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.

A375440 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(2*n, 2*k) * ([x^k] A(x)^n) for n >= 1.

Original entry on oeis.org

1, 1, 5, 65, 1593, 61953, 3476813, 265517441, 26492540401, 3349218907137, 523572565927509, 99215376614955457, 22415450137196941993, 5953820173628518544385, 1837040977427662958973341, 651657636773935012586716929, 263375512326578915885862469601, 120319850003020550647400856678401
Offset: 0

Views

Author

Paul D. Hanna, Sep 11 2024

Keywords

Comments

Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+k, 2*k) * ([x^k] C(x)^n) for n >= 1 is satisfied by the Catalan function C(x) = 1 + x*C(x)^2 (A000108), where coefficient [x^k] C(x)^n = binomial(n+2*k-1,k)*n/(n+k).

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 65*x^3 + 1593*x^4 + 61953*x^5 + 3476813*x^6 + 265517441*x^7 + 26492540401*x^8 + ...
RELATED TABLES.
The table of coefficients of x^k in A(x)^n begins:
  n=1: [1, 1,  5,  65,  1593,  61953,  3476813, ...];
  n=2: [1, 2, 11, 140,  3341, 127742,  7097687, ...];
  n=3: [1, 3, 18, 226,  5259, 197637, 10869476, ...];
  n=4: [1, 4, 26, 324,  7363, 271928, 14799444, ...];
  n=5: [1, 5, 35, 435,  9670, 350926, 18895290, ...];
  n=6: [1, 6, 45, 560, 12198, 434964, 23165174, ...];
  ...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(2*n, 2*k) * ([x^k] A(x)^n).
Using the coefficients in the table above, we see that
  n=1: 0 = 1*1 - 1*1;
  n=2: 0 = 1*1 - 6*2 + 1*11;
  n=3: 0 = 1*1 - 15*3 + 15*18 - 1*226;
  n=4: 0 = 1*1 - 28*4 + 70*26 - 28*324 + 1*7363;
  n=5: 0 = 1*1 - 45*5 + 210*35 - 210*435 + 45*9670 - 1*350926;
  n=6: 0 = 1*1 - 66*6 + 495*45 - 924*560 + 495*12198 - 66*434964 + 1*23165174;
  ...
The triangle A086645(n,k) = binomial(2*n, 2*k) begins:
  n=0: 1;
  n=1: 1,  1;
  n=2: 1,  6,   1;
  n=3: 1, 15,  15,   1;
  n=4: 1, 28,  70,  28,  1;
  n=5: 1, 45, 210, 210,  45,  1;
  n=6: 1, 66, 495, 924, 495, 66, 1;
  ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1],m); for(i=1, n, A=concat(A, 0); m=#A-1;
    A[m+1] = sum(k=0, m, (-1)^(m-k+1) * binomial(2*m, 2*k) * polcoef(Ser(A)^m, k) )/m ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

a(n) ~ c * 2^(4*n) * n^(2*n + 1/2) / (Pi^(2*n) * exp(2*n)), where c = 7.23682343848882192289996... - Vaclav Kotesovec, Sep 12 2024

A375441 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(3*n-2*k, k) * ([x^k] A(x)^n) for n >= 1.

Original entry on oeis.org

1, 1, 3, 27, 520, 17461, 924123, 72565316, 8092491188, 1237738519836, 252223179026317, 66826143258640668, 22547253821858556366, 9516722710620123685768, 4947980149610424524104600, 3126554347854062023615490499, 2372761989077914432028426533800, 2140079932579758851404630992687571
Offset: 0

Views

Author

Paul D. Hanna, Sep 11 2024

Keywords

Comments

Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+2*k, 3*k) * ([x^k] D(x)^n) for n >= 1 is satisfied by the function D(x) = 1 + x*D(x)^3 (g.f. of A001764), where coefficient [x^k] D(x)^n = binomial(n+3*k-1, k)*n/(n+2*k).

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 27*x^3 + 520*x^4 + 17461*x^5 + 924123*x^6 + 72565316*x^7 + 8092491188*x^8 + ...
RELATED TABLES.
The table of coefficients of x^k in A(x)^n begins:
  n=1: [1, 1,  3,  27,  520,  17461,  924123, ...];
  n=2: [1, 2,  7,  60, 1103,  36124, 1887017, ...];
  n=3: [1, 3, 12, 100, 1758,  56097, 2890755, ...];
  n=4: [1, 4, 18, 148, 2495,  77500, 3937572, ...];
  n=5: [1, 5, 25, 205, 3325, 100466, 5029880, ...];
  n=6: [1, 6, 33, 272, 4260, 125142, 6170284, ...];
  ...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(3*n-2*k, k) * ([x^k] A(x)^n).
Using the coefficients in the table above, we see that
  n=1: 0 = 1*1 - 1*1;
  n=2: 0 = 1*1 - 4*2 + 1*7;
  n=3: 0 = 1*1 - 7*3 + 10*12 - 1*100;
  n=4: 0 = 1*1 - 10*4 + 28*18 - 20*148 + 1*2495;
  n=5: 0 = 1*1 - 13*5 + 55*25 - 84*205 + 35*3325 - 1*100466;
  n=6: 0 = 1*1 - 16*6 + 91*33 - 220*272 + 210*4260 - 56*125142 + 1*6170284;
  ...
The triangle A193636(n,k) = binomial(3*n-2*k, k) begins:
  n=0: 1;
  n=1: 1, 1;
  n=2: 1, 4, 1;
  n=3: 1, 7, 10, 1;
  n=4: 1, 10, 28, 20, 1;
  n=5: 1, 13, 55, 84, 35, 1;
  n=6: 1, 16, 91, 220, 210, 56, 1;
  ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1],m); for(i=1, n, A=concat(A, 0); m=#A-1;
    A[#A] = sum(k=0, m, (-1)^(m-k+1) * binomial(3*m-2*k, k) * polcoef(Ser(A)^m, k) )/m ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

