OFFSET
0,1
COMMENTS
The number of Pythagorean quadruples mod 3^n is given by a(n) 3^(2n-1). See A096018.
LINKS
FORMULA
a(n) = 3^(n+1)-(3^n-(-1)^n)/2.
a(n) = 2*a(n-1)+3*a(n-2). G.f.: (3+x)/((1+x)*(1-3*x)). - Colin Barker, Mar 26 2012
From Klaus Purath, May 28 2026: (Start)
a(n) = 10*3^(n-1) - a(n-1).
a(n) = 20*3^(n-2) + a(n-2).
a(n) = (5*3^n + (-1)^n)/2.
a(n) = A321573(n+1) + (-1)^n.
a(n) = A137340(n) + 2*(-1)^n. (End)
MATHEMATICA
LinearRecurrence[{2, 3}, {3, 7}, 30] (* Harvey P. Dale, Feb 10 2024 *)
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
T. D. Noe, Jun 15 2004
STATUS
approved
