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 1-4 of 4 results.

A356500 Coefficients T(n,k) of x^n*y^k in A(x,y) for n >= 0, k = 0..3*n+1, where A(x,y) satisfies: y = Sum_{n=-oo..+oo} (-x)^(n^2) * A(x,y)^((n-1)^2), as an irregular triangle read by rows.

Original entry on oeis.org

0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 4, 0, 0, 0, 4, 0, 0, 6, 0, 0, 0, 28, 0, 0, 0, 22, 0, 3, 0, 0, 0, 84, 0, 0, 0, 219, 0, 0, 0, 140, 0, 0, 0, 0, 135, 0, 0, 0, 981, 0, 0, 0, 1807, 0, 0, 0, 969, 0, 0, 0, 120, 0, 0, 0, 2568, 0, 0, 0, 10764, 0, 0, 0, 15368, 0, 0, 0, 7084, 0, 0, 54, 0, 0, 0, 4284, 0, 0, 0, 38896, 0, 0, 0, 114240, 0, 0, 0, 133266, 0, 0, 0, 53820, 0, 9, 0, 0, 0, 4662, 0, 0, 0, 94390, 0, 0, 0, 525980, 0, 0, 0, 1187433, 0, 0, 0, 1171390, 0, 0, 0, 420732
Offset: 0

Views

Author

Paul D. Hanna, Aug 09 2022

Keywords

Comments

T(n, 3*n+1) = [x^n*y^(3*n+1)] A(x,y) = binomial(4*n, n)/(3*n + 1) = A002293(n) for n >= 0, where g.f. G(x) of A002293 satisfies: G(x) = 1 + x*G(x)^4.
T(4*n, 1) = A000716(n) for n >= 0 (nonzero terms in column 1).
T(4*n+3, 2) = [x^(4*n+3)*y^2] A(x,y) = 2 * A354655(n+1) for n >= 0, where A354655 equals column 2 of triangle A354650.
T(4*n+2, 3) = [x^(4*n+2)*y^3] A(x,y) = 4 * A354656(n+1) for n >= 0, where A354656 equals column 3 of triangle A354650.
T(2*n, 2*n+1) = A356504(n), for n >= 0.
T(2*n+1, 2*n) = A356505(n) for n >= 0.
T(3*n, n+1) = A356506(n) for n >= 0.
T(3*n+1, n) = A355365(n) where A355365 is the central terms of A355360 = A355360(2*n,n).
Sum_{k=0..3*n+1} T(n, k) = A354248(n) for n >= 0 (row sums).
Sum_{k=0..3*n+1} T(n, k) * 2^k = A356502(n) for n >= 0.
Sum_{k=0..3*n+1} T(n, k) * 3^k = A356503(n) for n >= 0.
Sum_{k=0..3*n+1} T(4*n+1-k, k) = A355872(n+1) for n >= 0 (nonzero antidiagonal sums).
SPECIFIC VALUES.
(V.1) 1 = A(x,y) at x = -exp(-Pi) and y = Pi^(1/4)/gamma(3/4).
(V.2) 1 = A(x,y) at x = -exp(-2*Pi) and y = Pi^(1/4)/gamma(3/4) * (6 + 4*sqrt(2))^(1/4)/2.
(V.3) 1 = A(x,y) at x = -exp(-3*Pi) and y = Pi^(1/4)/gamma(3/4) * (27 + 18*sqrt(3))^(1/4)/3.
(V.4) 1 = A(x,y) at x = -exp(-4*Pi) and y = Pi^(1/4)/gamma(3/4) * (8^(1/4) + 2)/4.
(V.5) 1 = A(x,y) at x = -exp(-sqrt(3)*Pi) and y = gamma(4/3)^(3/2)*3^(13/8)/(Pi*2^(2/3)).

