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.

A266797 a(n) = (6^n + 4^n + 3*2^n)/8.

Original entry on oeis.org

2, 8, 38, 200, 1112, 6368, 37088, 218240, 1292672, 7689728, 45874688, 274196480, 1640978432, 9829081088, 58907353088, 353175633920, 2117979963392, 12703584616448, 76204327436288, 457157244354560, 2742668586647552, 16454912005111808, 98725073977868288
Offset: 1

Views

Author

Nathaniel Johnston, Jan 03 2016

Keywords

Comments

Gives the number of ways that the product of the values on n different 6-sided dice can be a perfect square. Thus a(n)/6^n is the probability that the product of n different 6-sided dice is a perfect square.

Examples

			a(1) = 2 because there are two ways for one die to be a perfect square: if its value is 1 or 4.
a(2) = 8 because there are eight ways for the product of the values on two dice to result in perfect squares: 1*1, 1*4, 2*2, 3*3, 4*1, 4*4, 5*5, 6*6.
		

Programs

  • Maple
    seq((6^n+4^n+3*2^n)/8, n = 1 .. 40);
  • PARI
    a(n) = 2^(n-3)*(2^n+3^n+3) \\ Colin Barker, Jan 08 2016
    
  • PARI
    Vec(2*x*(1-3*x)*(1-5*x)/((1-2*x)*(1-4*x)*(1-6*x)) + O(x^30)) \\ Colin Barker, Jan 08 2016

Formula

From Colin Barker, Jan 08 2016: (Start)
a(n) = 2^(n - 3)*(2^n + 3^n + 3).
a(n) = 12*a(n-1) - 44*a(n-2) + 48*a(n-3) for n>3.
G.f.: 2*x*(1 - 3*x)*(1 - 5*x) / ((1 - 2*x)*(1 - 4*x)*(1 - 6*x)).
(End)