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%I #14 Mar 13 2021 00:48:26
%S 2357,2582,3334,4714,5774,6667,8165,8819,9428,10541,10542,10543,10544,
%T 10545,14907,14908,14909,18257,18258,18259,21081,21082,21083,23570,
%U 23571,25819,25820,27888,27889,29813,29814,31622,33332,33333
%N Numbers whose square starts with 4 identical digits.
%H Robert Israel, <a href="/A132391/b132391.txt">Table of n, a(n) for n = 1..10000</a>
%e Example: 2357^2 = 5555449.
%p R:= NULL: count:= 0:
%p for d from 1 while count < 100 do
%p for i from 1 to 9 do
%p L:= i*1111*10^d;
%p X:= [$ceil(sqrt(L)) .. floor(sqrt(L+10^d-1))];
%p m:= nops(X);
%p if m > 0 then
%p count:= count+nops(X);
%p R:= R, op(X);
%p fi
%p od od:
%p R; # _Robert Israel_, Mar 12 2021
%t Select[Range[10, 50000], Length[Union[Take[IntegerDigits[ #^2], 4]]] == 1 & ]
%t (* or *)
%t (* Here's a more generic Mathematica program that calculates the first q terms of squares starting with n identical digits *)
%t n=4; q=30; t=Table[(10^n-1)*i/9, {i,1,9}]; u=Sqrt[Union[t,10*t]];
%t v=Sqrt[Union[t+1, 10*(t+1)]]; k=1; While[s=Sort[Flatten[Table[Union
%t [Table[Range[Ceiling[10^j*u[[i]]], f=10^j*v[[i]]; If[IntegerQ[f],
%t f=f-1]; Floor[f]], {i,1,18}]], {j,0,k}]]]; Length[s]<q, k++ ]; Take[s,q]
%t (* _Hans Havermann_, Aug 30 2007 *)
%o (Python)
%o def aupto(limit):
%o alst = []
%o for m in range(34, limit+1):
%o if len(set(str(m*m)[:4])) == 1: alst.append(m)
%o return alst
%o print(aupto(33333)) # _Michael S. Branicky_, Mar 12 2021
%Y Cf. A119887, A119511, A131573, A119866, A133183.
%K nonn,base,look
%O 1,1
%A _Jonathan Vos Post_, Aug 29 2007
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