A204502 Numbers such that floor[a(n)^2 / 9] is a square.

0, 1, 2, 3, 4, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174, 177

Offset: 1

Author

M. F. Hasler, Jan 15 2012

Keywords

• nonn

Comments

Or, numbers n such that n^2, with its last base-9 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 9.)

Links

• M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
• Index to sequences related to truncating digits of squares.

Crossrefs

The squares are in A204503, the squares with last base-9 digit dropped in A204504, and the square roots of the latter in A028310.

Cf. A031149=sqrt(A023110) (base 10), A204514=sqrt(A055872) (base 8), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

Programs

Mathematica

Select[Range[0,200],IntegerQ[Sqrt[Floor[#^2/9]]]&] (* Harvey P. Dale, May 05 2018 *)

PARI

b=9;for(n=0,200,issquare(n^2\b) & print1(n","))

Formula

Conjecture: a(n) = 3*n-12 for n>5. G.f.: x^2*(x^2+x+1)*(x^3-x+1)/(x-1)^2. [Colin Barker, Nov 23 2012]