cp's OEIS Frontend

A115531 Numbers k such that the concatenation of k with 3*k gives a square.

%I A115531 #15 Aug 08 2019 07:57:26
%S A115531 816326530612244897959183673469388,1836734693877551020408163265306123,
%T A115531 3265306122448979591836734693877552,
%U A115531 3746097814776274713839750260145681581685744016649323621228
%N A115531 Numbers k such that the concatenation of k with 3*k gives a square.
%C A115531 If 3+10^m is not squarefree, say 3+10^m = u^2*v where v is squarefree, then the terms with length m are t^2*v where 10^m > 3*t^2*v >= 10^(m-1).  The first m for which 3+10^m is not squarefree are 34, 59, 60, 61, 67. - _Robert Israel_, Aug 07 2019
%C A115531 Since 3+10^m is divisible by 7^2 for m = 34 + 42*k, the sequence contains 4*(3+10^m)/49, 9*(3+10^m)/49 and 16*(3+10^m)/49 for such m, and in particular is infinite. - _Robert Israel_, Aug 08 2019
%H A115531 Robert Israel, <a href="/A115531/b115531.txt">Table of n, a(n) for n = 1..113</a>
%p A115531 Res:= NULL:
%p A115531 for m from 1 to 67 do
%p A115531 if not numtheory:-issqrfree(3+10^m) then
%p A115531    F:= select(t -> t[2]=1, ifactors(3+10^m)[2]);
%p A115531    v:= mul(t[1], t=F);
%p A115531    Res:= Res, seq(t^2*v, t = ceil(sqrt(10^(m-1)/(3*v))) .. floor(sqrt(10^m/(3*v))))
%p A115531 fi
%p A115531 od:
%p A115531 Res;  # _Robert Israel_, Aug 07 2019
%Y A115531 Cf. A102567, A106497, A115527 - A115556.
%K A115531 nonn,base
%O A115531 1,1
%A A115531 _Giovanni Resta_, Jan 25 2006