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 1-8 of 8 results.

A302703 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n+1))^(n+1) for n>=0.

Original entry on oeis.org

1, 1, 3, 21, 235, 3470, 61933, 1274893, 29423331, 747440115, 20636072811, 613611700946, 19517927805840, 660667692682175, 23699856058131981, 897955765812058192, 35832679277251514074, 1502303284645831488072, 66031982339561373164915, 3036884343153028302140119, 145885192794643951791449387
Offset: 0

Views

Author

Paul D. Hanna, Apr 16 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 235*x^4 + 3470*x^5 + 61933*x^6 + 1274893*x^7 + 29423331*x^8 + 747440115*x^9 + 20636072811*x^10 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 4*x^2 + 31*x^3 + 356*x^4 + 5291*x^5 + 94592*x^6 + 1948763*x^7 + 45025516*x^8 + 1145651239*x^9 + 31696223593*x^10 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(n+1))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that b(n) begin:
[1, 2/2, 12/3, 124/4, 1780/5, 31746/6, 662144/7, 15590104/8, ...]
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  3,  21,  235,  3470,  61933,  1274893, ...];
n=1: [1, 2,  7,  48,  521,  7536, 132657,  2704342, ...];
n=2: [1, 3, 12,  82,  867, 12288, 213282,  4304877, ...];
n=3: [1, 4, 18, 124, 1283, 17828, 305056,  6094832, ...];
n=4: [1, 5, 25, 175, 1780, 24271, 409380,  8094540, ...];
n=5: [1, 6, 33, 236, 2370, 31746, 527824, 10326546, ...];
n=6: [1, 7, 42, 308, 3066, 40397, 662144, 12815839, ...];
n=7: [1, 8, 52, 392, 3882, 50384, 814300, 15590104, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(n+1))^(n+1):
n=0: [1, 1,  1,   3,   21,   235,    3470,    61933, ...];
n=1: [1, 2,  5,  18,  114,  1166,   16355,   283142, ...];
n=2: [1, 3, 12,  55,  354,  3372,   44463,   739917, ...];
n=3: [1, 4, 22, 124,  857,  7908,   98244,  1558788, ...];
n=4: [1, 5, 35, 235, 1780, 16501,  195980,  2955095, ...];
n=5: [1, 6, 51, 398, 3321, 31746,  368032,  5294250, ...];
n=6: [1, 7, 70, 623, 5719, 57302,  662144,  9182013, ...];
n=7: [1, 8, 92, 920, 9254, 98088, 1149804, 15590104, ...]; ...
to see that the main diagonals of the tables are the same.
		

Crossrefs

Programs

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

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n+1))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(n+1))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alfa, where d = A360279 = 2.1246065836242897918278825..., alfa = 2.2013296851505132606640400434738193121994558898350865326..., c = 0.026186121837027622395555466054900245177877028741031867... - Vaclav Kotesovec, Oct 06 2020, updated Feb 05 2023

A360231 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n-1))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 1, 6, 53, 628, 9167, 156309, 3021720, 64960004, 1532234825, 39270176511, 1085601040372, 32185085432757, 1018593646880447, 34279111177431666, 1222648239226278333, 46084480032637208699, 1830881732391546532475, 76488074741796221197580, 3352854778050665597014436
Offset: 0

