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.

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A361550 Expansion of g.f. A(x,y) satisfying x*y = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x,y)^(3*n) - 1/A(x,y)^(3*n+1)), as a triangle read by rows.

Original entry on oeis.org

1, 0, 1, 0, 5, 1, 0, 18, 10, 1, 0, 55, 61, 20, 1, 0, 149, 290, 215, 35, 1, 0, 371, 1172, 1660, 555, 56, 1, 0, 867, 4212, 10311, 5850, 1254, 84, 1, 0, 1923, 13833, 54688, 47460, 17773, 2555, 120, 1, 0, 4086, 42262, 256815, 319409, 188300, 46844, 4810, 165, 1, 0, 8374, 121625, 1093790, 1864445, 1621116, 621915, 111348, 8505, 220, 1, 0, 16634, 332764, 4297370, 9717550, 11913160, 6557572, 1818022, 243795, 14290, 286, 1
Offset: 0

Views

Author

Paul D. Hanna, Mar 19 2023

Keywords

Comments

A359920(n) = Sum_{k=0..n} T(n,k) for n >= 0.
A361552(n) = Sum_{k=0..n} T(n,k) * 2^k for n >= 0.
A361553(n) = Sum_{k=0..n} T(n,k) * 3^k for n >= 0.
A361554(n) = Sum_{k=0..n} T(n,k) * 4^k for n >= 0.
A361555(n) = Sum_{k=0..n} T(n,k) * 5^k for n >= 0.
A361556(n) = T(2*n,n) for n >= 0.
A360191(n) = T(n+1,1) for n >= 0.
A361535(n) = T(n+2,2) for n >= 0.

Examples

			G.f.: A(x,y) = 1 + y*x + (5*y + y^2)*x^2 + (18*y + 10*y^2 + y^3)*x^3 + (55*y + 61*y^2 + 20*y^3 + y^4)*x^4 + (149*y + 290*y^2 + 215*y^3 + 35*y^4 + y^5)*x^5 + (371*y + 1172*y^2 + 1660*y^3 + 555*y^4 + 56*y^5 + y^6)*x^6 + (867*y + 4212*y^2 + 10311*y^3 + 5850*y^4 + 1254*y^5 + 84*y^6 + y^7)*x^7 + (1923*y + 13833*y^2 + 54688*y^3 + 47460*y^4 + 17773*y^5 + 2555*y^6 + 120*y^7 + y^8)*x^8 + (4086*y + 42262*y^2 + 256815*y^3 + 319409*y^4 + 188300*y^5 + 46844*y^6 + 4810*y^7 + 165*y^8 + y^9)*x^9 + (8374*y + 121625*y^2 + 1093790*y^3 + 1864445*y^4 + 1621116*y^5 + 621915*y^6 + 111348*y^7 + 8505*y^8 + 220*y^9 + y^10)*x^10 + ...
This triangle of coefficients T(n,k) of x^n*y^k in g.f. A(x,y) begins:
1;
0, 1;
0, 5, 1;
0, 18, 10, 1;
0, 55, 61, 20, 1;
0, 149, 290, 215, 35, 1;
0, 371, 1172, 1660, 555, 56, 1;
0, 867, 4212, 10311, 5850, 1254, 84, 1;
0, 1923, 13833, 54688, 47460, 17773, 2555, 120, 1;
0, 4086, 42262, 256815, 319409, 188300, 46844, 4810, 165, 1;
0, 8374, 121625, 1093790, 1864445, 1621116, 621915, 111348, 8505, 220, 1;
0, 16634, 332764, 4297370, 9717550, 11913160, 6557572, 1818022, 243795, 14290, 286, 1;
0, 32152, 871641, 15771148, 46148620, 77162284, 58002140, 23152872, 4811721, 499180, 23012, 364, 1;
...

Links

Crossrefs

Cf. A359920 (y=1), A361552 (y=2), A361553 (y=3), A361554 (y=4), A361555 (y=5).
Cf. A360191 (column 1), A361535 (column 2), A361556 (central terms).
Cf. A361050 (related triangle).

