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A115432
Numbers k such that the concatenation of k with k-4 gives a square.
14
65, 6653, 9605, 218413, 283720, 996005, 58446925, 99960005, 6086712229, 7385370133, 8478948853, 9999600005, 120178240093, 161171620229, 358247912200, 426843573160, 893417179213, 999996000005, 23376713203604
OFFSET
1,1
COMMENTS
The terms of this sequence (k//k-4 = m*m), A116104 (k//k-8 = m*(m+4)) and A116121 (k//k-5 = m*(m+2)) agree as long as the two concatenated numbers k and k-x have the same length. This condition is satisfied for the given terms of all three sequences. - Georg Fischer, Sep 12 2022
From Robert Israel, Sep 13 2023: (Start)
Numbers k of the form (y^2+4)/(10^d + 1) where 10^(d-1) <= k - 4 < 10^d and y is a square root of -4 mod (10^d + 1).
Includes 10^(2*d) - 4*10^d + 5 for all d >= 1, as the concatenation of this with 10^(2*d) - 4*10^d + 1 is 10^(4*d) - 4 * 10^(3*d) + 6 * 10^(2*d) - 4 * 10^d + 1 = (10^d - 1)^4.
This is the same sequence as A116104 and A116121. The only possible differences would be if 10^(d-1) + 4 <= k <= 10^(d-1) + 7 or 10^d + 4 <= k <= 10^d + 7, so that k - 4 and k - 8 have different numbers of digits.
But in none of those cases can (10^d + 1)*k - 4 be a square:
If k = 10^(d-1) + 4 or 10^d + 4, (10^d + 1)*k - 4 == 6 (mod 9).
If k = 10^(d-1) + 5 or 10^d + 5, (10^d + 1)*k - 4 == 2 (mod 3).
If k = 10^(d-1) + 6 or 10^d + 6, (10^d + 1)*k - 4 == 2 (mod 10).
If k = 10^(d-1) + 7 or 10^d + 7, (10^d + 1)*k - 4 == 3 (mod 10). (End)
LINKS
EXAMPLE
9605_9601 = 9801^2.
MAPLE
f:= proc(d) uses NumberTheory; local m, r;
m:= 10^d + 1;
if QuadraticResidue(-4, m) = -1 then return NULL fi;
r:= ModularSquareRoot(-4, m);
op(sort(select(t -> t >= 10^(d-1)+4 and t < 10^d+4, map(t -> ((r*t mod m)^2+4)/m, convert(RootsOfUnity(2, m), list)))))
end proc:
map(f, [$1..20]); # Robert Israel, Sep 12 2023
KEYWORD
base,nonn
AUTHOR
Giovanni Resta, Jan 25 2006
STATUS
approved