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%I A052049 #19 May 12 2023 12:21:17
%S A052049 88,478,577,583,715,836,880,881,893,3362,3386,3911,4077,4780,5077,
%T A052049 5239,5369,5770,5784,5789,5830,5858,6523,6756,6772,6926,6941,7107,
%U A052049 7150,7359,7535,7827,8043,8196,8229,8360,8525,8810,8930,8989,9251,9701,9764,9786
%N A052049 a(n)^2 is a square whose digits occur with an equal minimum frequency of 2.
%C A052049 There are A225428(10) = 597959 terms in this sequence. The last term is 9994363488, whose square is 99887301530267526144 = A052050(597959). - _Hugo Pfoertner_, May 12 2023
%H A052049 Nathaniel Johnston, <a href="/A052049/b052049.txt">Table of n, a(n) for n = 1..5000</a>
%H A052049 Patrick De Geest, <a href="http://www.worldofnumbers.com/samedigits.htm">Numbers whose digits occur with same frequency</a>
%e A052049 577^2 = 332929, which contains each of its digits (2, 3, and 9) twice, so 577 is in this sequence.
%p A052049 isA052049 := proc(n) local d, k, fr, eqfr: d:=convert(n^2, base, 10): eqfr:=true: fr:=numboccur(d[1], d): if(fr=1)then return false: fi: for k from 0 to 9 do if(not member(numboccur(k, d), {fr, 0}))then eqfr:=false: break: fi: od: return eqfr: end: seq(`if`(isA052049(n), n, NULL), n=1..9800); # _Nathaniel Johnston_, Jun 02 2011
%t A052049 ta[n_]:=DeleteDuplicates[Transpose[Tally[IntegerDigits[n^2]]][[2]]]; t ={}; Do[If[Length[x=ta[n]]==1 && x[[1]]>=2, AppendTo[t,n]],{n,9800}]; t (* _Jayanta Basu_, May 11 2013 *)
%Y A052049 Cf. A045540, A052046, A052047, A052048, A052050, A052051, A052052, A225428.
%K A052049 nonn,base,fini
%O A052049 1,1
%A A052049 _Patrick De Geest_, Dec 15 1999

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