A031149 Numbers whose square with its last digit deleted is also a square.

0, 1, 2, 3, 4, 7, 13, 16, 19, 38, 57, 136, 253, 487, 604, 721, 1442, 2163, 5164, 9607, 18493, 22936, 27379, 54758, 82137, 196096, 364813, 702247, 870964, 1039681, 2079362, 3119043, 7446484, 13853287, 26666893, 33073696, 39480499, 78960998, 118441497, 282770296

Offset: 1

Author

Patrick De Geest

Keywords

• nonn
• base

Comments

Square root of A023110(n).

For the first 4 terms, the square has only one digit, but in analogy to the sequences in other bases (A204502, A204512, A204514, A204516, A204518, A204520, A004275, A055793, A055792), it is understood that deleting this digit yields 0.

From Robert Israel, Feb 16 2016: (Start)

Solutions x to x^2 = 10*y^2 + j, j in {0,1,4,6,9}, in increasing order of x.

j=0 occurs only for x=0.

Let M be the 2 X 2 matrix [19, 60; 6, 19].

Solutions of x^2 = 10*y^2 + 1 are (x,y)^T = M^k (1,0)^T for k >= 0.

Solutions of x^2 = 10*y^2 + 4 are (x,y)^T = M^k (2,0)^T for k >= 0.

Solutions of x^2 = 10*y^2 + 6 are (x,y)^T = M^k (4,1)^T and M^k (16,5)^T for k >= 0.

Solutions of x^2 = 10*y^2 + 9 are (x,y)^T = M^k (3,0)^T, M^k (7,2)^T, M^k (13,4)^T for k >= 0.

Since (1,0)^T <= (2,0)^T <= (3,0)^T <= (4,1)^T <= (7,2)^T <= (13,4)^T <= (16,5)^T <= (19,6)^T = M (1,0)^T (element-wise) and M has positive entries, we see that the terms always occur in this order, for successive k.

The eigenvalues of M are 19 + 6*sqrt(10) and 19 - 6*sqrt(10).

From this follow my formulas below and the G.f. (End)

Examples

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

References

• R. K. Guy, Neg and Reg, preprint, Jan 2012. [From N. J. A. Sloane, Jan 12 2012]

Links

• Dmitry Petukhov and Robert Israel, Table of n, a(n) for n = 1..4400 (n = 1..67 from Dmitry Petukhov)
• M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
• Index to sequences related to truncating digits of squares.

Crossrefs

Cf. A000290, A023110, A030686, A030687, A031150.

Programs

Maple

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

Mathematica

fQ[n_] := IntegerQ@ Sqrt@ Quotient[n^2, 10]; Select[ Range[ 0, 40000000], fQ] (* Harvey P. Dale, Jun 15 2011 *) (* modified by Robert G. Wilson v, Jan 16 2012 *)

PARI

s=[]; for(n=0, 1e7, if(issquare(n^2\10), s=concat(s,n))); s \\ Colin Barker, Jan 17 2014; typo fixed by Zak Seidov, Jan 31 2014

Formula

Appears to satisfy: a(n)=38a(n-7)-a(n-14) which would require a(-k) to look like 3, 2, 1, 4, 7, 13, 16, 57, 38, 19, 136, ... for k>0. - Henry Bottomley, May 08 2001

Empirical g.f.: x^2*(1 + 2*x + 3*x^2 + 4*x^3 + 7*x^4 + 13*x^5 + 16*x^6 - 19*x^7 - 38*x^8 - 57*x^9 - 16*x^10 - 13*x^11 - 7*x^12 - 4*x^13) / ((1 - 38*x^7 + x^14)). - Colin Barker, Jan 17 2014

a(n) = 38*a(n-7) - a(n-14) for n>15 (conjectured). - Colin Barker, Dec 31 2017

With e1 = 19 + 6*sqrt(10) and e2 = 19 - 6*sqrt(10),

a(2+7k) = (e1^k + e2^k)/2,

a(3+7k) = e1^k + e2^k,

a(4+7k) = (3/2) (e1^k + e2^k),

a(5+7k) = (2+sqrt(10)/2) e1^k + (2-sqrt(10)/2) e2^k,

a(6+7k) = (7/2+sqrt(10)) e1^k + (7/2-sqrt(10)) e2^k,

a(7+7k) = (13/2+2 sqrt(10)) e1^k + (13/2-2 sqrt(10)) e2^k,

a(8+7k) = (8+5 sqrt(10)/2) e1^k + (8-5 sqrt(10)/2) e2^k. - Robert Israel, Feb 16 2016

Extensions

4 initial terms added by M. F. Hasler, Jan 15 2012

a(40) from Robert G. Wilson v, Jan 15 2012