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 1-10 of 13 results. Next

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

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

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

A367543 a(n) = Product_{i=1..n, j=1..n} (i^2 - i*j + j^2).

Original entry on oeis.org

1, 36, 777924, 51190934086656, 32435802373365731229926400, 483207398728525904876601066508152707481600, 350969035472356907726779584093506665415605824531908346799718400
Offset: 1

Views

Author

Vaclav Kotesovec, Nov 22 2023

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Product[Product[(i^2 - i*j + j^2), {i, 1, n}], {j, 1, n}], {n, 1, 10}]
  • Python
    from math import prod, factorial
    def A367543(n): return (prod(i*(i-j)+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) = A324426(n) / A079478(n).
a(n) ~ 3^(1/6) * Gamma(1/3)^2 * n^(2*n^2 - 1/3) / (2^(5/3) * Pi^(5/3) * exp(3*n^2 - (n^2 + n + 1/6)*Pi/sqrt(3))).

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

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

A367834 a(n) = Product_{i=1..n, j=1..n} (i^8 + j^8).

Original entry on oeis.org

1, 2, 67634176, 1775927682136440882473213952, 22495149450984565292579847926810488282934424886723006835982336
Offset: 0

Views

Author

Vaclav Kotesovec, Dec 02 2023

Keywords

Comments

Next term is too long to be included.
For m > 0, 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 - v)), where v = 0 if m > 1 and v = 1/6 if m = 1, 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).
c(1) = A / (2^(1/12) * exp(1/12) * sqrt(Pi)).
c(2) = exp(Pi/12) * Gamma(1/4) / (2^(5/4) * Pi^(5/4)).
c(3) = A * 3^(1/6) * exp(Pi/(6*sqrt(3)) - 1/12) * Gamma(1/3)^2 / (2^(7/4) * Pi^(13/6)), where A = A074962 is the Glaisher-Kinkelin constant.
c(4) = A306620.

Crossrefs

Cf. A079478 (m=1), A324403 (m=2), A324426 (m=3), A324437 (m=4), A324438 (m=5), A324439 (m=6), A324440 (m=7).

Programs

  • Mathematica
    Table[Product[i^8 + j^8, {i, 1, n}, {j, 1, n}], {n, 0, 6}]
  • Python
    from math import prod, factorial
    def A367834(n): return (prod(i**8+j**8 for i in range(1,n) for j in range(i+1,n+1))*factorial(n)**4)**2<Chai Wah Wu, Dec 02 2023

Formula

For n>0, a(n)/a(n-1) = A367833(n)^2 / (2*n^24).
a(n) ~ c * 2^(n*(n+1)) * (1 + 1/(sqrt(1 - 1/sqrt(2)) - 1/2))^(sqrt(2 + sqrt(2))*n*((n+1)/2)) * (1 + 1/(sqrt(1 + 1/sqrt(2)) - 1/2))^(sqrt(2 - sqrt(2))*n*((n+1)/2)) * (n^(8*n^2 - 2) / exp(12*n^2 - Pi*sqrt(1 + 1/sqrt(2))*n*(n+1))), where c = 0.043985703178712025347328240881106818917398444790454628282522057393529338998...

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

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

A272246 a(n) = Product_{k=0..n} (n^3 + k^3).

Original entry on oeis.org

0, 2, 1152, 1428840, 3488808960, 15044494500000, 105235903511101440, 1119277024472896248960, 17216259547948971039129600, 368066786222106315186876633600, 10591209807103301277597696000000000, 399472472359100444604916002033020774400
Offset: 0

Views

Author

Vaclav Kotesovec, Apr 23 2016

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Product[n^3+k^3,{k,0,n}],{n,0,12}]

Formula

a(n) ~ 2^(2*n + 1/2) * n^(3*n + 3) / exp((3 - Pi/sqrt(3))*n).

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

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

A368722 a(n) = Product_{i=1..n, j=1..n, k=1..n} (i^3 + j^3 + k^3).

Original entry on oeis.org

1, 3, 353736000, 4795587079853800623303366670123008000000
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 04 2024

Keywords

Comments

Next term is too long to be included.

Crossrefs

Programs

  • Mathematica
    Table[Product[i^3 + j^3 + k^3, {i, 1, n}, {j, 1, n}, {k, 1, n}], {n, 0, 5}]

Formula

Limit_{n->oo} a(n)^(1/(n^3)) / n^3 = exp(Integral_{x=0..1, y=0..1, z=0..1} log(x^3 + y^3 + z^3) dz dy dx) = 0.5334736919092639993380174031245...

A307209 Decimal expansion of Product_{i>=1, j>=1} (1 + 1/(i^3 + j^3)).

Original entry on oeis.org

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

Views

Author

Vaclav Kotesovec, Mar 28 2019

Keywords

Comments

Product_{i>=1, j>=1} (1 + 1/(i^2 + j^2)) is divergent.
A324443(n) / A324403(n) ~ c * n^(Pi/2), where c = A306398 * 2^(3/4) * exp(-Pi/12) * Pi^(1/4) * Gamma(3/4) = 0.36753062884677326134620846786416595535234038999313...
Product_{i=1..n, j=1..n} (1 + 1/(i + j)) = A324444(n) / A079478(n) ~ 2^(2*n + 1) / (sqrt(Pi)*n^(3/2)).

Examples

			3.50478299933972837589112057043806125583893247862712753541994626614058385...
		

Crossrefs

Programs

  • Mathematica
    (* The iteration cycle: *) $MaxExtraPrecision = 1000; funs[n_] := Product[1 + 1/(i^3 + j^3), {i, 1, n}, {j, 1, n}]; Do[Print[N[Sum[(-1)^(m + j)*funs[j*Floor[200/m]] * j^(m - 1)/(j - 1)!/(m - j)!, {j, 1, m}], 100]], {m, 10, 100, 10}]
  • PARI
    default(realprecision, 50); exp(sumalt(k=1, -(-1)^k/k*sumnum(i=1, sumnum(j=1, 1/(i^3+j^3)^k)))) \\ 15 decimals correct

Formula

Equals limit_{n->infinity} A307210(n) / A324426(n).
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