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.

A263394 a(n) = Product_{i=1..n} (3^i - 2^i).

Original entry on oeis.org

1, 5, 95, 6175, 1302925, 866445125, 1784010512375, 11248186280524375, 215638979183932793125, 12512451767147700321078125, 2190917791975795178520458609375, 1155369543009475708416871245360859375, 1832567448623162714866960405275465241328125
Offset: 1

Views

Author

Bob Selcoe, Mar 02 2016

Keywords

Comments

Generally, for sequences of the form a(n) = Product_{i=1..n} j^i-k^i, where j>k>=1 and n>=1: given probability p=(k/j)^n that an outcome will occur at the n-th stage of an infinite process, then r = 1 - a(n)/j^((n^2+n)/2) is the probability that the outcome has occurred up to and including the n-th iteration. Here, j=3 and k=2, so p=(2/3)^n and r = 1-a(n)/A047656(n+1). The limiting ratio of r ~ 0.9307279.

Crossrefs

Cf. sequences of the form Product_{i=1..n}(j^i - 1): A005329 (j=2), A027871 (j=3), A027637 (j=4), A027872 (j=5), A027873 (j=6), A027875 (j=7),A027876 (j=8), A027877 (j=9), A027878 (j=10), A027879 (j=11), A027880 (j=12).
Cf. sequences of the form Product_{i=1..n}(j^i - k^1), k>1: A269576 (j=4, k=3), A269661 (j=5, k=4).

Programs

Formula

a(n) = Product_{i=1..n} A001047(i).
a(n) ~ c * 3^(n*(n+1)/2), where c = QPochhammer(2/3) = 0.0692720728018644... . - Vaclav Kotesovec, Oct 10 2016

A269576 a(n) = Product_{i=1..n} (4^i - 3^i).

Original entry on oeis.org

1, 7, 259, 45325, 35398825, 119187843775, 1692109818073675, 99792176520894983125, 24195710911432718503470625, 23942309231057283642583777144375, 96180015123706384385790918441966041875
Offset: 1

Views

Author

Bob Selcoe, Mar 02 2016

Keywords

Comments

In general, for sequences of the form a(n) = Product_{i=1..n} j^i-k^i, where j>k>=1 and n>=1: given probability p=(k/j)^n that an outcome will occur at the n-th stage of an infinite process, then r = 1 - a(n)/j^((n^2+n)/2) is the probability that the outcome has occurred at or before the n-th iteration. Here j=4 and k=3, so p=(3/4)^n and r = 1-a(n)/A053763(n+1). The limiting ratio of r is ~ 0.9844550.

Crossrefs

Cf. sequences of the form Product_{i=1..n}(j^i - 1): A005329 (j=2), A027871 (j=3), A027637 (j=4), A027872 (j=5), A027873 (j=6), A027875 (j=7),A027876 (j=8), A027877 (j=9), A027878 (j=10), A027879 (j=11), A027880 (j=12).
Cf. sequences of the form Product_{i=1..n}(j^i - k^1), k>1: A263394 (j=3, k=2), A269661 (j=5, k=4).

Programs

  • Maple
    seq(mul(4^i-3^i,i=1..n),n=0..20); # Robert Israel, Jun 01 2023
  • Mathematica
    Table[Product[4^i - 3^i, {i, n}], {n, 11}] (* Michael De Vlieger, Mar 07 2016 *)
    FoldList[Times,Table[4^n-3^n,{n,20}]] (* Harvey P. Dale, Jul 30 2018 *)
  • PARI
    a(n) = prod(k=1, n, 4^k-3^k); \\ Michel Marcus, Mar 05 2016

Formula

a(n) = Product_{i=1..n} A005061(i).
a(n) ~ c * 2^(n*(n+1)), where c = QPochhammer(3/4) = 0.015545038845451847... . - Vaclav Kotesovec, Oct 10 2016
a(n+3)/a(n+2) - 7 * a(n+2)/a(n+1) + 12 * a(n+1)/a(n) = 0. - Robert Israel, Jun 01 2023
Showing 1-2 of 2 results.