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 1-5 of 5 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

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

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.
Showing 1-5 of 5 results.