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%I #18 Feb 27 2021 04:19:34
%S 121,144,169,268,274,280,319,400,441,484,519,574,900,961,1000,1191,
%T 1331,8000,10000,14641,100000,1000000
%N n^2 in base b is n in base 10 for some base b < n.
%C The corresponding bases b are 120, 142, 166, 188, 192, 196, 184, 200, 220, 241, 232, 256, 300, 320, 100, 112, 120, ...
%C The squares of the sequence are 121, 144, 169, 400, 441, 484, 900, ...
%C Includes 10^k for all k > 2, as 10^(2k) in base 100 is 10^k in base 10. - _Robert Israel_, Feb 26 2021
%e 319 is in the sequence because 319^2 in base 184 is 319 in base 10.
%p for n from 1 to 2000 do: x:=convert(n,base,10): for b from 2 to n-1 do: y:=convert(n^2,base,b): if x=y then printf ( "%d %d \n",n,b) fi:od:od:
%p # Alternative:
%p f:= proc(n) local L,k,x,P,S;
%p L:= convert(n,base,10);
%p P:= add(L[k]*x^(k-1),k=1..nops(L)) - n^2;
%p S:= map(t -> rhs(op(t)),[isolve(P)]);
%p ormap(t -> t > max(L) and t < n ,S);
%p end proc:
%p select(f, [$2..100000]); # _Robert Israel_, Feb 26 2021
%o (PARI) isok(m) = {for (b=2, m-1, my(d=digits(m^2, b)); if ((vecmax(d) < 10) && (fromdigits(d) == m), return (1)););} \\ _Michel Marcus_, Feb 27 2021
%K nonn,base,hard
%O 1,1
%A _Michel Lagneau_, May 12 2014
%E Definition corrected and more terms by _Robert Israel_, Feb 26 2021
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