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A210848 a(n) = (A048898(n)^2 + 1)/5^n, n >= 0.

%I A210848 #28 Feb 18 2025 05:57:11
%S A210848 1,1,2,26,53,1354,13562,26858,200965,40193,3859882,13496122,62298370,
%T A210848 12459674,4106065226,4044371993,69072101242,218014644394,
%U A210848 3137550252170,627510050434,66696011833378,280704828874769,2167389209973245,433477841994649,41870795375097221,40277856145834642
%N A210848 a(n) = (A048898(n)^2 + 1)/5^n, n >= 0.
%C A210848 a(n) is an 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 A210848 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 A210848 T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.
%H A210848 Paolo Xausa, <a href="/A210848/b210848.txt">Table of n, a(n) for n = 0..1000</a>
%F A210848 a(n) = (A048898(n)^2 + 1)/5^n.
%e A210848 a(0) = 1/1 = 1.
%e A210848 a(3) = (57^2 + 1)/5^3 = 26 (b(3) = 7^5 (mod 5^3) = 57).
%p A210848 b:=proc(n) option remember: if n=0 then 0 elif n=1 then 2
%p A210848 else modp(b(n-1)^5,5^n) fi: end proc:
%p A210848 [seq((b(n)^2+1)/5^n,n=0..29)];
%t A210848 Join[{1}, MapIndexed[(#^2 + 1)/5^#2[[1]] &, FoldList[PowerMod[#, 5, 5^#2] &, 2, Range[2, 25]]]] (* _Paolo Xausa_, Jan 14 2025 *)
%Y A210848 Cf. A048898, A048899, A210849 (companion sequence).
%K A210848 nonn,easy
%O A210848 0,3
%A A210848 _Wolfdieter Lang_, Apr 28 2012