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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.

Showing 1-4 of 4 results.

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

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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...

Links

Crossrefs

Cf. A359915, A359916, A359919, A359921, A359924, A359719.
Cf. A359914.

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

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

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) * ...

Links

Crossrefs

Cf. A359920, A359921, A359924, A359915, A359916.
Cf. A359914.

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) * ...

Links

Crossrefs

Cf. A359920, A359916, A359919, A359921, A359924.
Cf. A359914.

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.
Showing 1-4 of 4 results.

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