

A115431


Numbers k such that the concatenation of k with k2 gives a square.


22



6, 5346, 8083, 10578, 45531, 58626, 2392902, 2609443, 7272838, 51248898, 98009803, 159728062051, 360408196038, 523637103531, 770378933826, 998000998003, 1214959556998, 1434212848998, 3860012299771, 4243705560771
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OFFSET

1,1


COMMENTS

From Robert Israel, Feb 20 2019: (Start) The same as A116117 and A116135 (see link).
So there are two equivalent definitions: numbers k such that k concatenated with k6 gives the product of two numbers which differ by 4; and numbers k such that k concatenated with k3 gives the product of two numbers which differ by 2.
For each k >= 1, 10^(4*k)2*10^(3*k)+10^(2*k)2*10^k+3 is a term.
If k is a term and k2 has length m, then all prime factors of 10^m+1 must be congruent to 1 or 3 (mod 8). In particular, we can't have m == 2 (mod 4) or m == 3 (mod 6), as in those cases 10^m+1 would be divisible by 101 or 7 respectively. (End)


LINKS

Robert Israel, Table of n, a(n) for n = 1..1312
Robert Israel, Proof that A115431, A116117 and A116135 are the same


EXAMPLE

8083_8081 = 8991^2.
98009803_98009800 = 98999900 * 98999902, where _ denotes
concatenation


MAPLE

f:= proc(n) local S;
S:= map(t > rhs(op(t))^2 mod 10^n+2, [msolve(x^2+2, 10^n+1)]);
op(sort(select(t > t2 >= 10^(n1) and t2 < 10^n and issqr(t2 + t*10^n), S)))
end proc:
seq(f(n), n=1..20); # Robert Israel, Feb 20 2019


CROSSREFS

Cf. A030465, A102567, A115426, A115437, A115428, A115429, A115430, A115432, A115433, A115434, A115435, A115436, A115442.
Sequence in context: A066061 A028366 A305886 * A116117 A116135 A011788
Adjacent sequences: A115428 A115429 A115430 * A115432 A115433 A115434


KEYWORD

base,nonn


AUTHOR

Giovanni Resta, Jan 24 2006


STATUS

approved