Programs

  • PARI
    {T(n,k) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(x*y - prod(m=1, #A, (1 - x^m) * (1 - x^m*Ser(A)) * (1 - x^(m-1)/Ser(A)) * (1 - x^(2*m-1)*Ser(A)^2) * (1 - x^(2*m-1)/Ser(A)^2) ), #A-1) );
polcoeff(polcoeff(Ser(A),n,x),k,y)}
for(n=0, 12, for(k=0,n, print1(T(n,k), ", "));print(""))

Formula

G.f. A(x,y) = Sum_{n>=0} Sum_{k=0..n} T(n,k)*x^n*y^k satisfies the following.
(1) x*y = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x,y)^(3*n) - 1/A(x,y)^(3*n+1)).
(2) x*y = Sum_{n=-oo..+oo} x^(n*(3*n-1)/2) * A(x,y)^(3*n) * (x^n - 1/A(x,y)).
(3) x*y = Product_{n>=1} (1 - x^n) * (1 - x^n*A(x,y)) * (1 - x^(n-1)/A(x,y)) * (1 - x^(2*n-1)*A(x,y)^2) * (1 - x^(2*n-1)/A(x,y)^2), by the Watson quintuple product identity.
(4) Sum_{n>=0} T(n+1,1) * x^n = 1 / Product_{n>=1} (1 - x^n)^3 * (1 - x^(2*n-1))^2, which is the g.f. of A360191.
(5) Sum_{n>=0} T(n+2,2) * x^n = 1 / Product_{n>=1} (1 - x^n)^6 * (1 - x^(2*n-1))^4, which is the g.f. of A361535.

A359914 a(n) = coefficient of x^n in A(x) such that 2 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(3*n-1)/2) * A(x)^(3*n) * (1 + x^n*A(x)).

Original entry on oeis.org

1, 2, 4, 30, 154, 1078, 7046, 50766, 364268, 2713444, 20384884, 155954760, 1204192106, 9400024042, 73945396990, 586088682472, 4673927031694, 37484566094970, 302098932029282, 2445538771089012, 19875632898821430, 162118004651048048, 1326658157736876148
Offset: 0

Views

Author

Paul D. Hanna, Jan 23 2023

Keywords

Comments

Conjecture: a(n)/2 == A000041(n) (mod 2) for n >= 1; that is, a(n) is even, and a(n)/2 has the same parity as the number of partitions of n, for n >= 1.
Conjecture: a(n) == A361552(n+1) (mod 4) for n >= 0. - Paul D. Hanna, Mar 19 2023

Examples

			G.f.: A(x) = 1 + 2*x + 4*x^2 + 30*x^3 + 154*x^4 + 1078*x^5 + 7046*x^6 + 50766*x^7 + 364268*x^8 + 2713444*x^9 + 20384884*x^10 + ...
where A = A(x) satisfies the doubly infinite sum
2 = ... + x^26/A^12*(1 + 1/x^4*A) - x^15/A^9*(1 + 1/x^3*A) + x^7/A^6*(1 + 1/x^2*A) - x^2/A^3*(1 + 1/x^1*A) + x^0*A^0*(1 + x^0*A) - x^1*A^3*(1 + x^1*A) + x^5*A^6*(1 + x^2*A) - x^12*A^9*(1 + x^3*A) + x^22*A^12*(1 + x^4*A) + ... + (-1)^n*x^(n*(3*n+1)/2)*A^(3*n)*(1 + x^n*A) + ...
also, by the Watson quintuple product identity,
2 = (1-x)*(1+1*A)*(1+x/A)*(1-x*A^2)*(1-x/A^2) * (1-x^2)*(1+x*A)*(1+x^2/A)*(1-x^3*A^2)*(1-x^3/A^2) * (1-x^3)*(1+x^2*A)*(1+x^3/A)*(1-x^5*A^2)*(1-x^5/A^2) * (1-x^4)*(1+x^3*A)*(1+x^4/A)*(1-x^7*A^2)*(1-x^7/A^2) * ...
SPECIFIC VALUES.
A(x) at x = 100/876 diverges.
A(100/877) = 1.557056751214068970380867667963285879403994350720494...
A(1/9) = 1.450191456209956107571253359997937360795442585014595870...
A(1/10) = 1.324252801492679846747365280925526932201317768972870665...

Links

Crossrefs

Cf. A359920, A359915, A359916, A359919, A359921, A359924.
Cf. A361552, A040051, A000041.

