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 1-8 of 8 results.

A324403 a(n) = Product_{i=1..n, j=1..n} (i^2 + j^2).

Original entry on oeis.org

1, 2, 400, 121680000, 281324160000000000, 15539794609114833408000000000000, 49933566483104048708063697937367040000000000000000, 19323883089768863178599626514889213871887405416448000000000000000000000000
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 26 2019

Keywords

Comments

Next term is too long to be included.

Crossrefs

Cf. A079478, A324426, A324437, A324438, A324439, A324440, A367834.
Cf. A272244, A293290, A324402, A324443, A324425.

Programs

  • Maple
    a:= n-> mul(mul(i^2+j^2, i=1..n), j=1..n):
seq(a(n), n=0..7);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[Product[i^2+j^2, {i, 1, n}, {j, 1, n}], {n, 1, 10}]
  • PARI
    a(n) = prod(i=1, n, prod(j=1, n, i^2+j^2)); \\ Michel Marcus, Feb 27 2019
  • Python
    from math import prod, factorial
def A324403(n): return (prod(i**2+j**2 for i in range(1,n) for j in range(i+1,n+1))*factorial(n))**2<Chai Wah Wu, Nov 22 2023

Formula

a(n) ~ 2^(n*(n+1) - 3/4) * exp(Pi*n*(n+1)/2 - 3*n^2 + Pi/12) * n^(2*n^2 - 1/2) / (Pi^(1/4) * Gamma(3/4)).
a(n) = 2*n^2*a(n-1)*Product_{i=1..n-1} (n^2 + i^2)^2. - Chai Wah Wu, Feb 26 2019
For n>0, a(n)/a(n-1) = A272244(n)^2 / (2*n^6). - Vaclav Kotesovec, Dec 02 2023
a(n) = exp(2*Integral_{x=0..oo} (n^2/(x*exp(x)) - (cosh(n*x) - cos(n*x))/(x*exp((n + 1)*x)*(cosh(x) - cos(x)))) dx)/2^(n^2). - Velin Yanev, Jun 30 2025

Extensions

a(0)=1 prepended by Alois P. Heinz, Jun 24 2023

A324437 a(n) = Product_{i=1..n, j=1..n} (i^4 + j^4).

Original entry on oeis.org

1, 2, 18496, 189567553208832, 53863903330477722171391434817536, 4194051697335929481600368256016484482740174637152337920000, 530545265060440849231458462212366841894726534233233018777709463062563850450708386692464640000
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 28 2019

Keywords

Crossrefs

Cf. A079478, A324403, A324426, A324438, A324439, A324440, A367834.
Cf. A203677, A272247, A307215.

Programs

  • Maple
    a:= n-> mul(mul(i^4+j^4, i=1..n), j=1..n):
seq(a(n), n=0..7);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[Product[i^4 + j^4, {i, 1, n}, {j, 1, n}], {n, 1, 6}]
  • Python
    from math import prod, factorial
def A324437(n): return (prod(i**4+j**4 for i in range(1,n) for j in range(i+1,n+1))*factorial(n)**2)**2<Chai Wah Wu, Nov 26 2023

Formula

a(n) ~ c * 2^(n*(n+1)) * exp(Pi*n*(n+1)/sqrt(2) - 6*n^2) * (1 + sqrt(2))^(sqrt(2)*n*(n+1)) * n^(4*n^2 - 1), where c = A306620 = 0.23451584451404279281807143317500518660696293944961...
For n>0, a(n)/a(n-1) = A272247(n)^2 / (2*n^12). - Vaclav Kotesovec, Dec 01 2023

Extensions

a(0)=1 prepended by Alois P. Heinz, Jun 24 2023

A324426 a(n) = Product_{i=1..n, j=1..n} (i^3 + j^3).

Original entry on oeis.org

1, 2, 2592, 134425267200, 3120795915109442519040000, 180825857777547616919759624941086965760000000, 99356698720512072045648926659510730227553351200000695922065408000000000
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 27 2019

Keywords

Crossrefs

Cf. A079478, A324403, A324437, A324438, A324439, A324440, A367834.
Cf. A272246, A307210, A367517, A367543.