Examples

			G.f.: A(x,y) = y + x*(1 + y^4) + x^2*(4*y^3 + 4*y^7) + x^3*(6*y^2 + 28*y^6 + 22*y^10) + x^4*(3*y + 84*y^5 + 219*y^9 + 140*y^13) + x^5*(135*y^4 + 981*y^8 + 1807*y^12 + 969*y^16) + x^6*(120*y^3 + 2568*y^7 + 10764*y^11 + 15368*y^15 + 7084*y^19) + x^7*(54*y^2 + 4284*y^6 + 38896*y^10 + 114240*y^14 + 133266*y^18 + 53820*y^22) + x^8*(9*y + 4662*y^5 + 94390*y^9 + 525980*y^13 + 1187433*y^17 + 1171390*y^21 + 420732*y^25) + x^9*(3250*y^4 + 160965*y^8 + 1670942*y^12 + 6640711*y^16 + 12167001*y^20 + 10399545*y^24 + 3362260*y^28) + ...
such that A = A(x,y) satisfies
y = ... + x^16*A^25 - x^9*A^16 + x^4*A^9 - x*A^4 + A - x + x^4*A - x^9*A^4 + x^16*A^9 - x^25*A^16 +- ... + (-x)^(n^2) * A(x,y)^((n-1)^2) + ...
This irregular triangle of coefficients T(n,k) of x^n*y^k in A(x,y) for n >= 0, k = 0..3*n+1, begins:
  n = 0: [0, 1];
  n = 1: [1, 0, 0, 0, 1];
  n = 2: [0, 0, 0, 4, 0, 0, 0, 4];
  n = 3: [0, 0, 6, 0, 0, 0, 28, 0, 0, 0, 22];
  n = 4: [0, 3, 0, 0, 0, 84, 0, 0, 0, 219, 0, 0, 0, 140];
  n = 5: [0, 0, 0, 0, 135, 0, 0, 0, 981, 0, 0, 0, 1807, 0, 0, 0, 969];
  n = 6: [0, 0, 0, 120, 0, 0, 0, 2568, 0, 0, 0, 10764, 0, 0, 0, 15368, 0, 0, 0, 7084];
  n = 7: [0, 0, 54, 0, 0, 0, 4284, 0, 0, 0, 38896, 0, 0, 0, 114240, 0, 0, 0, 133266, 0, 0, 0, 53820];
  n = 8: [0, 9, 0, 0, 0, 4662, 0, 0, 0, 94390, 0, 0, 0, 525980, 0, 0, 0, 1187433, 0, 0, 0, 1171390, 0, 0, 0, 420732];
  n = 9: [0, 0, 0, 0, 3250, 0, 0, 0, 160965, 0, 0, 0, 1670942, 0, 0, 0, 6640711, 0, 0, 0, 12167001, 0, 0, 0, 10399545, 0, 0, 0, 3362260];
  ...
Reading this triangle by nonzero antidiagonals [x^(4*n+1-k)*y^k] A(x,y) for n >= 0, k = 0..3*n+1, yields triangle A356501:
  [1, 1];
  [0, 3, 6, 4, 1];
  [0, 9, 54, 120, 135, 84, 28, 4];
  [0, 22, 294, 1360, 3250, 4662, 4284, 2568, 981, 219, 22];
  [0, 51, 1260, 10120, 41405, 103020, 170324, 196172, 160965, 94390, 38896, 10764, 1807, 140];
  [0, 108, 4590, 58380, 368145, 1404102, 3587696, 6515712, 8715465, 8763645, 6684744, 3863496, 1670942, 525980, 114240, 15368, 969];
  ...
		

Crossrefs

Programs

  • PARI
    {T(n,k) = my(A=[y],M); for(i=1,n, A = concat(A,0); M = ceil(sqrt(n+1));
    A[#A] = -polcoeff( sum(m=-M,M, (-x)^(m^2)*Ser(A)^((m-1)^2)), #A-1)); polcoeff(A[n+1],k,y) }
    for(n=0,9, for(k=0,3*n+1, print1(T(n,k),", "));print(""))

Formula

G.f. A(x,y) = Sum_{n>=0} Sum_{k=0..3*n+1} T(n,k) * x^n * y^k satisfies:
(1) y = Sum_{n=-oo..+oo} (-x)^(n^2) * A(x,y)^((n+1)^2).
(2) y = A(x,y) * Product_{n>=1} (1 - x^(2*n)*A(x,y)^(2*n)) * (1 - x^(2*n-1)*A(x,y)^(2*n+1)) * (1 - x^(2*n-1)*A(x,y)^(2*n-3)), by the Jacobi triple product identity.
(3) y = (-x) * Product_{n>=1} (1 - x^(2*n)*A(x,y)^(2*n)) * (1 - x^(2*n+1)*A(x,y)^(2*n-1)) * (1 - x^(2*n-3)*A(x,y)^(2*n-1)), by the Jacobi triple product identity.
(4) y = A(x, F(x,y)) where F(x,y) = Sum_{n=-oo..+oo} (-x)^(n^2) * y^((n-1)^2).
(5) 1 = A(x, theta_4(x)) where theta_4(x) = 1 + 2*Sum_{n>=1} (-1)^n * x^(n^2) is a Jacobi theta function.

A356502 G.f. A(x) satisfies: 2 = Sum_{n=-oo..+oo} (-x)^(n^2) * A(x)^((n-1)^2).

Original entry on oeis.org

2, 17, 544, 24344, 1261702, 71159152, 4240009152, 262584135640, 16734002688722, 1090225325371424, 72285357987696768, 4861658409827006872, 330874470176939132844, 22744684876060771599568, 1576898258893213475814464, 110136698483814852518084528, 7742091796859524187452564262
Offset: 0

Views

Author

Paul D. Hanna, Aug 09 2022

Keywords

Examples

			G.f.: A(x) = 2 + 17*x + 544*x^2 + 24344*x^3 + 1261702*x^4 + 71159152*x^5 + 4240009152*x^6 + 262584135640*x^7 + 16734002688722*x^8 + ...
such that A = A(x) satisfies
2 = ... + x^16*A^25 - x^9*A^16 + x^4*A^9 - x*A^4 + A - x + x^4*A - x^9*A^4 + x^16*A^9 - x^25*A^16 +- ... + (-x)^(n^2) * A(x,y)^((n-1)^2) + ...
		

