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.

A360582 Expansion of A(x) satisfying [x^n] A(x) / (1 + x*A(x)^n) = 0 for n > 0.

Original entry on oeis.org

1, 1, 2, 8, 48, 382, 3793, 45208, 627957, 9928646, 175476102, 3420270423, 72789704826, 1678446235555, 41675807453127, 1108522434288617, 31444611938560078, 947522959703143140, 30225484159719768548, 1017558928058932606182, 36053690169955373601165, 1341103168079733579768368
Offset: 0

Views

Author

Paul D. Hanna, Mar 12 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 48*x^4 + 382*x^5 + 3793*x^6 + 45208*x^7 + 627957*x^8 + 9928646*x^9 + 175476102*x^10 + ...
The table of coefficients in the successive powers of g.f. A(x) begins:
n = 1: [1, 1,  2,   8,   48,  382,  3793,  45208, ...];
n = 2: [1, 2,  5,  20,  116,  892,  8606, 100298, ...];
n = 3: [1, 3,  9,  37,  210, 1566, 14687, 167280, ...];
n = 4: [1, 4, 14,  60,  337, 2448, 22340, 248580, ...];
n = 5: [1, 5, 20,  90,  505, 3591, 31935, 347120, ...];
n = 6: [1, 6, 27, 128,  723, 5058, 43919, 466410, ...];
n = 7: [1, 7, 35, 175, 1001, 6923, 58828, 610653, ...];
...
The table of coefficients in A(x)/(1 + x*A(x)^n) begins:
n = 1: [1, 0,  1,   5,  34,  293, 3066, 37900, ...];
n = 2: [1, 0,  0,   3,  25,  235, 2601, 33346, ...];
n = 3: [1, 0, -1,   0,  14,  167, 2055, 28049, ...];
n = 4: [1, 0, -2,  -4,   0,   89, 1432, 21994, ...];
n = 5: [1, 0, -3,  -9, -18,    0,  742, 15216, ...];
n = 6: [1, 0, -4, -15, -41, -102,    0,  7820, ...];
n = 7: [1, 0, -5, -22, -70, -220, -775,     0, ...];
...
in which the diagonal of all zeros illustrates that
[x^n] A(x) / (1 + x*A(x)^n) = 0 for n > 0.

Links

Crossrefs

Cf. A360581, A360583, A360584.

Programs

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

Formula

a(n) ~ c * n! * n^(2*LambertW(1) - 1) / LambertW(1)^n, where c = 0.11249164340900724981958... - Vaclav Kotesovec, Mar 13 2023

A360583 Expansion of A(x) satisfying [x^n] A(x) / (1 + x*A(x)^(n+1)) = 0 for n > 0.

Original entry on oeis.org

1, 1, 3, 17, 139, 1455, 18326, 267700, 4426686, 81455357, 1646941293, 36238989035, 861298646217, 21978627323651, 599195351716464, 17379759869328515, 534392854606942358, 17363705283005593096, 594513962872698955686, 21395116871674867310280, 807416275879430472577570
Offset: 0

Views

Author

Paul D. Hanna, Mar 12 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 17*x^3 + 139*x^4 + 1455*x^5 + 18326*x^6 + 267700*x^7 + 4426686*x^8 + 81455357*x^9 + 1646941293*x^10 + ...
The table of coefficients in the successive powers of g.f. A(x) begins:
n = 1: [1, 1,  3,  17,  139,  1455,  18326, ...];
n = 2: [1, 2,  7,  40,  321,  3290,  40685, ...];
n = 3: [1, 3, 12,  70,  555,  5583,  67827, ...];
n = 4: [1, 4, 18, 108,  851,  8424, 100624, ...];
n = 5: [1, 5, 25, 155, 1220, 11916, 140085, ...];
n = 6: [1, 6, 33, 212, 1674, 16176, 187372, ...];
n = 7: [1, 7, 42, 280, 2226, 21336, 243817, ...];
...
The table of coefficients in A(x)/(1 + x*A(x)^(n+1)) begins:
n = 1: [1, 0,  1,   9,   88, 1021, 13736, 209940, ...];
n = 2: [1, 0,  0,   5,   64,  821, 11670, 184622, ...];
n = 3: [1, 0, -1,   0,   35,  587,  9283, 155666, ...];
n = 4: [1, 0, -2,  -6,    0,  315,  6555, 122855, ...];
n = 5: [1, 0, -3, -13,  -42,    0,  3467,  86025, ...];
n = 6: [1, 0, -4, -21,  -92, -364,     0,  45079, ...];
n = 7: [1, 0, -5, -30, -151, -784, -3866,      0, ...];
...
in which the diagonal of all zeros illustrates that
[x^n] A(x) / (1 + x*A(x)^(n+1)) = 0 for n > 0.