Programs

  • Maple
    a:= n-> mul(mul(i^3+j^3, i=1..n), j=1..n):
seq(a(n), n=0..7);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[Product[i^3+j^3, {i, 1, n}, {j, 1, n}], {n, 1, 10}]
  • PARI
    a(n) = prod(i=1, n, prod(j=1, n, i^3+j^3)); \\ Michel Marcus, Feb 27 2019
  • Python
    from math import prod, factorial
def A324426(n): return prod(i**3+j**3 for i in range(1,n) for j in range(i+1,n+1))**2*factorial(n)**3<Chai Wah Wu, Nov 26 2023

Formula

a(n) ~ A * 2^(2*n*(n+1) + 1/4) * exp(Pi*(n*(n+1) + 1/6)/sqrt(3) - 9*n^2/2 - 1/12) * n^(3*n^2 - 3/4) / (3^(5/6) * Pi^(1/6) * Gamma(2/3)^2), where A is the Glaisher-Kinkelin constant A074962.
a(n) = A079478(n) * A367543(n). - Vaclav Kotesovec, Nov 22 2023
For n>0, a(n)/a(n-1) = A272246(n)^2 / (2*n^9). - Vaclav Kotesovec, Dec 02 2023

Extensions

a(0)=1 prepended by Alois P. Heinz, Jun 24 2023

A324438 a(n) = Product_{i=1..n, j=1..n} (i^5 + j^5).

Original entry on oeis.org

1, 2, 139392, 305013568273920000, 1174837791623127613548781790822400000000, 139642003782073074626249921818187528362524804267528306032640000000000000
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 28 2019

Keywords

Crossrefs

Cf. A079478, A324403, A324426, A324437, A324439, A324440, A367834.
Cf. A272248, A367679.

Programs

  • Maple
    a:= n-> mul(mul(i^5 + j^5, i=1..n), j=1..n):
seq(a(n), n=0..5);  # Alois P. Heinz, Nov 26 2023
  • Mathematica
    Table[Product[i^5 + j^5, {i, 1, n}, {j, 1, n}], {n, 1, 6}]
  • Python
    from math import prod, factorial
def A324438(n): return prod(i**5+j**5 for i in range(1,n) for j in range(i+1,n+1))**2*factorial(n)**5<Chai Wah Wu, Nov 26 2023

Formula

a(n) ~ c * 2^(2*n*(n+1)) * phi^(sqrt(5)*n*(n+1)) * exp(Pi*sqrt(phi)*n*(n+1)/5^(1/4) - 15*n^2/2) * n^(5*n^2 - 5/4), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio and c = 0.1574073828647726237455544898360432469056972905505624900871695...
a(n) = A367679(n) * A079478(n). - Vaclav Kotesovec, Nov 26 2023
For n>0, a(n)/a(n-1) = A272248(n)^2 / (2*n^15). - Vaclav Kotesovec, Dec 02 2023

Extensions

a(0)=1 prepended by Alois P. Heinz, Nov 26 2023

A324439 a(n) = Product_{i=1..n, j=1..n} (i^6 + j^6).

Original entry on oeis.org

1, 2, 1081600, 528465082730906880000, 29276520893554373473343522853366005760000000000, 5719545329208791496596894540018824083491259163047733746620041978183680000000000000000
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 28 2019

Keywords

Crossrefs

Cf. A079478, A324403, A324426, A324437, A324438, A324440, A367834.
Cf. A367668, A367823.

Programs

  • Maple
    a:= n-> mul(mul(i^6 + j^6, i=1..n), j=1..n):
seq(a(n), n=0..5);  # Alois P. Heinz, Nov 26 2023
  • Mathematica
    Table[Product[i^6 + j^6, {i, 1, n}, {j, 1, n}], {n, 1, 6}]
  • Python
    from math import prod, factorial
def A324439(n): return (prod(i**6+j**6 for i in range(1,n) for j in range(i+1,n+1))*factorial(n)**3)**2<Chai Wah Wu, Nov 26 2023

Formula

a(n) ~ c * 2^(n*(n+1)) * (2 + sqrt(3))^(sqrt(3)*n*(n+1)) * exp(Pi*n*(n+1) - 9*n^2) * n^(6*n^2 - 3/2), where c = 0.104143806044091748191387307161835081649...
a(n) = A324403(n) * A367668(n). - Vaclav Kotesovec, Dec 01 2023
For n>0, a(n)/a(n-1) = A367823^2 / (2*n^18). - Vaclav Kotesovec, Dec 02 2023

Extensions

a(n)=1 prepended by Alois P. Heinz, Nov 26 2023

A324440 a(n) = Product_{i=1..n, j=1..n} (i^7 + j^7).

Original entry on oeis.org

1, 2, 8520192, 956147263254051187507200, 790929096572487518050439299107158612099228070051840000, 266108022587896795750359251172229660295854509829286134803404773931312693787460334360985600000000000
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 28 2019

Keywords

Comments

For m>1, Product_{j=1..n, k=1..n} (j^m + k^m) ~ c(m) * exp(n*(n+1)*s(m) - m*n*(n-2)/2) * n^(m*(n^2 - 1/4)), where s(m) = Sum_{j>=1} (-1)^(j+1)/(j*(1 + m*j)) and c(m) is a constant (dependent only on m). Equivalently, s(m) = log(2) - HurwitzLerchPhi(-1, 1, 1 + 1/m). - Vaclav Kotesovec, Dec 01 2023

Crossrefs

Cf. A079478, A324403, A324426, A324437, A324438, A324439, A367834.

