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Numbers n such that sopfr(n) - (floor(sqrt(n))*bigomega(n)) = floor(sqrt(n)).
0

%I #10 Apr 29 2018 12:50:14

%S 2,14,26,155,287,329,474,498,803,1079,1157,1432,2786,3396,3473,3611,

%T 5597,9287,11357,12599,21394,31418,49706,54023,56978,61150,63923,

%U 69791,72203,77789,78431,80987,81178,86897,99794,106194,109994,110338,110824

%N Numbers n such that sopfr(n) - (floor(sqrt(n))*bigomega(n)) = floor(sqrt(n)).

%e Take the number 287. Find the floor of its square root: sqrt(287)=16.941074... it's 16. Now get the factors of 287 = 7*41. Subtract the first prime factor from the floor of the square root: 7-16 = -9. Now subtract the second prime factor from the floor of the square root: 41-16 = 25. Add those values together: -9+25 = 16. It's the same as the floor of the square root. But it doesn't always work out that way.

%K nonn

%O 1,1

%A _Jason Earls_, Sep 23 2010