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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A359920 a(n) = coefficient of x^n in A(x) such that 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, 6, 29, 137, 690, 3815, 22579, 138353, 862692, 5451339, 34911444, 226475135, 1485571965, 9833401534, 65578882177, 440170565711, 2971402946711, 20161828468803, 137434420403678, 940701180157773, 6462787501335564, 44550102080595910, 308041365014677804, 2135938633975050831
Offset: 0

Views

Author

Paul D. Hanna, Jan 22 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 6*x^2 + 29*x^3 + 137*x^4 + 690*x^5 + 3815*x^6 + 22579*x^7 + 138353*x^8 + 862692*x^9 + 5451339*x^10 + 34911444*x^11 + 226475135*x^12 + ...
where A = A(x) satisfies the doubly infinite sum
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,
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) * ...
SPECIFIC VALUES.
A(x) at x = 100/738 diverges.
A(100/739) = 1.680090298639836342808608867776256534712736768391...
A(1/8) = 1.40048762211279862753069563580599076131617792526323...
A(1/9) = 1.28067125711115350114265686789651886973848631068277...
		

Crossrefs

Programs

  • Mathematica
    (* Calculation of constant d: *) With[{k = 1}, 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, 1/7}, {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(x - 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]); for(i=1,n, A = concat(A,0);
    A[#A] = polcoeff(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) x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).
(2) x = Sum_{n=-oo..+oo} x^(n*(3*n-1)/2) * A(x)^(3*n) * (x^n - 1/A(x)).
(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.
a(n) ~ c * d^n / n^(3/2), where d = 7.388458151593... and c = 0.36167254645... - Vaclav Kotesovec, Mar 19 2023
Formula (3) can be rewritten as the functional equation 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)). - Vaclav Kotesovec, Jan 19 2024

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

A361050 Expansion of g.f. A(x,y) satisfying y/x = 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, 4, 0, 18, 40, 22, 0, 55, 244, 335, 140, 0, 149, 1160, 2924, 2875, 969, 0, 371, 4688, 19090, 32745, 25081, 7084, 0, 867, 16848, 103110, 272250, 352814, 221397, 53820, 0, 1923, 55332, 485356, 1839075, 3565548, 3709244, 1971775, 420732, 0, 4086, 169048, 2054520, 10674985, 28909300, 44146487, 38344384, 17682895, 3362260
Offset: 1

Views

Author

Paul D. Hanna, Mar 18 2023

Keywords

Comments

A359921(n) = Sum_{k=0..n-1} T(n,k) for n >= 1.
A359924(n) = Sum_{k=0..n-1} T(n,k) * 2^k for n >= 1.
A361051(n) = Sum_{k=0..n-1} T(n,k) * 3^k for n >= 1.
A361052(n) = Sum_{k=0..n-1} T(n,k) * 4^k for n >= 1.
A361538(n) = T(2*n-1,n-1) for n >= 1.
A360191(n) = T(n+2,1) for n >= 0.
A361535(n) = T(n+3,2)/4 for n >= 0.
A002293(n) = T(n+1,n) for n >= 0.

Examples

			G.f.: A(x,y) = x + y*x^2 + (5*y + 4*y^2)*x^3 + (18*y + 40*y^2 + 22*y^3)*x^4 + (55*y + 244*y^2 + 335*y^3 + 140*y^4)*x^5 + (149*y + 1160*y^2 + 2924*y^3 + 2875*y^4 + 969*y^5)*x^6 + (371*y + 4688*y^2 + 19090*y^3 + 32745*y^4 + 25081*y^5 + 7084*y^6)*x^7 + (867*y + 16848*y^2 + 103110*y^3 + 272250*y^4 + 352814*y^5 + 221397*y^6 + 53820*y^7)*x^8 + (1923*y + 55332*y^2 + 485356*y^3 + 1839075*y^4 + 3565548*y^5 + 3709244*y^6 + 1971775*y^7 + 420732*y^8)*x^9 + (4086*y + 169048*y^2 + 2054520*y^3 + 10674985*y^4 + 28909300*y^5 + 44146487*y^6 + 38344384*y^7 + 17682895*y^8 + 3362260*y^9)*x^10 + ...
This triangle of coefficients T(n,k) of x^n*y^k, n >= 1, k = 0..n-1, in g.f. A(x,y) begins:
1;
0, 1;
0, 5, 4;
0, 18, 40, 22;
0, 55, 244, 335, 140;
0, 149, 1160, 2924, 2875, 969;
0, 371, 4688, 19090, 32745, 25081, 7084;
0, 867, 16848, 103110, 272250, 352814, 221397, 53820;
0, 1923, 55332, 485356, 1839075, 3565548, 3709244, 1971775, 420732;
0, 4086, 169048, 2054520, 10674985, 28909300, 44146487, 38344384, 17682895, 3362260;
0, 8374, 486500, 7984667, 55085875, 199363606, 417661860, 525322468, 391561335, 159463876, 27343888;
0, 16634, 1331056, 28909580, 258486830, 1211896230, 3335033317, 5680806120, 6069336891, 3961602925, 1444601027, 225568798;
...
		

