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%I #33 Mar 15 2019 22:48:58
%S 16,36,64,81,100,121,144,196,225,256,324,400,441,484,576,625,676,729,
%T 784,900,1024,1089,1156,1225,1296,1444,1521,1600,1764,1936,2025,2116,
%U 2304,2401,2500,2601,2704,2916,3025,3136,3249,3364,3600,3844,3969,4096,4225,4356,4624,4761,4900,5184
%N Brazilian squares.
%C Trivially, all even squares > 4 will be in this sequence.
%C The only square of a prime which is Brazilian is 121. - _Bernard Schott_, May 01 2017
%C Intersection of A000290 and A125134. - _Felix Fröhlich_, May 01 2017
%C Conjecture: Let r(n) = (a(n) - 1)/(a(n) + 1); then Product_{n>=1} r(n) = (15/17) * (35/37) * (63/65) * (40/41) * (99/101) * (60/61) * (143/145) * (195/197) * ... = (150 * Pi) / (61 * sinh(Pi)) = 0.668923905.... - _Dimitris Valianatos_, Feb 27 2019
%H Vincenzo Librandi, <a href="/A253260/b253260.txt">Table of n, a(n) for n = 1..405</a>
%H Bernard Schott, <a href="/A253260/a253260.pdf">Les nombres brésiliens</a> Quadrature, no. 76, avril-juin 2010, théorème 5, page 37.
%e From _Bernard Schott_, May 01 2017: (Start)
%e a(1) = 16 = 4^2 = 22_7.
%e a(6) = 121 = 11^2 = 11111_3. (End)
%t fQ[n_]:=Module[{b=2, found=False}, While[b<n-1&&Length[Union[IntegerDigits[n, b]]]>1, b++]; b<n-1]; Select[Range[1, 80]^2, fQ] (* _Vincenzo Librandi_, May 02 2017 *)
%o (PARI) for(n=4, 10^4, for(b=2, n-2, d=digits(n, b); if(vecmin(d)==vecmax(d)&&issquare(n), print1(n, ", "); break)))
%Y Cf. A000290, A125134.
%K nonn,base
%O 1,1
%A _Derek Orr_, Apr 30 2015
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