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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.

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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];
  ...

Links

Crossrefs

Cf. A354248, A356502, A356503, A356504, A356505, A355872.
Cf. A354655, A354656, A355365, A000716, A002293.
Cf. related tables: A356501, A355350, A355360, A355870.

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.

A356501 Coefficients T(n,k) of x^(4*n+1-k)*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

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
Offset: 0

Views

Author

Paul D. Hanna, Aug 09 2022

Keywords

Comments

Row sums equal A355872.
Alternating row sums equals zero for all rows.
Rightmost border equals [x^n*y^(3*n+1)] A(x,y) = A002293(n) = binomial(4*n, n)/(3*n + 1).
This triangle may be formed from the nonzero antidiagonals of triangle A356500; see main entry A356500 for further formulas for the coefficients in g.f. A(x,y).

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 triangle of coefficients of x^(4*n+1-k)*y^k in A(x,y) for n >= 0, k = 0..3*n+1, begins:
n = 0: [1, 1];
n = 1: [0, 3, 6, 4, 1];
n = 2: [0, 9, 54, 120, 135, 84, 28, 4];
n = 3: [0, 22, 294, 1360, 3250, 4662, 4284, 2568, 981, 219, 22];
n = 4: [0, 51, 1260, 10120, 41405, 103020, 170324, 196172, 160965, 94390, 38896, 10764, 1807, 140];
n = 5: [0, 108, 4590, 58380, 368145, 1404102, 3587696, 6515712, 8715465, 8763645, 6684744, 3863496, 1670942, 525980, 114240, 15368, 969];
n = 6: [0, 221, 14952, 282948, 2578147, 14039250, 51126740, 133101836, 258436719, 384735141, 446971668, 409367712, 296679006, 169713208, 75904032, 26050408, 6640711, 1187433, 133266, 7084];
...

Links

Crossrefs

Cf. A355872 (row sums), A356500 (main entry), A002293 (right border), A000716 (column 1).

Programs

  • PARI
    {T(n,k) = my(A=[y],M); for(i=1,4*n+1, A = concat(A,0); M = ceil(sqrt(4*n+1));
A[#A] = -polcoeff( sum(m=-M,M, (-x)^(m^2)*Ser(A)^((m-1)^2)), #A-1)); polcoeff(A[4*n+2-k],k,y)}
for(n=0,7, 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.
Showing 1-2 of 2 results.

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