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 1-2 of 2 results.

A369680 a(n) = Product_{k=0..n} (2^k + 3^(n-k)).

Original entry on oeis.org

2, 12, 250, 19404, 5780918, 6691848108, 30261978906250, 535757771934053916, 37171553237849766044342, 10113067879819381109893992732, 10789224041146220828897229003906250, 45150513047221188662211059385153001179564, 741117672560101894851755994230829254062662140918
Offset: 0

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Author

Paul D. Hanna, Feb 06 2024

Keywords

Comments

For p > 1, q > 1, limit_{n->oo} (Product_{k=0..n} (p^k + q^(n-k)) )^(1/n^2) = q^(1/2)*p^(1/(2*(1 + log(q)/log(p)))) = p^(1/2)*q^(1/(2*(1 + log(p)/log(q)))). - Vaclav Kotesovec, Feb 07 2024
Equivalently, limit_{n->oo} ( Product_{k=0..n} (p^k + q^(n-k)) )^(1/n^2) = exp((1/2) * (log(p)^2 + log(p)*log(q) + log(q)^2) / log(p*q)). - Paul D. Hanna, Feb 08 2024

Examples

			a(0) = (1 + 1) = 2;
a(1) = (1 + 3)*(2 + 1) = 12;
a(2) = (1 + 3^2)*(2 + 3)*(2^2 + 1) = 250;
a(3) = (1 + 3^3)*(2 + 3^2)*(2^2 + 3)*(2^3 + 1) = 19404;
a(4) = (1 + 3^4)*(2 + 3^3)*(2^2 + 3^2)*(2^3 + 3)*(2^4 + 1) = 5780918;
a(5) = (1 + 3^5)*(2 + 3^4)*(2^2 + 3^3)*(2^3 + 3^2)*(2^4 + 3)*(2^5 + 1) = 6691848108;
...
RELATED SERIES.
Sum_{n>=0} Product_{k=0..n} (1/2^k + 1/3^(n-k)) = 2 + 12/6 + 250/6^3 + 19404/6^6 + 5780918/6^10 + 6691848108/6^15 + ... + a(n)/6^(n*(n+1)/2) + ... = 5.6846111010137973166832330595516662115250385271...
		

Crossrefs

Programs

  • Mathematica
    Table[Product[2^k+3^(n-k),{k,0,n}],{n,0,12}] (* James C. McMahon, Feb 07 2024 *)
  • PARI
    {a(n) = prod(k=0, n, 2^k + 3^(n-k))}
    for(n=0, 15, print1(a(n), ", "))

Formula

a(n) = Product_{k=0..n} (2^k + 3^(n-k)).
a(n) = 6^(n*(n+1)/2) * Product_{k=0..n} (1/2^k + 1/3^(n-k)).
a(n) = 3^(n*(n+1)/2) * Product_{k=0..n} (1 + 2^n/6^k).
a(n) = 2^(n*(n+1)/2) * Product_{k=0..n} (1 + 3^n/6^k).
a(n) = 2^(-n*(n+1)/2) * Product_{k=0..n} (2^n + 6^k).
a(n) = 3^(-n*(n+1)/2) * Product_{k=0..n} (3^n + 6^k).
a(n) = 2^(n*(n+1)/2)*QPochhammer(-3^n, 1/6, n + 1). - Stefano Spezia, Feb 07 2024
Limit_{n->oo} a(n)^(1/n^2) = 2^(1/(2*(1 + log(3)/log(2)))) * sqrt(3) = 3^(1/(2*(1 + log(2)/log(3)))) * sqrt(2) = 1.9805589654474717155611061670180902111915926... - Vaclav Kotesovec, Feb 07 2024
Equivalently, limit_{n->oo} a(n)^(1/n^2) = exp((1/2) * (log(2)^2 + log(2)*log(3) + log(3)^2) / log(6)). - Paul D. Hanna, Feb 08 2024

A323716 a(n) = Product_{k=0..n} (3^k + 1).

Original entry on oeis.org

2, 8, 80, 2240, 183680, 44817920, 32717081600, 71584974540800, 469740602936729600, 9246374028206585446400, 545998386365598870609920000, 96722522147893108730806108160000, 51402410615320609490117059732766720000
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 25 2019

Keywords

Crossrefs

Programs

  • Magma
    [(&*[3^j+1: j in [0..n]]): n in [0..20]]; // G. C. Greubel, Aug 30 2023
    
  • Mathematica
    Table[Product[3^k+1, {k, 0, n}], {n, 0, 12}]
    Table[QPochhammer[-1, 3, n+1], {n, 0, 12}]
  • PARI
    a(n) = prod(k=0, n, 3^k+1); \\ Michel Marcus, Jan 25 2019
    
  • SageMath
    [product(3^j+1 for j in range(n+1)) for n in range(21)] # G. C. Greubel, Aug 30 2023

Formula

a(n) ~ c * 3^(n*(n+1)/2), where c = A132323 = QPochhammer[-1, 1/3] = 3.12986803713402307587769821345767...
Showing 1-2 of 2 results.