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%I #15 Dec 07 2015 22:53:36
%S 1,1,2,26,53,1354,13562,26858,200965,40193,3859882,13496122,62298370,
%T 12459674,4106065226,4044371993,69072101242,218014644394,
%U 3137550252170,627510050434,66696011833378,280704828874769
%N a(n) = (A048898(n)^2 + 1)/5^n, n >= 0.
%C a(n) is integer (nonnegative) because b(n):=A048898(n) satisfies b(n)^2 + 1 == 0 (mod 5^n), n>=0. The solution of this congruence for n>=1, which satisfies also b(n) == 2 (mod 5), is b(n) = 2^(5^(n-1)) (mod 5^n), but this is inconvenient for computing b(n) for large n. Instead one can use the b(n) recurrence which follows immediately, and this is given in the formula field below. To prove that the given b(n) formula solves the first congruence one can analyze the binomial expansion of (5 - 1)^(5^(n-1)) + 1 and show that it is 0 (mod 5^n)) term by term. The second congruence reduces to b(n) == 2^(5^(n-1)) (mod 5) which follows for n>=1 by induction. Because b(n) = 5^n - A048899(n) one could also use the result A048899(n) == 3 (mod 5), once this has been proved.
%C The fact that X^2 + 1 == 0 (mod 5^n) has precisely two solutions for each n>=1, called x(n) and y(n), follows from the fact that X^2 + 1 == 0 (mod 5) has the two simple roots x(1) = 2 and y(1) = 3, and a theorem, given, e.g., in the Nagell reference as Theorem 50 on p. 87. From that same theorem, it also follows that one can choose all x(n) == 2 (mod 5) and all y(n) == 3 (mod 5).
%D T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.
%F a(n) = (b(n)^2+1)/5^n, n>=0, with b(n) = A048898(n) given by the recurrence b(n) = b(n-1)^5 (mod 5^n), n>=2, b(0):=0, b(1)=2 (see the Mathematica Program by _Jean-François Alcover_ at A048898).
%e a(0) = 1/1 = 1.
%e a(3) = (57^2 + 1)/5^3 = 26 (b(3) = 7^5 (mod 5^3) = 57).
%p b:=proc(n) option remember: if n=0 then 0 elif n=1 then 2
%p else modp(b(n-1)^5,5^n) fi: end proc:
%p [seq((b(n)^2+1)/5^n,n=0..29)];
%Y Cf. A048898, A048899, A210849 (companion sequence).
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Apr 28 2012
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