 (O")^1 (1")^3 + 4 (0")^2 (1")^1 2*(2")^1 + 1 (0")^1 3!*(3")^1 = 1 (O")^1 (1")^3 + 8 (0")^2 (1")^1 (2")^1 + 6 (0")^1 (3")^1, the fourth partition polynomial here. - Tom Copeland, Oct 17 2014

Examples

			Triangle begins:
1;
1;
2,1;
6,8,1;
24,42,16,22,1;
120,264,180,192,136,52,1;
720,1920,1248,540,1824,2304,272,732,720,114,1;
5040,15840,10080,8064,18720,22752,9612,7056,10224,17928,3968,2538,3072,240,1;
40320,146160,92160,70560,32256,207360,249120,193536,73728,61560,144720,246816, 101844,142704,7936,51048,110448,34304,8334,11616,494,1;
Example for n=5:
partitions of number 9 into  5 parts in lexicographic order:
[1,1,1,1,5]
[1,1,1,2,4]
[1,1,1,3,3]
[1,1,2,2,3]
[1,2,2,2,2]
a(5) = (24*g(0)^4*g(4) +42*g(0)^3*g(1)*g(3) +16*g(0)^3*g(2)^2 +22*g(0)^2*g(1)^2*g(2) +g(0)*g(1)^4)/5!.

Programs

  • Maxima
    /* array of triangle */
M:[1,1,2,1,6,8,1,24,42,16,22,1,120,264,180,192,136,52,1,720,1920,1248,540,1824,2304,272,732,720,114,1,5040,15840,10080,8064,18720,22752,9612,7056,10224,17928,3968,2538,3072,240,1,40320,146160,92160,70560,32256,207360,249120,193536,73728,61560,144720,246816,101844,142704,7936,51048,110448,34304,8334,11616,494,1];
/* function of triangle */
T(n,k):=M[sum(num_partitions(i),i,0,n-1)+k+1];
/* count number of partitions of n into m parts */
b(n,m):=if n
  • Maxima
    /* Find triangle */
Co(n,k):=if k=1  then a(n) else sum(a(i+1)*Co(n-i-1,k-1),i,0,n-k);
a(n):=if n=1 then 1 else 1/n*sum(Co(n-1,k)*x(k),k,1,n-1);
makelist(ratsimp(n!*a(n)),n,1,5);
/* Vladimir Kruchinin, Jun 15 2012 */
  • PARI
    serlaplace( serreverse( intformal( 1 / sum(n=0, 9, eval(Str("g"n)) * x^n, x * O(x^9))))) /* Michael Somos, Oct 22 2014 */

A376174 E.g.f. A(x) satisfies: A'(x) = 1 + A(x)*A'(x)^4.

Original entry on oeis.org

1, 1, 9, 165, 4629, 175689, 8424801, 488756205, 33292495341, 2605108910481, 230300167685049, 22701604019859765, 2468971586334241989, 293687751861227612889, 37930838135497768301841, 5286141242516127169100925, 790686768925097978507354781, 126349187477795223746580576801
Offset: 1

Views

Author

Paul D. Hanna, Sep 14 2024

Keywords

Comments

In general, if k > 1 and e.g.f. A(x) satisfies: A'(x) = 1 + A(x)*A'(x)^k, then a(n) ~ k^(k/2 + 1) * (k^(k+1)*(k+1))^(n - 3/2) * n^(n-2) / ((k-1)^((k+3)/2) * ((k-1)^(k+1) + k^k)^(n - 3/2) * exp(n)). - Vaclav Kotesovec, Sep 17 2024

Examples

			E.g.f.: A(x) = x + x^2/2! + 9*x^3/3! + 165*x^4/4! + 4629*x^5/5! + 175689*x^6/6! + 8424801*x^7/7! + 488756205*x^8/8! + 33292495341*x^9/9! + ...
where A'(x) = 1 + A(x)*A'(x)^4.
Also,
A'(x) = 1 + A(x) + 4*A(x)^2 + 22*A(x)^3 + 140*A(x)^4 + 969*A(x)^5 + 7084*A(x)^6 + 53820*A(x)^7 + ... + A002293(n)*A(x)^n + ...
RELATED SERIES.
Series D(x) = 1 + x*D(x)^4 begins
D(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 + 53820*x^7 + ... + A002293(n)*x^n + ...
where Integral( dx/D(x) ) = x - x^2/2! - 6*x^3/3! - 90*x^4/4! - 2184*x^5/5! - 73440*x^6/6! - 3160080*x^7/7! - 165765600*x^8/8! + ...
and A(x) = Series_Reversion( Integral( dx/D(x) ) ).
A'(x)^4 = 1 + 4*x + 48*x^2/2! + 1008*x^3/3! + 30672*x^4/4! + 1229616*x^5/5! + 61348752*x^6/6! + 3668121072*x^7/7! + ...

Links

Crossrefs

Cf. A180254, A376175, A002293.

