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%I #61 Aug 21 2016 01:58:58
%S 2,2,3,2,2,3,2,2,2,3,2,2,3,2,2,2,3,2,2,3,2,2,7,2,2,2,3,2,2,15,2,2,2,3,
%T 2,2,7,2,2,5,2,2,2,5,2,2,7,2,2,2,3,2,2,13,2,2,2,3,2,2,6,2,2,3,2,2,2,6,
%U 2,2,3,2,2,2,6,2,2,5,2,2,3,2,2,2,6,2
%N a(n) is the least integer k such that some multiple of sqrt(k) falls strictly between n and n+1.
%C a(n) is squarefree for all n.
%C Record values occur at a(1)=2, a(3)=3, a(23)=7, a(30)=15, a(184)=38, a(8091)=43, a(16060)=46, a(16907)=58, a(20993)=61, a(26286)=97, a(130375)=118, a(169819)=127, a(2135662)=130, a(2345213)=187, a(222125822)=210, a(257240414)=227, ... - _Jon E. Schoenfield_, Sep 07 2015
%C a(n) > 2 iff n is in A001952. - _Robert Israel_, Aug 18 2016
%H Peter Kagey, <a href="/A261865/b261865.txt">Table of n, a(n) for n = 1..10000</a>
%e a(40) = 5 because:
%e 40 * sqrt(1) = 40 and 41 * sqrt(1) = 41
%e 28 * sqrt(2) < 40 and 29 * sqrt(2) > 41
%e 23 * sqrt(3) < 40 and 24 * sqrt(3) > 41
%e 20 * sqrt(4) = 40 and 21 * sqrt(4) > 41
%e 40 < 18 * sqrt(5) < 41
%e Thus sqrt(5) is the least integer root with integer multiple between 40 and 41.
%p f:= proc(n) local k;
%p for k from 2 do
%p if ceil(sqrt((n+1)^2/k)) - floor(sqrt(n^2/k)) >= 2 then return k fi
%p od
%p end proc:
%p map(f, [$1..100]); # _Robert Israel_, Aug 18 2016
%t Table[k = 2; While[Ceiling[Sqrt[(n + 1)^2/k]] - Floor[Sqrt[n^2/k]] < 2, k++]; k, {n, 120}] (* _Michael De Vlieger_, Aug 18 2016, after Maple *)
%o (Ruby) def a(n); (1..n**2+1).find { |k| (n/k**0.5+1).to_i*k**0.5 < n+1 } end
%o (PARI) ok(n,k)=h = floor((n+1)/sqrt(k)); (n < h*sqrt(k)) && (h*sqrt(k)< (n+1));
%o a(n) = my(k=1); while (!ok(n,k), k++); k; \\ _Michel Marcus_, Sep 04 2015
%Y Cf. A001952, A005117, A262036, A273620.
%K nonn
%O 1,1
%A _Peter Kagey_, Sep 03 2015
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