Crossrefs

Cf. A360191 (column 1), A361535 (column 2), A002293 (diagonal), A361538 (central terms).
Cf. A359921 (y=1), A359924 (y=2), A361051 (y=3), A361052 (y=4).
Cf. A002293, A356500 (related table), A361550 (related triangle).

Programs

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

Formula

G.f. A(x,y) = Sum_{n>=1} Sum_{k=0..n-1} T(n,k)*x^n*y^k satisfies the following.
(1) y/x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x,y)^(3*n) - 1/A(x,y)^(3*n+1)).
(2) y/x = 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.
(3) Sum_{n>=0} T(n+2,1) * x^n = 1 / Product_{n>=1} (1 - x^n)^3 * (1 - x^(2*n-1))^2, which is the g.f. of A360191.
(4) Sum_{n>=0} T(n+3,2) * x^n = 4*F(x) where F(x) = 1/Product_{n>=1} (1 - x^n)^6 * (1 - x^(2*n-1))^4, which is the g.f. of A361535.
(5) Sum_{n>=0} T(n+1,n) * x^n = D(x) where D(x) = 1 + x*D(x)^4 is the g.f. of A002293.
(6) T(n+1,n) = binomial(4*n, n)/(3*n + 1) for n >= 0.

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.

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.

A361052 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, 84, 2120, 61404, 1934548, 64379980, 2226478604, 79225597516, 2881791020120, 106672402111192, 4005192227754984, 152168779157569376, 5839221480075313396, 225986788425426186532, 8810672964167893735292, 345722424894740010814784, 13642862904817471637398044
Offset: 1

Views

Author

Paul D. Hanna, Mar 18 2023

Keywords

Examples

			G.f.: A(x) = x + 4*x^2 + 84*x^3 + 2120*x^4 + 61404*x^5 + 1934548*x^6 + 64379980*x^7 + 2226478604*x^8 + 79225597516*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=[0, 1]); for(i=1, n, A = concat(A, 0);
    A[#A] = polcoeff(4/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(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-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) 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 = 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.
(3) a(n) = Sum_{k=0..n-1} A361050(n,k) * 4^k, for n >= 1.
a(n) ~ c * d^n / n^(3/2), where d = 43.15078920061551630152405195461463024566432382819246... and c = 0.0036458304883879627950854318861022051996596920296... - Vaclav Kotesovec, Mar 19 2023

A361051 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, 51, 1008, 22746, 558177, 14469999, 389827008, 10805735061, 306185433921, 8828873667975, 258229614694974, 7642514652514140, 228450735379271754, 6887262023421308658, 209169231039167908596, 6393531094406983438776, 196536271435928605186752, 6071932630099467279020415
Offset: 1

Views

Author

Paul D. Hanna, Mar 18 2023

Keywords

Examples

			G.f.: A(x) = x + 3*x^2 + 51*x^3 + 1008*x^4 + 22746*x^5 + 558177*x^6 + 14469999*x^7 + 389827008*x^8 + 10805735061*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

  • PARI
    /* Using the doubly infinite series */
    {a(n) = my(A=[0, 1]); for(i=1, n, A = concat(A, 0);
    A[#A] = polcoeff(3/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(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-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) 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 = 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.
(3) a(n) = Sum_{k=0..n-1} A361050(n,k) * 3^k, for n >= 1.
a(n) ~ c * d^n / n^(3/2), where d = 33.61307737482651437383925998526816971444845895805... and c = 0.004710392090243985254460721389434519943286349... - Vaclav Kotesovec, Mar 19 2023
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