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A238903 Integers k such that (k^2 + (k+1)^2) has no square proper substring.

%I A238903 #16 Dec 09 2024 23:22:32
%S A238903 0,1,3,6,11,18,36,43,56,61,106,136,168,181,206,411,431,511,518,536,
%T A238903 606,613,1056,1068,1388,1631,1636,1668,1686,1693,1806,1813,1956,1981,
%U A238903 2068,2081,3363,3411,3418,3631,3693,3763,4106,4331,5136,5318,5411,5606,5868,6011,6036,6236,6238,6256,6431,6456,6581,10568,10668,10813,11581,11588,11806,11888
%N A238903 Integers k such that (k^2 + (k+1)^2) has no square proper substring.
%C A238903 Inspired by (and program used from) A238334.
%C A238903 Note that (m^2+(m+1)^2), for m>0, always ends with 5. Any other patterns?
%C A238903 From _Robert Israel_, Dec 09 2024: (Start)
%C A238903 The last two digits of k^2 + (k+1)^2 (if more than 2 digits) are 01, 05, 13, 21, 25, 41, 45, 61, 65, 81, or 85.  The only ones of these that don't contain the squares 0, 1, 4, or 25 are 65 and 85, so all terms k > 3 of this sequence have k^2 + (k+1)^2 ending in 65 or 85. (End)
%H A238903 Robert Israel, <a href="/A238903/b238903.txt">Table of n, a(n) for n = 1..1000</a>
%e A238903 1^2 + 2^2 = 5, 3^2 + 4^2 = 25, 6^2 + 7^2 = 85.
%p A238903 filter:= proc(m) local n,i,j,S;
%p A238903   n:= m^2 + (m+1)^2;
%p A238903 S:= {seq(seq(floor((n mod 10^i)/10^j),j=0..i-1),i=1 .. ilog10(n)+1)} minus {n};
%p A238903   not ormap(issqr,S);
%p A238903 end proc:
%p A238903 select(filter, [$0..20000]); # _Robert Israel_, Dec 09 2024
%Y A238903 Cf. A001844, A130448, A238334, A238335.
%K A238903 nonn,base
%O A238903 1,3
%A A238903 _Zak Seidov_, Mar 07 2014