Views

Author

Paul D. Hanna, Feb 02 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + x^2 + 6*x^3 + 53*x^4 + 628*x^5 + 9167*x^6 + 156309*x^7 + 3021720*x^8 + 64960004*x^9 + 1532234825*x^10 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 2*x^2 + 10*x^3 + 86*x^4 + 1004*x^5 + 14507*x^6 + 246218*x^7 + 4753205*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(n-1))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that b(n) begin:
[1/1, 2/2, 6/3, 40/4, 430/5, 6024/6, 101549/7, 1969744/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  1,   6,   53,  628,   9167,  156309, ...];
n=1: [1, 2,  3,  14,  119, 1374,  19732,  332844, ...];
n=2: [1, 3,  6,  25,  201, 2259,  31891,  531933, ...];
n=3: [1, 4, 10,  40,  303, 3308,  45870,  756192, ...];
n=4: [1, 5, 15,  60,  430, 4551,  61930, 1008565, ...];
n=5: [1, 6, 21,  86,  588, 6024,  80373, 1292370, ...];
n=6: [1, 7, 28, 119,  784, 7770, 101549, 1611352, ...];
n=7: [1, 8, 36, 160, 1026, 9840, 125864, 1969744, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(n-1))^(n+1):
n=0: [1, 1, -1,   0,   -5,   -42,   -528,   -7939, ...];
n=1: [1, 2,  1,   0,    0,     0,      0,       0, ...];
n=2: [1, 3,  6,  10,   30,   207,   2266,   31824, ...];
n=3: [1, 4, 14,  40,  141,   808,   7694,  101288, ...];
n=4: [1, 5, 25, 100,  430,  2376,  19680,  235165, ...];
n=5: [1, 6, 39, 200, 1035,  6024,  45879,  490524, ...];
n=6: [1, 7, 56, 350, 2135, 13601, 101549,  988338, ...];
n=7: [1, 8, 76, 560, 3950, 27888, 213952, 1969744, ...]; ...
to see that the main diagonals of the tables are the same:
[1, 2, 6, 40, 430, 6024, 101549, 1969744, ...].
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(m-1))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n-1))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(n-1))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alfa, where d = A360279, alfa = 0.311338934287018467072138011497837... and c = 0.1932932528309324180094... - Vaclav Kotesovec, Feb 03 2023

A360346 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 6, 82, 1724, 47223, 1555047, 58892186, 2496826094, 116434989450, 5900151126856, 322048641354617, 18810964989814291, 1169843128503194025, 77145176721564799777, 5376524285402806746719, 394887654026596322701724, 30489608056346314234108286, 2469347798211941105406473481
Offset: 0

Views

Author

Paul D. Hanna, Feb 05 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 6*x^2 + 82*x^3 + 1724*x^4 + 47223*x^5 + 1555047*x^6 + 58892186*x^7 + 2496826094*x^8 + 116434989450*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 7*x^2 + 101*x^3 + 2161*x^4 + 59544*x^5 + 1965132*x^6 + 74504861*x^7 + 3161424763*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that b(n) begin:
[1/1, 2/2, 21/3, 404/4, 10805/5, 357264/6, 13755924/7, 596038888/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  6,   82,  1724,  47223,  1555047,  58892186, ...];
n=1: [1, 2, 13,  176,  3648,  98878,  3231952, 121743878, ...];
n=2: [1, 3, 21,  283,  5790, 155319,  5039055, 188787837, ...];
n=3: [1, 4, 30,  404,  8169, 216924,  6985240, 260270488, ...];
n=4: [1, 5, 40,  540, 10805, 284096,  9079965, 336452690, ...];
n=5: [1, 6, 51,  692, 13719, 357264, 11333293, 417610542, ...];
n=6: [1, 7, 63,  861, 16933, 436884, 13755924, 504036226, ...];
n=7: [1, 8, 76, 1048, 20470, 523440, 16359228, 596038888, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(2*n+2))^(n+1):
n=0: [1, 1,   2,   13,   176,   3648, 98878, 3231952, ...];
n=1: [1, 2,   9,   68,   884,  17386,   454318,  14493920, ...];
n=2: [1, 3,  21,  190,  2508,  47406,  1190949,  36928479, ...];
n=3: [1, 4,  38,  404,  5585, 103464,  2504568,  75227160, ...];
n=4: [1, 5,  60,  735, 10805, 200001,  4698210, 136509465, ...];
n=5: [1, 6,  87, 1208, 19011, 357264,  8227591, 231595008, ...];
n=6: [1, 7, 119, 1848, 31199, 602427, 13755924, 376756199, ...];
n=7: [1, 8, 156, 2680, 48518, 970712, 22218108, 596038888, ...]; ...
to see that the main diagonals of the tables are the same.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(2*m+2))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 2.153902930660..., c = 0.01676305987174... - Vaclav Kotesovec, Feb 06 2023

A360344 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 4, 45, 820, 19820, 582007, 19812744, 760177656, 32275309743, 1497313010037, 75208566398988, 4062020902196139, 234638046113989856, 14432573619909530980, 941883830760366274935, 65013065172020161949992, 4733236746727327140204578, 362575149419405494321544263
Offset: 0