Programs

  • Maple
    a:= n-> mul(mul(i^7 + j^7, i=1..n), j=1..n):
seq(a(n), n=0..5);  # Alois P. Heinz, Nov 26 2023
  • Mathematica
    Table[Product[i^7+j^7, {i, 1, n}, {j, 1, n}], {n, 1, 6}]
  • Python
    from math import prod, factorial
def A324440(n): return prod(i**7+j**7 for i in range(1,n) for j in range(i+1,n+1))**2*factorial(n)**7<Chai Wah Wu, Nov 26 2023

Formula

Limit_{n->oo} (a(n)^(1/n^2))/n^7 = 2^(3/2) * (cos(3*Pi/14) / tan(Pi/7))^sin(3*Pi/14) / ((cos(Pi/14)*tan(3*Pi/14))^sin(Pi/14) * (sin(Pi/7)*tan(Pi/14))^cos(Pi/7)) * exp((Pi/sin(Pi/7) - 21)/2) = 0.0334234967249533921390751418772468470887965377...
From Vaclav Kotesovec, Dec 01 2023: (Start)
a(n) ~ c * exp(n*(n+1)*s - 7*n*(n-2)/2) * n^(7*(n^2 - 1/4)), where
s = Sum_{j>=1} (-1)^(j+1)/(j*(1 + 7*j)) = Pi/(2*sin(Pi/7)) + 3*log(2)/2 - 7 - cos(Pi/7) * log(2*sin(Pi/14)^2) - log(2*sin(3*Pi/14)^2) * sin(Pi/14) + log(cos(3*Pi/14)*cos(Pi/7) / sin(Pi/7)) * sin(3*Pi/14) = 0.10150386842315637912206687298894641634315636548242136512503... and
c = 0.068056503846689328929612652207251071282623125565150941566636264805878144...
Equivalently, s = log(2) - HurwitzLerchPhi(-1, 1, 1 + 1/7). (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Nov 26 2023

A367833 a(n) = Product_{k=0..n} (n^8 + k^8).

Original entry on oeis.org

0, 2, 33685504, 3851231199376068, 2670340833224951538384896, 8442437541556000328575439453125000, 96982590530284083757788173242580213780447232, 3406021755102121469537208768079471859185253483110025744
Offset: 0

Views

Author

Vaclav Kotesovec, Dec 02 2023

Keywords

Crossrefs

Cf. A126804, A272244, A272246, A272247, A272248, A367823.
Cf. A367834.

Programs

  • Mathematica
    Table[Product[n^8 + k^8, {k, 0, n}], {n, 0, 10}]

Formula

a(n) ~ 2^(n + 1/2) * ((2 + sqrt(2 + sqrt(2)))/(2 - sqrt(2 + sqrt(2))))^(sqrt(2 + sqrt(2))*n/2) * (sqrt(2) - 1 + sqrt(4 - 2*sqrt(2)))^(sqrt(2 - sqrt(2))*n) * exp((Pi/sqrt(2 - sqrt(2)) - 8)*n) * n^(8*(n + 1)).

A306620 Decimal expansion of a constant related to the asymptotics of A324437.

Original entry on oeis.org

2, 3, 4, 5, 1, 5, 8, 4, 4, 5, 1, 4, 0, 4, 2, 7, 9, 2, 8, 1, 8, 0, 7, 1, 4, 3, 3, 1, 7, 5, 0, 0, 5, 1, 8, 6, 6, 0, 6, 9, 6, 2, 9, 3, 9, 4, 4, 9, 6, 1, 0, 3, 9, 5, 5, 3, 2, 4, 5, 8, 2, 3, 6, 8, 3, 6, 6, 1, 0, 9, 9, 4, 1, 7, 0, 2, 5, 3, 0, 3, 2, 4, 1, 6, 1, 4, 5, 1, 7, 7, 7, 4, 7, 0, 5, 4, 3, 0, 2, 6, 0, 4, 9, 6, 6, 0
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 01 2019

Keywords

Examples

			0.234515844514042792818071433175005186606962939449610395532458236836610994170253...

Links

Crossrefs

Cf. A324437, A367834.

Formula

Equals limit_{n->oo} A324437(n) / (2^(n*(n+1)) * exp(Pi*n*(n+1)/sqrt(2) - 6*n^2) * (1 + sqrt(2))^(sqrt(2)*n*(n+1)) * n^(4*n^2 - 1)).
Equals limit_{n->oo} n*(Product_{i=1..n, j=1..n} ((i/n)^4 + (j/n)^4)) / exp(6*n + n*(n+1)*Integral_{x=0..1, y=0..1} log(x^4 + y^4) dy dx). - Vaclav Kotesovec, Dec 04 2023
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