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 1-10 of 13 results. Next

A359921 a(n) = coefficient of x^n in A(x) such that 1/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, 1, 9, 80, 774, 8077, 89059, 1021106, 12048985, 145347965, 1784282449, 22217589408, 279934808090, 3562376922346, 45721210139842, 591139659619262, 7692224199601436, 100663182977093130, 1323944771879772911, 17491108974090887920, 232015023433972687373, 3088855705228007528177
Offset: 1

Views

Author

Paul D. Hanna, Jan 22 2023

Keywords

Examples

			G.f.: A(x) = x + x^2 + 9*x^3 + 80*x^4 + 774*x^5 + 8077*x^6 + 89059*x^7 + 1021106*x^8 + 12048985*x^9 + 145347965*x^10 + ...
where A = A(x) satisfies the doubly infinite sum
1/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,
1/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) * ...
		

Crossrefs

Programs

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

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following.
(1) 1/x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).
(2) 1/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.
a(n) = Sum_{k=0..n-1} A361050(n,k) for n >= 1. - Paul D. Hanna, Mar 19 2023
a(n) ~ c * d^n / n^(3/2), where d = 14.308864552026948863076624... and c = 0.01145810893741095458355... - Vaclav Kotesovec, Mar 19 2023

A359924 a(n) = coefficient of x^n in A(x) such that 2/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, 2, 26, 372, 6006, 105338, 1952102, 37598422, 745116966, 15094772444, 311183832004, 6507065710068, 137683172641240, 2942394474649322, 63418690179207242, 1376986195691108990, 30090726682472126472, 661292884776232386766, 14606177871231796042658, 324062328994910188622258
Offset: 1

Views

Author

Paul D. Hanna, Jan 22 2023

Keywords

Examples

			G.f.: A(x) = x + 2*x^2 + 26*x^3 + 372*x^4 + 6006*x^5 + 105338*x^6 + 1952102*x^7 + 37598422*x^8 + 745116966*x^9 + 15094772444*x^10 + ...
where A = A(x) satisfies the doubly infinite sum
2/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,
2/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) * ...
		

Crossrefs

Programs

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

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following.
(1) 2/x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).
(2) 2/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.
a(n) = Sum_{k=0..n-1} A361050(n,k) * 2^k for n >= 1. - Paul D. Hanna, Mar 19 2023
a(n) ~ c * d^n / n^(3/2), where d = 24.0303544191480291910560326469... and c = 0.0066619562786442340995706184... - Vaclav Kotesovec, Mar 14 2023

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;
...
		

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...
		

Crossrefs

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

A359919 a(n) = coefficient of x^n in A(x) such that x^2 = 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, 0, 1, 5, 19, 65, 211, 681, 2255, 7830, 28786, 111230, 443789, 1795972, 7284981, 29466755, 118834438, 479034654, 1936617163, 7872885832, 32226147305, 132808096158, 550444192577, 2291095125465, 9564074472264, 40005894288101, 167610376198140, 703308153554903
Offset: 0

Views

Author

Paul D. Hanna, Jan 22 2023

Keywords

Examples

			G.f.: A(x) = 1 + x^2 + 5*x^3 + 19*x^4 + 65*x^5 + 211*x^6 + 681*x^7 + 2255*x^8 + 7830*x^9 + 28786*x^10 + 111230*x^11 + 443789*x^12 + ...
where A = A(x) satisfies the doubly infinite sum
x^2 = ... + 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,
x^2 = (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) * ...
		

Crossrefs

Programs

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

A361552 Expansion of g.f. A(x) satisfying 2*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, 2, 14, 84, 530, 3770, 29446, 240302, 2003914, 17024332, 147306448, 1294859540, 11524690228, 103605031978, 939357512086, 8580744729478, 78898896072996, 729661925134886, 6782435427053490, 63332055630823770, 593793935288453260, 5587934788557993846
Offset: 0

Views

Author

Paul D. Hanna, Mar 19 2023

Keywords

Comments

Conjecture: a(n) == A359914(n-1) (mod 4) for n > 1.
Conjecture: a(n)/2 == A000041(n-1) (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-1, for n >= 1.

Examples

			G.f.: A(x) = 1 + 2*x + 14*x^2 + 84*x^3 + 530*x^4 + 3770*x^5 + 29446*x^6 + 240302*x^7 + 2003914*x^8 + 17024332*x^9 + ...
where A = A(x) satisfies the doubly infinite sum
2*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,
2*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) * ...
		

Crossrefs

Programs

  • Mathematica
    (* Calculation of constant d: *) With[{k = 2}, 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, 1/10}, {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(2*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(2*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) 2*x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).
(2) 2*x = Sum_{n=-oo..+oo} x^(n*(3*n-1)/2) * A(x)^(3*n) * (x^n - 1/A(x)).
(3) 2*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) * 2^k for n >= 0.
a(n) ~ c * d^n / n^(3/2), where d = 10.118828419885936478438... and c = 0.4250308979334609908... - Vaclav Kotesovec, Mar 19 2023

A359915 a(n) = coefficient of x^n in A(x) such that A(x) = Sum_{n=-oo..+oo} (x*A(x))^(n*(3*n+1)/2) * (1/x^(3*n) - x^(3*n+1)).