Links

Crossrefs

Cf. A360582, A360584.

Programs

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

Formula

a(n) ~ c * n! * n^(3*LambertW(1) - 1 + 1/(1 + LambertW(1))) / LambertW(1)^n, where c = 0.0685246538334926708088... - Vaclav Kotesovec, Mar 13 2023

A360581 Expansion of A(x) satisfying [x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.

Original entry on oeis.org

1, 1, 3, 17, 131, 1204, 12587, 149131, 2036675, 32358153, 587313706, 11761213199, 252859744189, 5785648936988, 141627609404793, 3737907237793369, 106414467836076985, 3241492594168333618, 104522041356412895455, 3541554178675758259947, 125782730912626755808358
Offset: 0

Views

Author

Paul D. Hanna, Mar 12 2023

Keywords

Examples

			G.f. A(x) = 1 + x + 3*x^2 + 17*x^3 + 131*x^4 + 1204*x^5 + 12587*x^6 + 149131*x^7 + 2036675*x^8 + 32358153*x^9 + 587313706*x^10 + ...
The table of coefficients in the successive powers of g.f. A(x) begins:
n = 1: [1, 1,  3,  17,  131,  1204,  12587,  149131, ...];
n = 2: [1, 2,  7,  40,  305,  2772,  28657,  335114, ...];
n = 3: [1, 3, 12,  70,  531,  4782,  48936,  565245, ...];
n = 4: [1, 4, 18, 108,  819,  7324,  74272,  848064, ...];
n = 5: [1, 5, 25, 155, 1180, 10501, 105650, 1193530, ...];
n = 6: [1, 6, 33, 212, 1626, 14430, 144208, 1613214, ...];
n = 7: [1, 7, 42, 280, 2170, 19243, 191254, 2120511, ...];
n = 8: [1, 8, 52, 360, 2826, 25088, 248284, 2730872, ...];
...
The table of coefficients in A(x)^n/(1 + x*A(x)^n)^n begins:
n = 1: [1, 0,   2,   12,  100,  955, 10258, 124565, ...];
n = 2: [1, 0,   2,   18,  161, 1606, 17757, 220834, ...];
n = 3: [1, 0,   0,   15,  168, 1806, 21000, 272856, ...];
n = 4: [1, 0,  -4,    0,  114, 1504, 19220, 270692, ...];
n = 5: [1, 0, -10,  -30,    0,  800, 12970, 215445, ...];
n = 6: [1, 0, -18,  -78, -165,    0,  4797, 123990, ...];
n = 7: [1, 0, -28, -147, -364, -329,     0,  32767, ...];
n = 8: [1, 0, -40, -240, -572,  696,  7472,      0, ...];
...
in which the diagonal of all zeros illustrates that
[x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.

Links

Crossrefs

Cf. A360582, A360583, A360584, A303063.

Programs

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

Formula

From Vaclav Kotesovec, Mar 13 2023: (Start)
a(n) ~ c * n! * n^alpha / LambertW(1)^n, where alpha = 0.33953... and c = 0.1881608377753...
Conjecture: alpha = 3*LambertW(1) - 2 + 1/(1 + LambertW(1)) = 0.33953361459446... (End)
Showing 1-3 of 3 results.

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