Programs

  • Mathematica
    nmax = 20; B[] = 1; Do[B[x] = 1 + Integrate[B[x], x]*B[x]^4 + O[x]^nmax // Normal, nmax]; A[x_] = Integrate[B[x], x]; CoefficientList[A[x]/x, x] * Range[nmax]!  (* Vaclav Kotesovec, Sep 17 2024 *)
  • PARI
    /* A'(x) = 1 + A(x)*A'(x)^4 */
{a(n) = my(A=x); for(i=1, n, A = intformal(1 + A*(A')^4 + x*O(x^n))); n!*polcoeff(A, n)}
for(n=1, 20, print1(a(n)", "))
  • PARI
    /* A(x) = Series_Reversion( Integral( dx/D ) ) where D = 1 + x*D^4 */
{a(n) = my(D = ((1/x)*serreverse(x/(1 + x + x*O(x^n))^4))^(1/4), A); A = serreverse(intformal(1/D)); n!*polcoeff(A, n)}
for(n=1, 20, print1(a(n)", "))

Formula

E.g.f. A(x) = Sum_{n>=1} a(n)*x^n/n! satisfies:
(1) A(x) = Series_Reversion( Integral( dx/D(x) ) ) where D(x) = 1 + x*D(x)^4 is the g.f. of A002293.
(2) A'(x) = Sum_{n>=0} A002293(n)*A(x)^n where A002293(n) = binomial(4*n,n)/(3*n+1).
(3) A'(x) = 1 + A(x)*A'(x)^4.
(4) A'(x) = 1/(1 - A(x)*A'(x)^3).
(5) A(x) = x + Sum_{n>=0} d^n/dx^n G(x)^(n+1)/(n+1)!, where G(x) = Integral( x*D(x)^3 dx ) and D(x) = 1 + x*D(x)^4.
(6) A(x - B(x)) = x where B(x) = Sum_{n>=0} binomial(4*n+2,n) * x^(n+2) / ((n+1)*(n+2)).
a(n) ~ 2^(10*n-9) * 5^(n - 3/2) * n^(n-2) / (3^(7/2) * exp(n) * 499^(n - 3/2)). - Vaclav Kotesovec, Sep 17 2024

A376175 E.g.f. A(x) satisfies: A'(x) = 1 + A(x)*A'(x)^3.

Original entry on oeis.org

1, 1, 7, 97, 2035, 57445, 2042215, 87651865, 4410770875, 254705483725, 16603869256975, 1206175463317825, 96627476254984675, 8463175473211383925, 804573717383525464375, 82513390092813146091625, 9080444173122231239204875, 1067360792025339122846660125, 133468583774114314367364097375
Offset: 1

Views

Author

Paul D. Hanna, Sep 13 2024

Keywords

Examples

			E.g.f.: A(x) = x + x^2/2! + 7*x^3/3! + 97*x^4/4! + 2035*x^5/5! + 57445*x^6/6! + 2042215*x^7/7! + 87651865*x^8/8! + 4410770875*x^9/9! + ...
where A'(x) = 1 + A(x)*A'(x)^3.
Also,
A'(x) = 1 + A(x) + 3*A(x)^2 + 12*A(x)^3 + 55*A(x)^4 + 273*A(x)^5 + 1428*A(x)^6 + 7752*A(x)^7 + ... + A001764(n)*A(x)^n + ...
RELATED SERIES.
Series D(x) = 1 + x*D(x)^3 begins
D(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 + ... + A001764(n)*x^n + ...
where Integral( dx/D(x) ) = x - x^2/2! - 4*x^3/3! - 42*x^4/4! - 720*x^5/5! - 17160*x^6/6! - 524160*x^7/7! - 19535040*x^8/8! + ...
and A(x) = Series_Reversion( Integral( dx/D(x) ) ).
A'(x)^3 = 1 + 3*x + 27*x^2/2! + 423*x^3/3! + 9567*x^4/4! + 284355*x^5/5! + 10499715*x^6/6! + 464006655*x^7/7! + ...

Links

Crossrefs

Cf. A180254, A376174, A001764.

Programs

  • PARI
    /* A'(x) = 1 + A(x)*A'(x)^3 */
{a(n) = my(A=x); for(i=1, n, A = intformal(1 + A*(A')^3 + x*O(x^n))); n!*polcoeff(A, n)}
for(n=1, 20, print1(a(n)", "))
  • PARI
    /* A(x) = Series_Reversion( Integral( dx/D ) ) where D = 1 + x*D^3 */
{a(n) = my(D = ((1/x)*serreverse(x/(1 + x + x*O(x^n))^3))^(1/3), A); A = serreverse(intformal(1/D)); n!*polcoeff(A, n)}
for(n=1, 20, print1(a(n)", "))

Formula

E.g.f. A(x) = Sum_{n>=1} a(n)*x^n/n! satisfies:
(1) A(x) = Series_Reversion( Integral( dx/D(x) ) ) where D(x) = 1 + x*D(x)^3 is the g.f. of A001764.
(2) A'(x) = Sum_{n>=0} A001764(n)*A(x)^n where A001764(n) = binomial(3*n,n)/(2*n+1).
(3) A'(x) = 1 + A(x)*A'(x)^3.
(4) A'(x) = 1/(1 - A(x)*A'(x)^2).
(5) A(x) = x + Sum_{n>=0} d^n/dx^n G(x)^(n+1)/(n+1)!, where G(x) = Integral( x*D(x)^2 dx ) and D(x) = 1 + x*D(x)^3.
(6) A(x - B(x)) = x where B(x) = Sum_{n>=0} binomial(3*n+1,n) * x^(n+2) / ((n+1)*(n+2)).
a(n) ~ 2^(2*n - 6) * 3^(4*n - 7/2) * n^(n-2) / (exp(n) * 43^(n - 3/2)). - Vaclav Kotesovec, Sep 14 2024
Showing 1-3 of 3 results.

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