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

1,3

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)

Links

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

Example

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

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

Crossrefs

Cf. A001844, A130448, A238334, A238335.

Sequence in context: A058053 A264923 A321381 * A271079 A091094 A332446

Adjacent sequences: A238900 A238901 A238902 * A238904 A238905 A238906

Keyword

nonn,base

Author

Zak Seidov, Mar 07 2014

Status

approved