Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #23 Jun 21 2024 15:20:05
%S 60,12,2,24,6,8,12,3,4,12,6,8,6,24,4,3,12,4,6,24,2,24,12,1
%N Greatest common divisor of the product of terms in positive Pythagorean n-tuples, over all possible such tuples.
%C A Pythagorean n-tuple consists of positive integers X1,...,Xn where X1^2 + X2^2 + ... + X(n-1)^2 = Xn^2. The product of those terms is P = X1*X2*...*Xn.
%C a(n) is the GCD of all possible products P arising this way.
%C If n is odd, then 2 | a(n).
%C If n mod 3 is 0 or 1, then 3 | a(n).
%C If n is divisible by 4, then 4 | a(n).
%C If n mod 8 is 1 or 3, then 4 | a(n).
%C If n mod 8 is 0 or 6, then 8 | a(n).
%C If n mod 24 is 2, then a(n)=1.
%C a(n+24)=a(n) for n >= 4.
%H Des MacHale and Christian van den Bosch, <a href="https://www.jstor.org/stable/23249521">Generalising a result about Pythagorean triples</a>, The Mathematical Gazette, Vol. 96, March 2012.
%e For n = 8, then 8 | a(n). Since 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 = 4^2 with P = 1*1*1*1*2*2*2*4 = 32 and 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2 = 5^2 with P = 1*1*1*1*1*2*4*5 = 40, then a(8) = gcd(32,40) = 8, and no larger number will divide the product of terms in every Pythagorean octuple.
%K nonn,more
%O 3,1
%A _Brian Almond_, May 28 2024