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

%I A031149 #72 Aug 13 2021 12:27:59
%S A031149 0,1,2,3,4,7,13,16,19,38,57,136,253,487,604,721,1442,2163,5164,9607,
%T A031149 18493,22936,27379,54758,82137,196096,364813,702247,870964,1039681,
%U A031149 2079362,3119043,7446484,13853287,26666893,33073696,39480499,78960998,118441497,282770296
%N A031149 Numbers whose square with its last digit deleted is also a square.
%C A031149 Square root of A023110(n).
%C A031149 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.
%C A031149 From _Robert Israel_, Feb 16 2016: (Start)
%C A031149 Solutions x to x^2 = 10*y^2 + j, j in {0,1,4,6,9}, in increasing order of x.
%C A031149 j=0 occurs only for x=0.
%C A031149 Let M be the 2 X 2 matrix [19, 60; 6, 19].
%C A031149 Solutions of x^2 = 10*y^2 + 1 are (x,y)^T = M^k (1,0)^T for k >= 0.
%C A031149 Solutions of x^2 = 10*y^2 + 4 are (x,y)^T = M^k (2,0)^T for k >= 0.
%C A031149 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.
%C A031149 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.
%C A031149 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.
%C A031149 The eigenvalues of M are 19 + 6*sqrt(10) and 19 - 6*sqrt(10).
%C A031149 From this follow my formulas below and the G.f. (End)
%D A031149 R. K. Guy, Neg and Reg, preprint, Jan 2012. [From _N. J. A. Sloane_, Jan 12 2012]
%H A031149 Dmitry Petukhov and Robert Israel, <a href="/A031149/b031149.txt">Table of n, a(n) for n = 1..4400</a> (n = 1..67 from Dmitry Petukhov)
%H A031149 M. F. Hasler, <a href="/wiki/M. F. Hasler/Truncated_squares">Truncated squares</a>, OEIS wiki, Jan 16 2012
%H A031149 <a href="/index/Sq#sqtrunc">Index to sequences related to truncating digits of squares</a>.
%F A031149 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
%F A031149 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
%F A031149 a(n) = 38*a(n-7) - a(n-14) for n>15 (conjectured). - _Colin Barker_, Dec 31 2017
%F A031149 With e1 = 19 + 6*sqrt(10) and e2 = 19 - 6*sqrt(10),
%F A031149    a(2+7k) = (e1^k + e2^k)/2,
%F A031149    a(3+7k) = e1^k + e2^k,
%F A031149    a(4+7k) = (3/2) (e1^k + e2^k),
%F A031149    a(5+7k) = (2+sqrt(10)/2) e1^k + (2-sqrt(10)/2) e2^k,
%F A031149    a(6+7k) = (7/2+sqrt(10)) e1^k + (7/2-sqrt(10)) e2^k,
%F A031149    a(7+7k) = (13/2+2 sqrt(10)) e1^k + (13/2-2 sqrt(10)) e2^k,
%F A031149 a(8+7k) = (8+5 sqrt(10)/2) e1^k + (8-5 sqrt(10)/2) e2^k. - _Robert Israel_, Feb 16 2016
%e A031149 364813^2 = 133088524969, 115364^2 = 13308852496.
%p A031149 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;
%t A031149 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 *)
%o A031149 (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
%Y A031149 Cf. A000290, A023110, A030686, A030687, A031150.
%K A031149 nonn,base
%O A031149 1,3
%A A031149 _Patrick De Geest_
%E A031149 4 initial terms added by _M. F. Hasler_, Jan 15 2012
%E A031149 a(40) from _Robert G. Wilson v_, Jan 15 2012