Programs

  • Mathematica
    (* Calculation of constant d: *) 1/r /. FindRoot[{r^3 * s^3 * QPochhammer[r] * QPochhammer[1/(r*s^2), r^2] * QPochhammer[-1/s, r] * QPochhammer[-s/r, r] *  QPochhammer[s^2/r, r^2] / ((1 + s)*(r + s)*(-r + s^2)*(-1 + r*s^2)) == -2, (-3*r^2 - 2*r*(1 + r)*s + r^3*s^2 - s^4 + 2*r*(1 + r)*s^5 + 3*r*s^6) * Log[r] + (1 + s)*(r + s)*(r - s^2)*(-1 + r*s^2) * (QPolyGamma[0, Log[-1/s]/Log[r], r] + QPolyGamma[0, -1/2 - Log[s]/Log[r], r^2] - QPolyGamma[0, -1/2 + Log[s]/Log[r], r^2] - QPolyGamma[0, Log[-s/r]/Log[r], r]) == 0}, {r, 1/8}, {s, 2}, WorkingPrecision -> 70] (* Vaclav Kotesovec, Jan 19 2024 *)
  • PARI
    /* Using the doubly infinite series */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(2 - sum(m=-#A, #A, (-1)^m * x^(m*(3*m-1)/2) * Ser(A)^(3*m) * (1 + x^m*Ser(A)) ), #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    /* Using the quintuple product */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(2 - prod(m=1, #A, (1 - x^m) * (1 + x^(m-1)*Ser(A)) * (1 + x^m/Ser(A)) * (1 - x^(2*m-1)*Ser(A)^2) * (1 - x^(2*m-1)/Ser(A)^2) ), #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following.
(1) 2 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(3*n-1)/2) * A(x)^(3*n) * (1 + x^n*A(x)).
(2) 2 = Product_{n>=1} (1 - x^n) * (1 + x^(n-1)*A(x)) * (1 + x^n/A(x)) * (1 - x^(2*n-1)*A(x)^2) * (1 - x^(2*n-1)/A(x)^2), by the Watson quintuple product identity.
a(n) ~ c * d^n / n^(3/2), where d = 8.762505108391427770669887... and c = 0.25785454119524349137288... - Vaclav Kotesovec, Jan 24 2023
Formula (2) can be rewritten as the functional equation QPochhammer(x) * QPochhammer(-y/x, x)/(1 + y/x) * QPochhammer(-1/y, x)/(1 + 1/y) * QPochhammer(y^2/x, x^2)/(1 - y^2/x) * QPochhammer(1/(x*y^2), x^2)/(1 - 1/(x*y^2)) = 2. - Vaclav Kotesovec, Jan 19 2024

A361553 Expansion of g.f. A(x) satisfying 3*x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).

Original entry on oeis.org

1, 3, 24, 171, 1335, 11940, 115773, 1160901, 11901537, 124726644, 1332688035, 14455451526, 158660036535, 1758835084221, 19667067522966, 221573079684087, 2512635069594897, 28656903391830291, 328500210705228867, 3782806859877522522, 43738575934977450465
Offset: 0

Views

Author

Paul D. Hanna, Mar 19 2023

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 24*x^2 + 171*x^3 + 1335*x^4 + 11940*x^5 + 115773*x^6 + 1160901*x^7 + 11901537*x^8 + 124726644*x^9 + ...
where A = A(x) satisfies the doubly infinite sum
3*x = ... + x^12*(1/A^9 - A^8) + x^5*(1/A^6 - A^5) + x*(1/A^3 - A^2) + (1 - 1/A) + x^2*(A^3 - 1/A^4) + x^7*(A^6 - 1/A^7) + x^15*(A^9 - 1/A^10) + ... + x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)) + ...
also, by the Watson quintuple product identity,
3*x = (1-x)*(1-x*A)*(1-1/A)*(1-x*A^2)*(1-x/A^2) * (1-x^2)*(1-x^2*A)*(1-x/A)*(1-x^3*A^2)*(1-x^3/A^2) * (1-x^3)*(1-x^3*A)*(1-x^2/A)*(1-x^5*A^2)*(1-x^5/A^2) * (1-x^4)*(1-x^4*A)*(1-x^3/A)*(1-x^7*A^2)*(1-x^7/A^2) * ...

Links

Crossrefs

Cf. A361550, A359920, A361552, A361554, A361555.