Original entry on oeis.org

1, 1, 5, 23, 121, 713, 4487, 29374, 197896, 1363770, 9570226, 68156319, 491347930, 3578755113, 26295477075, 194677798065, 1450833583380, 10875262975274, 81940144475296, 620223662770067, 4714016885082577, 35962615212212852, 275282740190267268, 2113705107245941938
Offset: 0

Views

Author

Paul D. Hanna, Jan 22 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 23*x^3 + 121*x^4 + 713*x^5 + 4487*x^6 + 29374*x^7 + 197896*x^8 + 1363770*x^9 + 9570226*x^10 + ...
where A = A(x) satisfies the doubly infinite series
A(x) = ... + (x*A)^12*(x^9 - 1/x^8) + (x*A)^5*(x^6 - 1/x^5) + (x*A)*(x^3 - 1/x^2) + (1 - x) + (x*A)^2*(1/x^3 - x^4) + (x*A)^7*(1/x^6 - x^7) + (x*A)^15*(1/x^9 - x^10) + ... + (x*A)^(n*(3*n+1)/2) * (1/x^(3*m) - x^(3*m+1)) + ...
also, by the Watson quintuple product identity,
A(x) = -x * (1-x*A)*(1-x^2*A)*(1-1/x)*(1-x^3*A)*(1-1/x*A) * (1-x^2*A^2)*(1-x^3*A^2)*(1-A)*(1-x^5*A^3)*(1-x*A^3) * (1-x^3*A^3)*(1-x^4*A^3)*(1-x*A^2)*(1-x^7*A^5)*(1-x^3*A^5) * (1-x^4*A^4)*(1-x^5*A^4)*(1-x^2*A^3)*(1-x^9*A^7)*(1-x^5*A^7) * ...
		

Crossrefs

Programs

  • PARI
    /* Using the doubly infinite series */
    {a(n) = my(A=[1]); for(i=1,n, A = concat(A,0);
    A[#A] = polcoeff(Ser(A) - sum(m=-#A, #A, (x*Ser(A))^(m*(3*m+1)/2) * (1/x^(3*m) - x^(3*m+1)) ),#A-2) ); 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(Ser(A) + x*prod(m=1,#A, (1 - x^m*Ser(A)^m) * (1 - x^(m+1)*Ser(A)^m) * (1 - x^(m-2)*Ser(A)^(m-1)) * (1 - x^(2*m+1)*Ser(A)^(2*m-1)) * (1 - x^(2*m-3)*Ser(A)^(2*m-1))),#A-2) ); 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) A(x) = Sum_{n=-oo..+oo} (x*A(x))^(n*(3*n+1)/2) * (1/x^(3*n) - x^(3*n+1)).
(2) A(x) = -x * Product_{n>=1} (1 - x^n*A(x)^n) * (1 - x^(n+1)*A(x)^n) * (1 - x^(n-2)*A(x)^(n-1)) * (1 - x^(2*n+1)*A(x)^(2*n-1)) * (1 - x^(2*n-3)*A(x)^(2*n-1)), by the Watson quintuple product identity.

A359916 a(n) = coefficient of x^n in A(x) such that A(x) = 1 + Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (1/A(x)^(3*n) - A(x)^(3*n+1)).

Original entry on oeis.org

1, 1, 7, 48, 349, 2718, 22403, 192375, 1701544, 15389227, 141643233, 1322344998, 12491424723, 119177917679, 1146750961711, 11115577075944, 108437559699613, 1063849149587086, 10489551647580027, 103891138998923739, 1033113794091793406, 10310925888014393461
Offset: 1

Views

Author

Paul D. Hanna, Jan 22 2023

Keywords

Examples

			G.f.: A(x) = x + x^2 + 7*x^3 + 48*x^4 + 349*x^5 + 2718*x^6 + 22403*x^7 + 192375*x^8 + 1701544*x^9 + 15389227*x^10 + ...
where A = A(x) satisfies the doubly infinite series
A(x) - 1 = ... + x^12*(A^9 - 1/A^8) + x^5*(A^6 - 1/A^5) + x*(A^3 - 1/A^2) + (1 - A) + x^2*(1/A^3 - A^4) + x^7*(1/A^6 - A^7) + x^15*(1/A^9 - 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,
-1 = (1-x^1)*(1-x^1*A)*(1-x^1/A)*(1-x^1*A^2)*(1-x^1/A^2) * (1-x^2)*(1-x^2*A)*(1-x^2/A)*(1-x^3*A^2)*(1-x^3/A^2) * (1-x^3)*(1-x^3*A)*(1-x^3/A)*(1-x^5*A^2)*(1-x^5/A^2) * (1-x^4)*(1-x^4*A)*(1-x^4/A)*(1-x^7*A^2)*(1-x^7/A^2) * ...
		

Crossrefs

Programs

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

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following.
(1) A(x) = 1 + Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (1/A(x)^(3*n) - A(x)^(3*n+1)).
(2) -1 = Product_{n>=1} (1 - x^n) * (1 - x^n*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.

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) * ...
		

Crossrefs

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) * ...
		

Crossrefs

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.
Showing 1-10 of 13 results. Next