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 1-2 of 2 results.

A365516 Expansion of g.f. A(x) satisfying A(x) = (1 + 2*x*A(x))^2 / (1 + 2*x*A(x) - 2*x^3*A(x)^3).

Original entry on oeis.org

1, 2, 4, 10, 32, 112, 400, 1464, 5520, 21296, 83456, 331136, 1328320, 5379200, 21959936, 90271904, 373347840, 1552438016, 6486311680, 27217331456, 114649525760, 484640538112, 2055185596416, 8740711936000, 37273693649920, 159340373710848, 682708771254272, 2931290431277056
Offset: 0

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Author

Paul D. Hanna, Sep 07 2023

Keywords

Comments

Related identities for the Catalan function C(x) = 1 + x*C(x)^2 (A000108):
(1) [x^(n-1)] (1 + n*x*C(x))^n / C(x)^n = n^(n-1) for n >= 1.
(2) [x^(n-1)] (1 + (n+1)*x*C(x)^2)^n / C(x)^(2*n) = n^(n-1) for n >= 1.
Related identity: [x^(n-1)] (1 + (n+1)*x*B(x))^n / B(x)^n = n*(n-1)^(n-1) for n >= 1 when B(x) = 1/(1 - 2*x).
Related identity: F(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n * F(x)^n / (1 + (n+1)*x*F(x))^(n+1), which holds formally for all Maclaurin series F(x). - Paul D. Hanna, Oct 03 2023

Examples

			G.f.: A(x) = 1 + 2*x + 4*x^2 + 10*x^3 + 32*x^4 + 112*x^5 + 400*x^6 + 1464*x^7 + 5520*x^8 + 21296*x^9 + 83456*x^10 + ...
where A(x) = (1 + 2*x*A(x))^2/(1 + 2*x*A(x) - 2*x^3*A(x)^3).
RELATED SERIES.
B(x) = A(x/B(x)) where B(x) = (1 + 2*x)^2/(1 + 2*x - 2*x^3) begins
B(x) = 1 + 2*x + 2*x^3 + 4*x^6 - 8*x^7 + 16*x^8 - 24*x^9 + 32*x^10 - 32*x^11 + 16*x^12 + 32*x^13 - 128*x^14 + ...
RELATED TABLE.
The table of coefficients of x^k in (1 + (n+1)*x*A(x))^n/A(x)^n begins:
n=1: [1,  0,    0,    -2,     -8,    -24,     -76,    -272, ...];
n=2: [1,  2,    1,    -4,    -20,    -64,    -196,    -664, ...];
n=3: [1,  6,   12,     2,    -48,   -192,    -600,   -1896, ...];
n=4: [1, 12,   54,   100,    -23,   -600,   -2224,   -6944, ...];
n=5: [1, 20,  160,   630,   1080,   -696,   -8660,  -30960, ...];
n=6: [1, 30,  375,  2488,   9027,  14406,  -15371, -144852, ...];
n=7: [1, 42,  756,  7546,  44800, 153552,  229376, -342728, ...];
n=8: [1, 56, 1372, 19192, 167222, 921400, 3026332, 4251528, ...]; ...
in which the main diagonal equals n*(n+1)^(n-2) for n > 1.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Module[{A = {1}, m, x}, Do[AppendTo[A, 0]; m = Length[A]; A[[-1]] = Coefficient[(1 + (m + 1)*x*SeriesData[x, 0, A, 0, m + 1, 1])^m/SeriesData[x, 0, A, 0, m + 1, 1]^m, x, m - 1]/m - (m + 1)^(m - 2), {i, 1, n}]; A[[n + 1]]]; Table[a[n], {n, 0, 27}] (* Robert P. P. McKone, Sep 07 2023 *)
  • PARI
    /* Using series reversion */
    {a(n) = my(A = (1/x)*serreverse(x*(1 + 2*x - 2*x^3)/(1 + 2*x +x*O(x^n))^2));
    polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    /* Formula [x^(n-1)] (1 + (n+1)*x*A(x))^n/A(x)^n = n*(n+1)^(n-2) */
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0); m=#A;
    A[#A] = polcoeff( (1 + (m+1)*x*Ser(A))^m / Ser(A)^m , m-1)/m - (m+1)^(m-2) );A[n+1]}
    for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x) = (1 + 2*x*A(x))^2 / (1 + 2*x*A(x) - 2*x^3*A(x)^3).
(2) A(x/B(x)) = B(x) where B(x) = (1 + 2*x)^2 / (1 + 2*x - 2*x^3).
(3) A(x) = (1/x) * Series_Reversion( x*(1 + 2*x - 2*x^3)/(1 + 2*x)^2 ).
(4) [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n+1)^(n-2) for n > 1.
(5) [x^(n-1)] (1 + (n-1)*x*A(x))^n / A(x)^n = -n*(n-3)^(n-2) for n > 1.
(6) [x^(n-1)] (1 + n*x*A(x))^n / A(x)^n = (n^(n-1) - n*(n-2)^(n-2))/2 for n >= 1.
(7) A(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n * A(x)^n / (1 + (n+1)*x*A(x))^(n+1). - Paul D. Hanna, Oct 03 2023

