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A031149
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Numbers whose square with its last digit deleted is also a square.
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22
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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
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OFFSET
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1,3
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COMMENTS
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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.
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)
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REFERENCES
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R. K. Guy, Neg and Reg, preprint, Jan 2012. [From N. J. A. Sloane, Jan 12 2012]
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LINKS
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FORMULA
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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
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EXAMPLE
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364813^2 = 133088524969, 115364^2 = 13308852496.
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MAPLE
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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;
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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