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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.

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) * ...

Links

Crossrefs

Cf. A359920, A359924, A361050, A361052, A361538.

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) * ...

Links

Crossrefs

Cf. A359920, A359921, A361050, A361052, A361538.

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;
...

Links

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.

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) * ...

Links

Crossrefs

Cf. A361050, A359921, A359924, A361052, A361538.

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
Showing 1-4 of 4 results.

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