a(n) ~ c * d^n * n!^3 * n^alpha, where d = 0.1579852929267375678916376580224..., alpha = 2.6601429516008505168108..., c = 0.86048778713891683578001... - Vaclav Kotesovec, Sep 12 2024

A375442 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(4*n-3*k, k) * ([x^k] A(x)^n) for n >= 1.

Original entry on oeis.org

1, 1, 4, 58, 2040, 141471, 17414299, 3548205700, 1133401180600, 542919997056985, 375875913182499803, 364645747772536556398, 482860134802496098766740, 853352199545631748454314991, 1973943714691108570803508891357, 5875562740836479708875221002947800, 22169669117423382437537944364347343144
Offset: 0

Views

Author

Paul D. Hanna, Sep 12 2024

Keywords

Comments

Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+k, 2*k) * ([x^k] C(x)^n) for n >= 1 is satisfied by the Catalan function C(x) = 1 + x*C(x)^2 (A000108), where coefficient [x^k] C(x)^n = binomial(n+2*k-1,k)*n/(n+k).
Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+2*k, 3*k) * ([x^k] D(x)^n) for n >= 1 is satisfied by the function D(x) = 1 + x*D(x)^3 (g.f. of A001764), where coefficient [x^k] D(x)^n = binomial(n+3*k-1, k)*n/(n+2*k).

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 58*x^3 + 2040*x^4 + 141471*x^5 + 17414299*x^6 + 3548205700*x^7 + 1133401180600*x^8 + ...
RELATED TABLES.
The table of coefficients of x^k in A(x)^n begins:
  n=1: [1, 1,  4,  58,  2040, 141471,  17414299, ...];
  n=2: [1, 2,  9, 124,  4212, 287486,  35131224, ...];
  n=3: [1, 3, 15, 199,  6528, 438267,  53158351, ...];
  n=4: [1, 4, 22, 284,  9001, 594052,  71503584, ...];
  n=5: [1, 5, 30, 380, 11645, 755096,  90175175, ...];
  n=6: [1, 6, 39, 488, 14475, 921672, 109181745, ...];
  ...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(4*n-3*k, k) * ([x^k] A(x)^n).
Using the coefficients in the table above, we see that
  n=1: 0 = 1*1 - 1*1;
  n=2: 0 = 1*1 - 5*2 + 1*9;
  n=3: 0 = 1*1 - 9*3 + 15*15 - 1*199;
  n=4: 0 = 1*1 - 13*4 + 45*22 - 35*284 + 1*9001;
  n=5: 0 = 1*1 - 17*5 + 91*30 - 165*380 + 70*11645 - 1*755096;
  n=6: 0 = 1*1 - 21*6 + 153*39 - 455*488 + 495*14475 - 126*921672 + 1*109181745;
  ...
The triangle T(n,k) = binomial(4*m-3*k, k) begins:
  n=0: 1;
  n=1: 1, 1;
  n=2: 1, 5, 1;
  n=3: 1, 9, 15, 1;
  n=4: 1, 13, 45, 35, 1;
  n=5: 1, 17, 91, 165, 70, 1;
  n=6: 1, 21, 153, 455, 495, 126, 1;
  ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1],m); for(i=1, n, A=concat(A, 0); m=#A-1;
    A[#A] = sum(k=0, m, (-1)^(m-k+1) * binomial(4*m-3*k, k) * polcoef(Ser(A)^m, k) )/m ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

From Vaclav Kotesovec, Sep 12 2024: (Start)
a(n) ~ c * d^n * n!^4 * n^alpha, where d = 4/Pi^4, alpha = 3*(1 + Pi*tanh(Pi/2))/2 = 5.821978559932543777976797..., c = 0.0936210705524085151687305...
Equivalently, a(n) ~ c * 2^(2*n) * n^(4*n + (7 + 3*Pi*tanh(Pi/2))/2) / (exp(4*n) * Pi^(4*n)), where c = 3.69601171983499372807019... (End)
Showing 1-3 of 3 results.