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A359846 a(n) = (((5 - (n mod 2))*10^(3 + n*(9/2) - (n mod 2)*(5/2)))^2 + 2)/81.

%I A359846 #49 Jan 23 2025 12:37:59
%S A359846 308642,1975308642,308641975308641975308642,
%T A359846 1975308641975308641975308642,
%U A359846 308641975308641975308641975308641975308642,1975308641975308641975308641975308641975308642,308641975308641975308641975308641975308641975308641975308642
%N A359846 a(n) = (((5 - (n mod 2))*10^(3 + n*(9/2) - (n mod 2)*(5/2)))^2 + 2)/81.
%C A359846 Also numbers of the form ((d*10^k)^2 + 2)/9^2 that are not squares, where d is a single-digit number.
%C A359846 The square roots of these numbers show runs of equal digits, see the link to Schizophrenic numbers.
%H A359846 Paolo Xausa, <a href="/A359846/b359846.txt">Table of n, a(n) for n = 0..100</a>
%H A359846 K. S. Brown, <a href="http://www.mathpages.com/home/kmath404.htm">Schizophrenic numbers</a>
%H A359846 Mathematics StackExchange, <a href="https://math.stackexchange.com/questions/2134903/is-there-any-mathematical-reason-for-this-digit-repetition-show">Is there any mathematical reason for this 'digit-repetition-show'?</a>
%H A359846 Wikipedia, <a href="http://wikipedia.org/wiki/Schizophrenic_number">Schizophrenic Number</a>
%H A359846 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1000000000000000000,-1000000000000000000).
%F A359846 G.f.: (2/81)*(1/(1-x)+6249960/(1+1000000000*x)+6250040/(1-1000000000*x)).
%F A359846 a(n) = a(n-1) + 10^18*a(n-2) - 10^18*a(n-3).
%F A359846 a(2*n) = (25*10^(6 + 18*n) + 2)/81.
%F A359846 a(2*n + 1) = (16*10^(10 + 18*n) + 2)/81.
%F A359846 We use in the next formulas a special notation for real numbers where (x) after a digit denotes a run of length x for this digit. Example: 3(4).2(3) is 3333.2222 .
%F A359846 sqrt(a(2*n)) = 5(3+9*n).5(4+9*n)7(8+18*n)3(7+18*n)5(1)1(6+18*n)0(1)2(7+18*n)7(1)1(1)9(5+18*n)7(1)0(1)1(1)3(4+18*n)... .
%F A359846 sqrt(a(2*n+1)) = 4(5+9*n).4(6+9*n)7(1)2(10+18*n)1(1)3(1)5(1)4(1)1(1)6(6+18*n)7(1)2(1)0(1)9(1)2(1)0(1)1(1)3(1)8(4+18*n)... .
%F A359846 sqrt(1/a(2*n)) = 0.0(2+9*n)1(1)7(1)9(6+18*n)2(1)8(1)0(5+18*n)... .
%F A359846 sqrt(1/a(2*n+1)) = 0.0(4+9*n)2(2)4(1)9(8+18*n)8(1)5(1)9(1)3(1)7(1)5(1)0(5+18*n)... .
%F A359846 sqrt(a(2*n)-(2/81)) = 10^(4+9*n)/18.
%F A359846 sqrt(a(2*n+1)-(2/81)) = 10^(7+9*n)/225.
%e A359846 a(0) = 308642 and sqrt(a(0)) = 555.555577777777333333351111110222222271999997013333521... .
%e A359846 a(1) = 1975308642 and sqrt(a(1)) = 44444.4444447222222222213541666666720920138888465033637... .
%t A359846 A359846[n_] := (10^(9*n - 5*# + 6)*(# - 5)^2 + 2)/81 & [Mod[n, 2]]; Array[A359846, 10, 0] (* or *)
%t A359846 LinearRecurrence[{1, 10^18, -10^18}, {308642, 1975308642, 308641975308641975308642}, 10] (* _Paolo Xausa_, Jan 23 2025 *)
%o A359846 (PARI) a(n) = (((5-(n%2))*10^(3+n*(9/2)-(n%2)*(5/2)))^2+2)/81
%Y A359846 Cf. A014824.
%K A359846 nonn,easy,base
%O A359846 0,1
%A A359846 _Thomas Scheuerle_, Jan 15 2023