|
%I #39 Nov 19 2022 21:58:24
%S 0,1,1,2,2,2,3,4,4,4,5,4,5,6,6,8,8,7,8,7,8,10,11,9,10,11,12,12,13,11,
%T 12,14,14,15,16,14,15,16,17,16,17,16,17,18,18,20,21,19,21,21,21,22,23,
%U 22,23,23,24,26,27,22,23,24,25,28,28,28,29,29,30,30,31,27
%N Number of prime powers in the sequence of the floor of n/k for k <= n, A010766.
%C Conjecture: a(n+1) - a(n) has all its record values at n = A135972(k) and the record values of a(n-1) - a(n) are a subsequence of A025487. This was verified for n = 1..20000. - _Thomas Scheuerle_, Oct 06 2022
%H R. Heyman, <a href="http://math.colgate.edu/~integers/w59/w59.pdf">Primes in floor function sets</a>, Integers 22(2022), #A59.
%F a(n) = c*n + O(n^(1/2)), where c is the sum of 1/(q*(q+1)) over all prime powers q.
%e For n=8 we have floor(8/1) = 8 = 2^3, a prime power; floor(8/2) = 4 = 2^2, a prime power; floor(8/3) = floor(8/4) = 2 = 2^1, a prime power. Each remaining term of the sequence is 1, which is not a prime power, so a(8) = 4.
%o (MATLAB)
%o function a = A357604( max_n )
%o for n = 1:max_n
%o s = floor(n./[1:n]); c = 0;
%o for m = 1:n-1
%o f = factor(s(m));
%o if s(m) > 1 && length(unique(f)) == 1
%o c = c+1;
%o end
%o end
%o a(n) = c;
%o end
%o end % _Thomas Scheuerle_, Oct 06 2022
%o (PARI) a(n) = sum(k=1,n, isprimepower(n\k)!=0); \\ _Thomas Scheuerle_, Oct 07 2022
%Y Cf. A000961, A025487, A135972, A010766.
%K nonn
%O 1,4
%A _Randell Heyman_, Oct 06 2022
%E a(12)-a(72) from _Thomas Scheuerle_, Oct 06 2022
|