Programs

  • Mathematica
    (* Calculation of constant d: *) With[{k = 3}, 1/r /. FindRoot[{r^3*s^3 * QPochhammer[r] * QPochhammer[1/(r*s^2), r^2] *  QPochhammer[1/(r*s), r] * QPochhammer[s, r] * QPochhammer[s^2/r, r^2] / ((-1 + s)*(-1 + r*s)*(-r + s^2)*(-1 + r*s^2)) == k*r, 1/(-1 + s) + 1/(s*(-1 + r*s)) + (2*s)/(-r + s^2) - 2/(s - r*s^3) + (-QPolyGamma[0, -Log[r*s]/Log[r], r] + QPolyGamma[0, Log[s]/Log[r], r] - QPolyGamma[0, -Log[r*s^2]/Log[r^2], r^2] + QPolyGamma[0, Log[s^2/r]/Log[r^2], r^2]) / (s*Log[r]) == 0}, {r, 1/12}, {s, 2}, WorkingPrecision -> 70]] (* Vaclav Kotesovec, Jan 18 2024 *)
  • PARI
    /* Using the doubly infinite series */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(3*x - sum(m=-#A, #A, x^(m*(3*m-1)/2) * Ser(A)^(3*m-1) * (x^m*Ser(A) - 1) ) , #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    /* Using the quintuple product */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(3*x - prod(m=1, #A, (1 - x^m) * (1 - x^m*Ser(A)) * (1 - x^(m-1)/Ser(A)) * (1 - x^(2*m-1)*Ser(A)^2) * (1 - x^(2*m-1)/Ser(A)^2) ), #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following.
(1) 3*x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).
(2) 3*x = Sum_{n=-oo..+oo} x^(n*(3*n-1)/2) * A(x)^(3*n) * (x^n - 1/A(x)).
(3) 3*x = Product_{n>=1} (1 - x^n) * (1 - x^n*A(x)) * (1 - x^(n-1)/A(x)) * (1 - x^(2*n-1)*A(x)^2) * (1 - x^(2*n-1)/A(x)^2), by the Watson quintuple product identity.
(4) a(n) = Sum_{k=0..n} A361550(n,k) * 3^k for n >= 0.
a(n) ~ c * d^n / n^(3/2), where d = 12.47776743014414138089586... and c = 0.474320402676760199022... - Vaclav Kotesovec, Mar 29 2023

A361554 Expansion of g.f. A(x) satisfying 4*x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).

Original entry on oeis.org

1, 4, 36, 296, 2732, 28980, 329996, 3872908, 46575260, 573472248, 7197096168, 91640952360, 1180636398320, 15364364313588, 201691201775092, 2667523242203932, 35510152549696208, 475424653523498396, 6397601663340197268, 86481499341290372804, 1173813146742741571560
Offset: 0

Views

Author

Paul D. Hanna, Mar 19 2023

Keywords

Examples

			G.f.: A(x) = 1 + 4*x + 36*x^2 + 296*x^3 + 2732*x^4 + 28980*x^5 + 329996*x^6 + 3872908*x^7 + 46575260*x^8 + 573472248*x^9 + ...
where A = A(x) satisfies the doubly infinite sum
4*x = ... + x^12*(1/A^9 - A^8) + x^5*(1/A^6 - A^5) + x*(1/A^3 - A^2) + (1 - 1/A) + x^2*(A^3 - 1/A^4) + x^7*(A^6 - 1/A^7) + x^15*(A^9 - 1/A^10) + ... + x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)) + ...
also, by the Watson quintuple product identity,
4*x = (1-x)*(1-x*A)*(1-1/A)*(1-x*A^2)*(1-x/A^2) * (1-x^2)*(1-x^2*A)*(1-x/A)*(1-x^3*A^2)*(1-x^3/A^2) * (1-x^3)*(1-x^3*A)*(1-x^2/A)*(1-x^5*A^2)*(1-x^5/A^2) * (1-x^4)*(1-x^4*A)*(1-x^3/A)*(1-x^7*A^2)*(1-x^7/A^2) * ...

Links

Crossrefs

Cf. A361550, A359920, A361552, A361553, A361555.

Programs

  • PARI
    /* Using the doubly infinite series */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(4*x - sum(m=-#A, #A, x^(m*(3*m-1)/2) * Ser(A)^(3*m-1) * (x^m*Ser(A) - 1) ) , #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    /* Using the quintuple product */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(4*x - prod(m=1, #A, (1 - x^m) * (1 - x^m*Ser(A)) * (1 - x^(m-1)/Ser(A)) * (1 - x^(2*m-1)*Ser(A)^2) * (1 - x^(2*m-1)/Ser(A)^2) ), #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following.
(1) 4*x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).
(2) 4*x = Sum_{n=-oo..+oo} x^(n*(3*n-1)/2) * A(x)^(3*n) * (x^n - 1/A(x)).
(3) 4*x = Product_{n>=1} (1 - x^n) * (1 - x^n*A(x)) * (1 - x^(n-1)/A(x)) * (1 - x^(2*n-1)*A(x)^2) * (1 - x^(2*n-1)/A(x)^2), by the Watson quintuple product identity.
(4) a(n) = Sum_{k=0..n} A361550(n,k) * 4^k for n >= 0.

A361555 Expansion of g.f. A(x) satisfying 5*x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).

