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

0, 1, 3, 6, 11, 18, 36, 43, 56, 61, 106, 136, 168, 181, 206, 411, 431, 511, 518, 536, 606, 613, 1056, 1068, 1388, 1631, 1636, 1668, 1686, 1693, 1806, 1813, 1956, 1981, 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

Offset: 1

Author

Zak Seidov, Mar 07 2014

Keywords

• nonn
• base

Comments

Inspired by (and program used from) A238334.

Note that (m^2+(m+1)^2), for m>0, always ends with 5. Any other patterns?

From Robert Israel, Dec 09 2024: (Start)

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)

Examples

			1^2 + 2^2 = 5, 3^2 + 4^2 = 25, 6^2 + 7^2 = 85.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A001844, A130448, A238334, A238335.

Programs

Maple

filter:= proc(m) local n,i,j,S;
  n:= m^2 + (m+1)^2;
S:= {seq(seq(floor((n mod 10^i)/10^j),j=0..i-1),i=1 .. ilog10(n)+1)} minus {n};
  not ormap(issqr,S);
end proc:
select(filter, [$0..20000]); # Robert Israel, Dec 09 2024