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.

A352236 G.f. A(x) satisfies: A(x) = 1 + x*A(x)^2 / (A(x) - 2*x*A'(x)).

Original entry on oeis.org

1, 1, 3, 19, 185, 2353, 36075, 638115, 12683761, 278485217, 6674259667, 173097575603, 4826128088489, 143896870347793, 4568544366818747, 153883892657000259, 5481761893234193889, 205939077652874352577, 8138639816942009694627, 337568614331296733526867
Offset: 0

Views

Author

Paul D. Hanna, Mar 08 2022

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 19*x^3 + 185*x^4 + 2353*x^5 + 36075*x^6 + 638115*x^7 + 12683761*x^8 + ...
such that A(x) = 1 + x*A(x)^2/(A(x) - 2*x*A'(x)).
Related table.
The table of coefficients of x^k in A(x)^(2*n+1) begins:
n=0: [1,  1,   3,   19,   185,   2353,   36075, ...];
n=1: [1,  3,  12,   76,   705,   8595,  127680, ...];
n=2: [1,  5,  25,  165,  1490,  17506,  252050, ...];
n=3: [1,  7,  42,  294,  2632,  30016,  419454, ...];
n=4: [1,  9,  63,  471,  4239,  47295,  643017, ...];
n=5: [1, 11,  88,  704,  6435,  70785,  939312, ...];
n=6: [1, 13, 117, 1001,  9360, 102232, 1329016, ...]; ...
in which the following pattern holds:
[x^n] A(x)^(2*n+1) = [x^(n-1)] (2*n+1) * A(x)^(2*n+1), n >= 1,
as illustrated by
[x^1] A(x)^3 = 3 = [x^0] 3*A(x)^3 = 3*1;
[x^2] A(x)^5 = 25 = [x^1] 5*A(x)^5 = 5*5;
[x^3] A(x)^7 = 294 = [x^2] 7*A(x)^7 = 7*42;
[x^4] A(x)^9 = 4239 = [x^3] 9*A(x)^9 = 9*471;
[x^5] A(x)^11 = 70785 = [x^4] 11*A(x)^11 = 11*6435;
[x^6] A(x)^13 = 1329016 = [x^5] 13*A(x)^13 = 13*102232; ...
Also, compare the above terms along the diagonal to the series
B(x) = A(x*B(x)^2) = 1 + x + 5*x^2 + 42*x^3 + 471*x^4 + 6435*x^5 + 102232*x^6 + 1837630*x^7 + ... + A317352(n)*x^n + ...
where B(x)^2 = (1/x) * Series_Reversion( x/A(x)^2 ).
		

Crossrefs

Programs

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

Formula

G.f. A(x) satisfies:
(1) [x^n] A(x)^(2*n+1) = [x^(n-1)] (2*n+1) * A(x)^(2*n+1) for n >= 1.
(2) A(x) = 1 + x*A(x)^2/(A(x) - 2*x*A'(x)).
(3) A'(x) = A(x) * (1 + x*A(x)/(1 - A(x))) / (2*x).
(4) A(x) = exp( Integral (1 + x*A(x)/(1 - A(x)))/(2*x) dx ).
a(n) ~ c * 2^n * n! * n^(3/2), where c = 0.06926688933886004638602492... - Vaclav Kotesovec, Nov 16 2023

A352237 G.f. A(x) satisfies: A(x) = 1 + x*A(x)^2 / (A(x) - 3*x*A'(x)).

Original entry on oeis.org

1, 1, 4, 37, 532, 9994, 226252, 5910445, 173581060, 5634589906, 199792389160, 7671942375898, 316936631324368, 14011781050744984, 660054967923455212, 33008607551445324157, 1746771084107236755604, 97536010045722766992778, 5731874036042145864368824
Offset: 0

Views

Author

Paul D. Hanna, Mar 08 2022

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 37*x^3 + 532*x^4 + 9994*x^5 + 226252*x^6 + 5910445*x^7 + 173581060*x^8 + ...
such that A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)).
Related table.
The table of coefficients of x^k in A(x)^(3*n+1) begins:
n=0: [1,  1,   4,   37,   532,   9994,   226252, ...];
n=1: [1,  4,  22,  200,  2717,  48788,  1069122, ...];
n=2: [1,  7,  49,  462,  6069, 104664,  2219784, ...];
n=3: [1, 10,  85,  850, 11020, 183832,  3777355, ...];
n=4: [1, 13, 130, 1391, 18083, 294203,  5869734, ...];
n=5: [1, 16, 184, 2112, 27852, 445632,  8659920, ...];
n=6: [1, 19, 247, 3040, 41002, 650161, 12353059, ...]; ...
in which the following pattern holds:
[x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1) * A(x)^(3*n+1), n >= 1,
as illustrated by
[x^1] A(x)^4 = 4 = [x^0] 4*A(x)^4 = 4*1;
[x^2] A(x)^7 = 49 = [x^1] 7*A(x)^7 = 7*7;
[x^3] A(x)^10 = 850 = [x^2] 10*A(x)^10 = 10*85;
[x^4] A(x)^13 = 18083 = [x^3] 13*A(x)^13 = 13*1391;
[x^5] A(x)^16 = 445632 = [x^4] 16*A(x)^16 = 16*27852;
[x^6] A(x)^19 = 12353059 = [x^5] 19*A(x)^19 = 19*650161; ...
Also, compare the above terms along the diagonal to the series
B(x) = A(x*B(x)^3) = 1 + x + 7*x^2 + 85*x^3 + 1391*x^4 + 27852*x^5 + 650161*x^6 + 17204220*x^7 + ...
where B(x)^3 = (1/x) * Series_Reversion( x/A(x)^3 ).
		

