cp's OEIS Frontend

A274188 Number n such that there is a smaller positive number j == n (mod 5) such that sqrt(j*n) is an integer.

%I A274188 #36 Jan 09 2025 02:51:45
%S A274188 9,16,18,20,27,32,36,40,45,48,49,54,60,63,64,72,80,81,90,96,98,99,100,
%T A274188 108,112,117,120,121,125,126,128,135,140,144,147,153,160,162,169,171,
%U A274188 176,180,189,192,196,198,200,207,208,216,220,224,225,234,240,242,243,245,250
%N A274188 Number n such that there is a smaller positive number j == n (mod 5) such that sqrt(j*n) is an integer.
%C A274188 Or numbers n>=9 having a divisor t^2 > 1, where t=k/m, 1<= m < k, such that n == n/t^2  (mod 5).
%C A274188 Or positive numbers n such that if n == 0 (mod 5), then it divisible by 5^3 or by the square of some other prime; otherwise n divisible by k^2, such that there is a k_1, 0< k_1 <k with the condition k_1^2 == k^2 (mod 5).
%C A274188 A generalization see in our seqfan list from Jun 13 (correction Jun 14) 2016.
%H A274188 Charles R Greathouse IV, <a href="/A274188/b274188.txt">Table of n, a(n) for n = 1..10000</a>
%F A274188 Let A(x) be the number of a(n)<=x. Then A(x)/x ~ 1 - 149/(20*Pi^2) = 1 - 7.45/Pi^2 = 0.24515718... as x goes to infinity. - _Vladimir Shevelev_, Jun 15 2016; corrected by _Charles R Greathouse IV_, Jun 15 2016
%e A274188 9 is member, since 4 == 9 (mod 5) and 4*9 is a square;
%e A274188 32 is member, since 2 == 32 (mod 5) and 2*32 is a square;
%e A274188 45 is member, since 5 == 45 (mod 5) and 5*45 is a square.
%t A274188 Select[Range@ 250, Function[n, Count[Sqrt[# n] & /@ Select[Range[n - 1], Mod[#, 5] == Mod[n, 5] &], k_ /; IntegerQ@ k] > 0]] (* _Michael De Vlieger_, Jun 14 2016 *)
%o A274188 (PARI) isok(n)=j = n-5; while (j >0, if (issquare(j*n), return (1)); j -= 5;); 0; \\ _Michel Marcus_, Jun 14 2016
%o A274188 (PARI) is(n)=!issquarefree(n/if(n%5, if(n%4, 1, 4), 5)) \\ _Charles R Greathouse IV_, Jun 15 2016
%Y A274188 Cf. A046790, A274141.
%K A274188 nonn,easy
%O A274188 1,1
%A A274188 _Vladimir Shevelev_, Jun 12 2016
%E A274188 Corrected and extended by _Michel Marcus_, Jun 14 2016