Crossrefs

Programs

  • Mathematica
    (* Calculation of constant d: *) 1/r /. FindRoot[{k == r^4*s^2 * QPochhammer[1/(r^3*s), r^2*s^2] * QPochhammer[r/s, r^2*s^2] * QPochhammer[r^2*s^2, r^2*s^2]/((r - s)*(-1 + r^3*s)), 1/r^3*(k*(1 + r^4 - 2*r/s) + 2*r^6*s^3*QPochhammer[r/s, r^2*s^2] * QPochhammer[r^2*s^2, r^2*s^2]* Derivative[0, 1][QPochhammer][1/(r^3*s), r^2*s^2] + 2*k*r^2*(r - s)*s*(-1 + r^3*s) * Derivative[0, 1][QPochhammer][r/s, r^2*s^2]/ QPochhammer[r/s, r^2*s^2] + 1/s*k*(r - s)*(-1 + r^3*s) * (1/ Log[r^2*s^2]*(-2*QPolyGamma[0, 1, r^2*s^2] + QPolyGamma[0, Log[1/(r^3*s)] / Log[r^2*s^2], r^2*s^2] + QPolyGamma[0, Log[r/s] / Log[r^2*s^2], r^2*s^2]) + 2*r^2*s^2 * Derivative[0, 1][QPochhammer][r^2*s^2, r^2*s^2] / QPochhammer[r^2*s^2, r^2*s^2])) == 0} /. k -> 2, {r, 1/75}, {s, 2}, WorkingPrecision -> 70] (* Vaclav Kotesovec, Jan 18 2024 *)
  • PARI
    {a(n) = my(A=[2],M); for(i=1,n, A = concat(A,0); M = ceil(sqrt(n+1));
    A[#A] = -polcoeff( sum(m=-M,M, (-x)^(m^2)*Ser(A)^((m-1)^2)), #A-1)); H=A; A[n+1]}
    for(n=0,20,print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) 2 = Sum_{n=-oo..+oo} (-x)^(n^2) * A(x)^((n+1)^2).
(2) 2 = A(x) * Product_{n>=1} (1 - x^(2*n)*A(x)^(2*n)) * (1 - x^(2*n-1)*A(x)^(2*n+1)) * (1 - x^(2*n-1)*A(x)^(2*n-3)), by the Jacobi triple product identity.
(3) 2 = (-x) * Product_{n>=1} (1 - x^(2*n)*A(x)^(2*n)) * (1 - x^(2*n+1)*A(x)^(2*n-1)) * (1 - x^(2*n-3)*A(x)^(2*n-1)), by the Jacobi triple product identity.
a(n) ~ c * d^n / n^(3/2), where d = 77.309779325704779292317107617559471210592218708634855530355675234... and c = 0.31219183409397424726366930735250286274022579073644627976468... - Vaclav Kotesovec, Mar 19 2023

A356505 a(n) = A356500(2*n+1, 2*n) for n >= 0.

Original entry on oeis.org

1, 6, 135, 4284, 160965, 6684744, 296679006, 13805453160, 665611197093, 32988925715610, 1671463040525586, 86231285273788180, 4516133521439246825, 239551205985729110664, 12846081444122599438850, 695428535332816056597520, 37960416340437631597631877
Offset: 0

Views

Author

Paul D. Hanna, Aug 09 2022

Keywords

Comments

Triangle A356500 describes the coefficients in G(x,y) that satisfies: y = Sum_{n=-oo..+oo} (-x)^(n^2) * G(x,y)^((n-1)^2).

Crossrefs

Programs

  • PARI
    {A356500(n,k) = my(A=[y],M); for(i=1,n, A = concat(A,0); M = ceil(sqrt(n+1));
    A[#A] = -polcoeff( sum(m=-M,M, (-x)^(m^2)*Ser(A)^((m-1)^2)), #A-1)); polcoeff(A[n+1],k,y) }
    for(n=0,20, print1( A356500(2*n+1, 2*n),", "))

A356506 a(n) = A356500(3*n, n+1) for n >= 0.

Original entry on oeis.org

1, 6, 120, 3250, 103020, 3587696, 133101836, 5167564380, 207615129579, 8567305854998, 361201849117032, 15498967122249676, 674906101555736960, 29757755664623031984, 1326196334421645347368, 59655785739373960058296, 2705420198806474232850741
Offset: 0

Views

Author

Paul D. Hanna, Aug 09 2022

Keywords

Comments

Triangle A356500 describes the coefficients in G(x,y) that satisfies: y = Sum_{n=-oo..+oo} (-x)^(n^2) * G(x,y)^((n-1)^2).

Crossrefs

Programs

  • PARI
    {A356500(n,k) = my(A=[y],M); for(i=1,n, A = concat(A,0); M = ceil(sqrt(n+1));
    A[#A] = -polcoeff( sum(m=-M,M, (-x)^(m^2)*Ser(A)^((m-1)^2)), #A-1)); polcoeff(A[n+1],k,y) }
    for(n=0,20, print1( A356500(3*n, n+1),", "))
Showing 1-4 of 4 results.