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Start of first run of n successive numbers in which the arithmetic derivative of the i-th number has exactly i prime factors, for i = 1..n.
1

%I #7 Dec 14 2016 09:15:23

%S 6,105,1001,2945,240485,1671414,22551962,22551962

%N Start of first run of n successive numbers in which the arithmetic derivative of the i-th number has exactly i prime factors, for i = 1..n.

%e 2945' = 839 that is a prime number;

%e 2946' = 2461 = 23*107;

%e 2947' = 428 = 2*2*107;

%e 2948' = 3260 = 2*2*5*163.

%e No other number < 2945 has this property and therefore a(4) = 2945.

%p with(numtheory): P:=proc(q) local a,b,d,i,j,k,ok,n; d:=1;

%p for k from 1 to q do for n from d to q do ok:=1; for j from 1 to k do

%p b:=ifactors(sigma(n+j-1))[2]; if add(b[i][2],i=1..nops(b))<>j then ok:=0; break; fi; od;

%p if ok=1 then d:=n; print(n); break; fi; od; od; end: P(10^12);

%Y Cf. A003415, A086560, A072875, A279518.

%K nonn,more

%O 1,1

%A _Paolo P. Lava_, Dec 14 2016

%E a(7)-a(8) from _Giovanni Resta_, Dec 14 2016