Views

Author

Paul D. Hanna, Feb 05 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 45*x^3 + 820*x^4 + 19820*x^5 + 582007*x^6 + 19812744*x^7 + 760177656*x^8 + 32275309743*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 5*x^2 + 58*x^3 + 1057*x^4 + 25471*x^5 + 746143*x^6 + 25364298*x^7 + 972602305*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(2*n))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that b(n) begin:
[1/1, 2/2, 15/3, 232/4, 5285/5, 152826/6, 5223001/7, 202914384/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  4,  45,   820,  19820,  582007,  19812744, ...];
n=1: [1, 2,  9,  98,  1746,  41640, 1212239,  41021862, ...];
n=2: [1, 3, 15, 160,  2790,  65643, 1894300,  63714729, ...];
n=3: [1, 4, 22, 232,  3965,  92028, 2632070,  87984416, ...];
n=4: [1, 5, 30, 315,  5285, 121011, 3429725, 113930075, ...];
n=5: [1, 6, 39, 410,  6765, 152826, 4291758, 141657348, ...];
n=6: [1, 7, 49, 518,  8421, 187726, 5223001, 171278801, ...];
n=7: [1, 8, 60, 640, 10270, 225984, 6228648, 202914384, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(2*n))^(n+1):
n=0: [1, 1,   0,    0,     0,      0,       0,         0, ...];
n=1: [1, 2,   5,   22,   218,   3724,   87245,   2516506, ...];
n=2: [1, 3,  15,   91,   888,  13929,  308182,   8594133, ...];
n=3: [1, 4,  30,  232,  2397,  35712,  742902,  19860536, ...];
n=4: [1, 5,  50,  470,  5285,  77631, 1530000,  38965400, ...];
n=5: [1, 6,  75,  830, 10245, 152826, 2900808,  70300080, ...];
n=6: [1, 7, 105, 1337, 18123, 280140, 5223001, 121085308, ...];
n=7: [1, 8, 140, 2016, 29918, 485240, 9053576, 202914384, ...]; ...
to see that the main diagonals of the tables are the same.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(2*m))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(2*n))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 1.1169011279372..., c = 0.052820142023857... - Vaclav Kotesovec, Feb 06 2023

A360347 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+3))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 7, 105, 2366, 68776, 2390230, 95166058, 4228436480, 206090296497, 10887958126763, 618187895371965, 37479711430699245, 2414492049517400164, 164626564026042206780, 11841796830661101527618, 896184803460067359995232, 71189783172592806474895908
Offset: 0

Views

Author

Paul D. Hanna, Feb 05 2023

Keywords

Comments

Sequences with g.f. A(x,k) such that [x^n] A(x,k)^(n+1) = [x^n] (1 + x*A(x,k)^(2*n+k))^(n+1) have a rate of growth: a(n) ~ c(k) * d^n * n! * n^alpha(k), where d = 3.93464558322824528799... (independent on k) and alpha(k) = 1.1169011279372... + k*0.518500901361... - Vaclav Kotesovec, Feb 06 2023

Examples

			G.f.: A(x) = 1 + x + 7*x^2 + 105*x^3 + 2366*x^4 + 68776*x^5 + 2390230*x^6 + 95166058*x^7 + 4228436480*x^8 + 206090296497*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 8*x^2 + 127*x^3 + 2927*x^4 + 85892*x^5 + 2998264*x^6 + 119665415*x^7 + 5325877575*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(2*n+3))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that b(n) begin:
[1/1, 2/2, 24/3, 508/4, 14635/5, 515352/6, 20987848/7, 957323320/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  7,  105,  2366,  68776,  2390230,  95166058, ...];
n=1: [1, 2, 15,  224,  4991, 143754,  4962161, 196572300, ...];
n=2: [1, 3, 24,  358,  7896, 225396,  7727644, 304572936, ...];
n=3: [1, 4, 34,  508, 11103, 314192, 10699244, 419541832, ...];
n=4: [1, 5, 45,  675, 14635, 410661, 13890275, 541873525, ...];
n=5: [1, 6, 57,  860, 18516, 515352, 17314836, 671984280, ...];
n=6: [1, 7, 70, 1064, 22771, 628845, 20987848, 810313190, ...];
n=7: [1, 8, 84, 1288, 27426, 751752, 24925092, 957323320, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(2*n+3))^(n+1):
n=0: [1, 1,   3,   24,   358,   7896,   225396,   7727644, ...];
n=1: [1, 2,  11,  100,  1465,  31070,   859367,  28808972, ...];
n=2: [1, 3,  24,  253,  3780,  77994,  2089024,  68277867, ...];
n=3: [1, 4,  42,  508,  7915, 161316,  4196916, 133476480, ...];
n=4: [1, 5,  65,  890, 14635, 298981,  7602705, 235213110, ...];
n=5: [1, 6,  93, 1424, 24858, 515352, 12914214, 389369448, ...];
n=6: [1, 7, 126, 2135, 39655, 842331, 20987848, 619044602, ...];
n=7: [1, 8, 164, 3048, 60250, 1320480, 32998388, 957323320, ...]; ...
to see that the main diagonals of the tables are the same.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(2*m+3))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+3))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(2*n+3))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 2.672403832022..., c = 0.0085435225111... - Vaclav Kotesovec, Feb 06 2023