Original entry on oeis.org

1, 5, 50, 465, 4925, 59870, 776155, 10364135, 142082065, 1995371980, 28549274995, 414327073520, 6084353526535, 90258375062245, 1350607531232830, 20361436162127965, 308964002231172075, 4715119823819824535, 72324133311820587435, 1114404268419043050750
Offset: 0

Views

Author

Paul D. Hanna, Mar 19 2023

Keywords

Examples

			G.f.: A(x) = 1 + 5*x + 50*x^2 + 465*x^3 + 4925*x^4 + 59870*x^5 + 776155*x^6 + 10364135*x^7 + 142082065*x^8 + 1995371980*x^9 + ...
where A = A(x) satisfies the doubly infinite sum
5*x = ... + x^12*(1/A^9 - A^8) + x^5*(1/A^6 - A^5) + x*(1/A^3 - A^2) + (1 - 1/A) + x^2*(A^3 - 1/A^4) + x^7*(A^6 - 1/A^7) + x^15*(A^9 - 1/A^10) + ... + x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)) + ...
also, by the Watson quintuple product identity,
5*x = (1-x)*(1-x*A)*(1-1/A)*(1-x*A^2)*(1-x/A^2) * (1-x^2)*(1-x^2*A)*(1-x/A)*(1-x^3*A^2)*(1-x^3/A^2) * (1-x^3)*(1-x^3*A)*(1-x^2/A)*(1-x^5*A^2)*(1-x^5/A^2) * (1-x^4)*(1-x^4*A)*(1-x^3/A)*(1-x^7*A^2)*(1-x^7/A^2) * ...

Links

Crossrefs

Cf. A361550, A359920, A361552, A361553, A361554.

Programs

  • Mathematica
    (* Calculation of constant d: *) 1/r /. FindRoot[{r^2 * s^3 * QPochhammer[r] * QPochhammer[1/(r*s^2), r^2] * QPochhammer[1/(r*s), r] * QPochhammer[s, r] *(QPochhammer[s^2/r, r^2]/ ((-1 + s)*(-1 + r*s)*(-r + s^2)*(-1 + r*s^2))) == 5, (3*r - 2*r*(1 + r)*s - s^2 + r^3*s^4 + 2*r*(1 + r)*s^5 - 3*r^2*s^6)*Log[r] + (-1 + s)*(-1 + r*s)*(r - s^2)*(-1 + r*s^2) * (QPolyGamma[0, Log[s]/Log[r], r] - QPolyGamma[0, -1/2 - Log[s]/Log[r], r^2] + QPolyGamma[0, -1/2 + Log[s]/Log[r], r^2] - QPolyGamma[0, -Log[r*s]/Log[r], r]) == 0}, {r, 1/16}, {s, 2}, WorkingPrecision -> 70] (* Vaclav Kotesovec, Feb 01 2024 *)
  • PARI
    /* Using the doubly infinite series */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(4*x - sum(m=-#A, #A, x^(m*(3*m-1)/2) * Ser(A)^(3*m-1) * (x^m*Ser(A) - 1) ) , #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    /* Using the quintuple product */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(5*x - prod(m=1, #A, (1 - x^m) * (1 - x^m*Ser(A)) * (1 - x^(m-1)/Ser(A)) * (1 - x^(2*m-1)*Ser(A)^2) * (1 - x^(2*m-1)/Ser(A)^2) ), #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following.
(1) 5*x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).
(2) 5*x = Sum_{n=-oo..+oo} x^(n*(3*n-1)/2) * A(x)^(3*n) * (x^n - 1/A(x)).
(3) 5*x = Product_{n>=1} (1 - x^n) * (1 - x^n*A(x)) * (1 - x^(n-1)/A(x)) * (1 - x^(2*n-1)*A(x)^2) * (1 - x^(2*n-1)/A(x)^2), by the Watson quintuple product identity.
(4) a(n) = Sum_{k=0..n} A361550(n,k) * 5^k for n >= 0.
From Vaclav Kotesovec, Feb 01 2024: (Start)
Formula (3) can be rewritten as the functional equation 5*x = QPochhammer(x) * QPochhammer(y, x)/(1 - y) * QPochhammer(1/(x*y), x)/(1 - 1/(x*y)) * QPochhammer(y^2/x, x^2)/(1-y^2/x) * QPochhammer(1/(x*y^2), x^2)/(1-1/(x*y^2)).
a(n) ~ c * d^n / n^(3/2), where d = 16.695183607901729043700484293708659594719464935528330676878595927048... and c = 0.5534958293134675625595273281664529583363592593727800077222126752653... (End)
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