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