A360342 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n-2))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 2, 20, 316, 6686, 173379, 5255624, 180911070, 6938866748, 292678301988, 13446616806957, 668017569348751, 35678261176871802, 2038906890461704040, 124171127134721710130, 8030684434410398312840, 549848454475826567644385, 39744302449387229743134043
Offset: 0

Views

Author

Paul D. Hanna, Feb 05 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 20*x^3 + 316*x^4 + 6686*x^5 + 173379*x^6 + 5255624*x^7 + 180911070*x^8 + 6938866748*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 3*x^2 + 27*x^3 + 417*x^4 + 8727*x^5 + 225018*x^6 + 6800714*x^7 + 233778499*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(2*n-2))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that b(n) begin:
[1/1, 2/2, 9/3, 108/4, 2085/5, 52362/6, 1575126/7, 54405712/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  2,  20,  316,  6686,  173379,  5255624, ...];
n=1: [1, 2,  5,  44,  676, 14084,  361794, 10897390, ...];
n=2: [1, 3,  9,  73, 1086, 22266,  566441, 16950588, ...];
n=3: [1, 4, 14, 108, 1553, 31312,  788620, 23442284, ...];
n=4: [1, 5, 20, 150, 2085, 41311, 1029745, 30401460, ...];
n=5: [1, 6, 27, 200, 2691, 52362, 1291355, 37859166, ...];
n=6: [1, 7, 35, 259, 3381, 64575, 1575126, 45848685, ...];
n=7: [1, 8, 44, 328, 4166, 78072, 1882884, 54405712, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(2*n-2))^(n+1):
n=0: [1, 1,  -2,   -1,   -32,   -519,  -11490,  -305967, ...];
n=1: [1, 2,   1,    0,     0,      0,       0,        0, ...];
n=2: [1, 3,   9,   28,   180,   2379,   47111,  1182009, ...];
n=3: [1, 4,  22,  108,   745,   8556,  153292,  3658316, ...];
n=4: [1, 5,  40,  265,  2085,  22706,  366450,  8157230, ...];
n=5: [1, 6,  63,  524,  4743,  52362,  781973, 16041192, ...];
n=6: [1, 7,  91,  910,  9415, 109536, 1575126, 29886445, ...];
n=7: [1, 8, 124, 1448, 16950, 211840, 3042820, 54405712, ...]; ...
to see that the main diagonals of the tables are the same.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(2*m-2))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n-2))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(2*n-2))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 0.0798993252137..., c = 0.118957192149397... - Vaclav Kotesovec, Feb 06 2023

A360343 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n-1))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 3, 31, 526, 11907, 328980, 10580531, 384937042, 15549217485, 688430225102, 33096289502982, 1715499922758709, 95339852384471586, 5655337634718941111, 356683962066445400017, 23840465113068534382248, 1683771696557415075462436, 125327912444852044066759399
Offset: 0