A365574 Expansion of g.f. A(x) satisfying [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n+2)^(n-2) for n > 1.

Original entry on oeis.org

1, 2, 3, 4, 16, 104, 515, 2090, 8170, 34704, 160014, 751282, 3479758, 16012684, 74362915, 350282602, 1665651094, 7952638460, 38067823370, 182874936368, 882344022104, 4274341269824, 20773195676078, 101228332620524, 494521566769160, 2421729829067636, 11886902458813596
Offset: 0

Views

Author

Paul D. Hanna, Sep 11 2023

Keywords

Comments

Conjecture 1: a(k) is odd iff k = 2^n - 2 for n >= 1.
Conjecture 2: a(2^n - 2) == 3 (mod 16) for n > 1.
Is there a closed formula for the g.f. of this sequence? Compare to the g.f. of A365516.
Related identities for the Catalan function C(x) = 1 + x*C(x)^2 (A000108):
(1) [x^(n-1)] (1 + n*x*C(x))^n / C(x)^n = n^(n-1) for n >= 1.
(2) [x^(n-1)] (1 + (n+1)*x*C(x)^2)^n / C(x)^(2*n) = n^(n-1) for n >= 1.
Related identity: F(x) = (3/x) * Sum{n>=1} n*(n+2)^(n-2) * x^n * F(x)^n / (1 + (n+2)*x*F(x))^(n+1), which holds formally for all Maclaurin series F(x). - Paul D. Hanna, Oct 03 2023

Examples

			G.f.: A(x) = 1 + 2*x + 3*x^2 + 4*x^3 + 16*x^4 + 104*x^5 + 515*x^6 + 2090*x^7 + 8170*x^8 + 34704*x^9 + 160014*x^10 + 751282*x^11 + 3479758*x^12 + ...
RELATED TABLE.
The table of coefficients of x^k in (1 + (n+1)*x*A(x))^n/A(x)^n begins:
n=1: [1,  0,    1,     0,    -11,     -54,    -182,    -594, ...];
n=2: [1,  2,    3,     2,    -21,    -130,    -494,   -1660, ...];
n=3: [1,  6,   15,    20,    -18,    -288,   -1391,   -5070, ...];
n=4: [1, 12,   58,   144,    151,    -468,   -3934,  -17376, ...];
n=5: [1, 20,  165,   720,   1715,    1274,   -8960,  -60530, ...];
n=6: [1, 30,  381,  2650,  10824,   24576,   10623, -176034, ...];
n=7: [1, 42,  763,  7812,  49084,  191016,  413343,   49818, ...];
n=8: [1, 56, 1380, 19600, 176242, 1033664, 3873296, 8000000, ...]; ...
in which the main diagonal equals n*(n+2)^(n-2) for n > 1.
		

Crossrefs

Programs

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

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n+2)^(n-2) for n > 1.
(2) [x^(n-1)] (1 + (n-2)*x*A(x))^n / A(x)^n = -2*n*(n-4)^(n-2) for n > 1.
(3) [x^(n-1)] (1 + n*x*A(x))^n / A(x)^n = 2*n*((n+1)^(n-2) - (n-2)^(n-2))/3 for n > 1.
(4) A(x) = (3/x) * Sum{n>=1} n*(n+2)^(n-2) * x^n * A(x)^n / (1 + (n+2)*x*A(x))^(n+1). - Paul D. Hanna, Oct 03 2023
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