Crossrefs

Programs

  • PARI
    /* Using A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)) */
    {a(n) = my(A=1); for(i=1,n, A = 1 + x*A^2/(A - 3*x*A' + x*O(x^n)) );
    polcoeff(A,n)}
    for(n=0,20, print1(a(n),", "))
    
  • PARI
    /* Using [x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1)*A(x)^(3*n+1) */
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
    A[#A] = polcoeff((x*Ser(A)^(3*(#A)-2) - Ser(A)^(3*(#A)-2)/(3*(#A)-2)),#A-1));A[n+1]}
    for(n=0,20, print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) [x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1) * A(x)^(3*n+1) for n >= 1.
(2) A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)).
(3) A'(x) = A(x) * (1 + x*A(x)/(1 - A(x))) / (3*x).
(4) A(x) = exp( Integral (1 + x*A(x)/(1 - A(x)))/(3*x) dx ).
a(n) ~ c * 3^n * n! * n^(4/3), where c = 0.0543186200722307001992331... - Vaclav Kotesovec, Nov 16 2023

A352238 G.f. A(x) satisfies: A(x) = 1 + x*A(x)^2 / (A(x) - 4*x*A'(x)).

Original entry on oeis.org

1, 1, 5, 61, 1161, 28857, 864141, 29861749, 1160382737, 49854838897, 2340623599381, 119051103325613, 6516915195123097, 381912592990453545, 23856225840952434333, 1582482450123627473637, 111113139625779846025761, 8234335766045466358238433
Offset: 0

Views

Author

Paul D. Hanna, Mar 08 2022

Keywords

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 61*x^3 + 1161*x^4 + 28857*x^5 + 864141*x^6 + 29861749*x^7 + 1160382737*x^8 + ...
such that A(x) = 1 + x*A(x)^2/(A(x) - 4*x*A'(x)).
Related table.
The table of coefficients of x^k in A(x)^(4*n+1) begins:
n=0: [1,  1,   5,   61,   1161,   28857,   864141, ...];
n=1: [1,  5,  35,  415,   7430,  176286,  5107530, ...];
n=2: [1,  9,  81,  993,  17127,  389583, 10916559, ...];
n=3: [1, 13, 143, 1859,  31564,  693212, 18802212, ...];
n=4: [1, 17, 221, 3077,  52309, 1118549, 29427153, ...];
n=5: [1, 21, 315, 4711,  81186, 1704906, 43640030, ...];
n=6: [1, 25, 425, 6825, 120275, 2500555, 62513875, ...]; ...
in which the following pattern holds:
[x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1) * A(x)^(4*n+1), n >= 1,
as illustrated by
[x^1] A(x)^5 = 5 = [x^0] 5*A(x)^5 = 5*1;
[x^2] A(x)^9 = 81 = [x^1] 9*A(x)^9 = 9*9;
[x^3] A(x)^13 = 1859 = [x^2] 13*A(x)^13 = 13*143;
[x^4] A(x)^17 = 52309 = [x^3] 17*A(x)^17 = 17*3077;
[x^5] A(x)^21 = 1704906 = [x^4] 21*A(x)^21 = 21*81186;
[x^6] A(x)^25 = 62513875 = [x^5] 25*A(x)^25 = 25*2500555; ...
Also, compare the above terms along the diagonal to the series
B(x) = A(x*B(x)^4) = 1 + x + 9*x^2 + 143*x^3 + 3077*x^4 + 81186*x^5 + 2500555*x^6 + 87388600*x^7 + ...
where B(x)^4 = (1/x) * Series_Reversion( x/A(x)^4 ).
		

Crossrefs

Programs

  • PARI
    /* Using A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)) */
    {a(n) = my(A=1); for(i=1,n, A = 1 + x*A^2/(A - 4*x*A' + x*O(x^n)) );
    polcoeff(A,n)}
    for(n=0,20, print1(a(n),", "))
    
  • PARI
    /* Using [x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1)*A(x)^(4*n+1) */
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
    A[#A] = polcoeff((x*Ser(A)^(4*(#A)-3) - Ser(A)^(4*(#A)-3)/(4*(#A)-3)),#A-1));A[n+1]}
    for(n=0,20, print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) [x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1) * A(x)^(4*n+1) for n >= 1.
(2) A(x) = 1 + x*A(x)^2/(A(x) - 4*x*A'(x)).
(3) A'(x) = A(x) * (1 + x*A(x)/(1 - A(x))) / (4*x).
(4) A(x) = exp( Integral (1 + x*A(x)/(1 - A(x)))/(4*x) dx ).
a(n) ~ c * 4^n * n! * n^(5/4), where c = 0.0440035900116077498469559... - Vaclav Kotesovec, Nov 16 2023
Showing 1-3 of 3 results.