1,2

Square root of 'Squares from A023110 with last digit removed'.

One could include an initial '0', and even list it with multiplicity 3 or 4, since 00, 01, 04 and 09 are all perfect squares: In analogy to corresponding sequences for other bases, this sequence could be defined as sqrt(floor[A023110/10]), see A204512 [base 8], A204517 (base 7), A204519 (base 6), A204521 [base 5], A001353 [base 3], A001542 [base 2]. (For bases 4 and 9, the corresponding sequence contains all integers.) - M. F. Hasler, Jan 16 2012

R. K. Guy, Neg and Reg, preprint, Jan 2012.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012

Joshua Stucky, Pell's Equation and Truncated Squares, Number Theory Seminar, Kansas State University, Feb 19 2018.

Index to sequences related to truncating digits of squares.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,38,0,0,0,0,0,0,-1).

G.f.: x*(x^10+2*x^9+4*x^8+5*x^7+18*x^6+12*x^5+6*x^4+5*x^3+4*x^2+2*x+1) / (x^14-38*x^7+1). - Colin Barker, Jan 30 2013

5^2 = 25 and 16^2 = 256, so 5 is in the sequence.

115364^2 = 13308852496, 364813^2 = 133088524969.

for i from 1 to 150000 do if (floor(sqrt(10 * i^2 + 9)) > floor(sqrt(10 * i^2))) then print(i) end if end do;

CoefficientList[Series[(x^10 + 2 x^9 + 4 x^8 + 5 x^7 + 18 x^6 + 12 x^5 + 6 x^4 + 5 x^3 + 4 x^2 + 2 x + 1)/(x^14 - 38 x^7 + 1), {x, 0, 50}], x] (* Vincenzo Librandi, Oct 19 2013 *)

LinearRecurrence[{0, 0, 0, 0, 0, 0, 38, 0, 0, 0, 0, 0, 0, -1}, {1, 2, 4, 5, 6, 12, 18, 43, 80, 154, 191, 228, 456, 684}, 40] (* Harvey P. Dale, Jun 09 2017 *)

Cf. A023110, A030686, A030687, A053784.

See A202303 for the resulting squares.

Sequence in context: A276001 A182109 A006539 * A125775 A331624 A191165

Adjacent sequences: A031147 A031148 A031149 * A031151 A031152 A031153

nonn,base,easy

approved