A266797 - OEIS

%I #14 Jan 08 2016 09:34:43

%S 2,8,38,200,1112,6368,37088,218240,1292672,7689728,45874688,274196480,

%T 1640978432,9829081088,58907353088,353175633920,2117979963392,

%U 12703584616448,76204327436288,457157244354560,2742668586647552,16454912005111808,98725073977868288

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

%C 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.

%H Colin Barker, <a href="/A266797/b266797.txt">Table of n, a(n) for n = 1..1000</a>

%H Math.StackExchange, <a href="http://math.stackexchange.com/questions/1592140/when-the-product-of-dice-rolls-yields-a-square">When the product of dice rolls yields a square</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (12,-44,48).

%F From _Colin Barker_, Jan 08 2016: (Start)

%F a(n) = 2^(n - 3)*(2^n + 3^n + 3).

%F a(n) = 12*a(n-1) - 44*a(n-2) + 48*a(n-3) for n>3.

%F G.f.: 2*x*(1 - 3*x)*(1 - 5*x) / ((1 - 2*x)*(1 - 4*x)*(1 - 6*x)).

%F (End)

%e a(1) = 2 because there are two ways for one die to be a perfect square: if its value is 1 or 4.

%e 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.

%p seq((6^n+4^n+3*2^n)/8, n = 1 .. 40);

%o (PARI) a(n) = 2^(n-3)*(2^n+3^n+3) \\ _Colin Barker_, Jan 08 2016

%o (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

%K nonn,easy

%O 1,1

%A _Nathaniel Johnston_, Jan 03 2016