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Number of m such that floor(prime(m)/m) = n.
0

%I #13 Nov 04 2024 21:48:19

%S 3,8,19,41,117,254,616,1642,3766,9461,24183,60252,151368,385600,

%T 979844,2507393,6428977,16513542,42642649,110283280,285776799,

%U 742428731,1932223170,5038580446,13159683245,34423463648,90173540312

%N Number of m such that floor(prime(m)/m) = n.

%e Only m = 2,3,4 give [p(m)/m] = 1, so a(1) = 3.

%e There are 8 values of m giving floor(prime(m)/m) = 2, namely m = 1,5,6,7,8,9,10,11, so a(2) = 8.

%t a(n_) := Length[Cases[Table[Floor[Prime[m]/m], {m, 1, 1000000}], n]]

%Y Cf. A062742, A102281.

%K nonn

%O 1,1

%A _Zak Seidov_ Aug 11 2002

%E a(16)-a(27) from _Farideh Firoozbakht_, Sep 13 2005

%E Typo corrected by _David W. Wilson_, Oct 22 2005