Views

Author

Paul D. Hanna, Feb 05 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 31*x^3 + 526*x^4 + 11907*x^5 + 328980*x^6 + 10580531*x^7 + 384937042*x^8 + 15549217485*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 4*x^2 + 41*x^3 + 687*x^4 + 15433*x^5 + 424524*x^6 + 13620842*x^7 + 495005025*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(2*n-1))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that {b(n)} begins:
[1/1, 2/2, 12/3, 164/4, 3435/5, 92598/6, 2971668/7, 108966736/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
  n=0: [1, 1,  3,  31,  526,  11907,  328980,  10580531, ...];
  n=1: [1, 2,  7,  68, 1123,  25052,  685891,  21923076, ...];
  n=2: [1, 3, 12, 112, 1800,  39555, 1072896,  34076544, ...];
  n=3: [1, 4, 18, 164, 2567,  55548, 1492336,  47093172, ...];
  n=4: [1, 5, 25, 225, 3435,  73176, 1946745,  61028770, ...];
  n=5: [1, 6, 33, 296, 4416,  92598, 2438866,  75942984, ...];
  n=6: [1, 7, 42, 378, 5523, 113988, 2971668,  91899578, ...];
  n=7: [1, 8, 52, 472, 6770, 137536, 3548364, 108966736, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(2*n-1))^(n+1):
  n=0: [1, 1,  -1,   -2,   -26,   -463,  -10778,   -303048, ...];
  n=1: [1, 2,   3,    8,    69,   1120,   24937,    683012, ...];
  n=2: [1, 3,  12,   55,   444,   6351,  132492,   3504654, ...];
  n=3: [1, 4,  26,  164,  1411,  18560,  357624,   9024812, ...];
  n=4: [1, 5,  45,  360,  3435,  43926,  785715,  18700710, ...];
  n=5: [1, 6,  69,  668,  7134,  92598, 1570420,  35086104, ...];
  n=6: [1, 7,  98, 1113, 13279, 179816, 2971668,  62645353, ...];
  n=7: [1, 8, 132, 1720, 22794, 327032, 5403036, 108966736, ...]; ...
to see that the main diagonals of the tables are the same.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(2*m-1))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n-1))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(2*n-1))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 0.5984002265754..., c = 0.08321697608093... - Vaclav Kotesovec, Feb 06 2023

A360337 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(3*n+1))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 7, 124, 3446, 125706, 5540958, 282129207, 16148101259, 1020687876920, 70377734170699, 5246775452965364, 420104327765022458, 35937961751407922101, 3270668852260460283730, 315546031669853942486219, 32173855061751806476275665, 3457696770952845858846954590
Offset: 0

Views

Author

Paul D. Hanna, Feb 06 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 7*x^2 + 124*x^3 + 3446*x^4 + 125706*x^5 + 5540958*x^6 + 282129207*x^7 + 16148101259*x^8 + 1020687876920*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 8*x^2 + 146*x^3 + 4083*x^4 + 149077*x^5 + 6569555*x^6 + 334401750*x^7 + 19137707066*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(3*n+1))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that b(n) begin:
[1/1, 2/2, 24/3, 584/4, 20415/5, 894462/6, 45986885/7, 2675214000/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  7,  124,  3446,  125706,  5540958,  282129207, ...];
n=1: [1, 2, 15,  262,  7189,  260040, 11396948,  577954822, ...];
n=2: [1, 3, 24,  415, 11250,  403521, 17583859,  888063051, ...];
n=3: [1, 4, 34,  584, 15651,  556696, 24118370, 1213065672, ...];
n=4: [1, 5, 45,  770, 20415,  720141, 31017985, 1553601145, ...];
n=5: [1, 6, 57,  974, 25566,  894462, 38301069, 1910335764, ...];
n=6: [1, 7, 70, 1197, 31129, 1080296, 45986885, 2283964852, ...];
n=7: [1, 8, 84, 1440, 37130, 1278312, 54095632, 2675214000, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(3*n+1))^(n+1):
n=0: [1, 1,   1,    7,   124,    3446,   125706,    5540958, ...];
n=1: [1, 2,   9,   76,  1252,   32742,  1150522,   49515052, ...];
n=2: [1, 3,  24,  253,  4179,  103866,  3499510,  146421240, ...];
n=3: [1, 4,  46,  584, 10061,  242520,  7836278,  317454824, ...];
n=4: [1, 5,  75, 1115, 20415,  487566, 15193230,  594390940, ...];
n=5: [1, 6, 111, 1892, 37119,  894462, 27139545, 1025356992, ...];
n=6: [1, 7, 154, 2961, 62412, 1538698, 45986885, 1682957396, ...];
n=7: [1, 8, 204, 4368, 98894, 2519232, 75032832, 2675214000, ...]; ...
to see that the main diagonals of the tables are the same.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(3*m+1))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(3*n+1))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(3*n+1))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alpha, where d = 5.7189630165873859806..., alpha = 1.4141427006501..., c = 0.027880568114272... - Vaclav Kotesovec, Feb 06 2023
Showing 1-8 of 8 results.