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 1-5 of 5 results.

A360345 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, 5, 62, 1214, 31269, 973485, 34993597, 1412846469, 62926155294, 3053566438307, 160005640085764, 8992869671470675, 539298198547460797, 34364052537634696986, 2318526571023659653665, 165143229278977841236029, 12385688813185721332861730, 975844100444710104444582984
Offset: 0

Views

Author

Paul D. Hanna, Feb 05 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 62*x^3 + 1214*x^4 + 31269*x^5 + 973485*x^6 + 34993597*x^7 + 1412846469*x^8 + 62926155294*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 6*x^2 + 78*x^3 + 1543*x^4 + 39810*x^5 + 1239252*x^6 + 44537587*x^7 + 1798314384*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) begin:
[1/1, 2/2, 18/3, 312/4, 7715/5, 238860/6, 8674764/7, 356300696/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  5,  62,  1214,  31269,   973485,  34993597, ...];
n=1: [1, 2, 11, 134,  2577,  65586,  2025492,  72397390, ...];
n=2: [1, 3, 18, 217,  4104, 103212,  3161648, 112357788, ...];
n=3: [1, 4, 26, 312,  5811, 144428,  4387978, 155030276, ...];
n=4: [1, 5, 35, 420,  7715, 189536,  5710930, 200579975, ...];
n=5: [1, 6, 45, 542,  9834, 238860,  7137401, 249182232, ...];
n=6: [1, 7, 56, 679, 12187, 292747,  8674764, 301023241, ...];
n=7: [1, 8, 68, 832, 14794, 351568, 10330896, 356300696, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(2*n+1))^(n+1):
n=0: [1, 1,   1,    5,    62,   1214,    31269,    973485, ...];
n=1: [1, 2,   7,   42,   479,   8750,   216258,   6562156, ...];
n=2: [1, 3,  18,  136,  1560,  26895,   633608,  18631701, ...];
n=3: [1, 4,  34,  312,  3767,  62888,  1412530,  40031684, ...];
n=4: [1, 5,  55,  595,  7715, 128041,  2763270,  75234930, ...];
n=5: [1, 6,  81, 1010, 14172, 238860,  5016947, 131313798, ...];
n=6: [1, 7, 112, 1582, 24059, 418166,  8674764, 219340759, ...];
n=7: [1, 8, 148, 2336, 38450, 696216, 14466592, 356300696, ...]; ...
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 = 1.635402029299..., c = 0.0308525091280143... - Vaclav Kotesovec, Feb 06 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
Showing 1-5 of 5 results.