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%I #6 Apr 16 2015 06:27:35
%S 1,2,1,4,1,6,1,2,1,10,1,12,1,2,1,5,1,3,1,2,1,22,1,7,1,2,1,3,1,30,1,2,
%T 1,34,1,36,1,2,1,40,1,11,1,2,1,4,1,3,1,2,1,52,1,13,1,2,1,3,1,4,1,2,1,
%U 64,1,66,1,2,1,70,1,72,1,2,1,17,1,3,1,2,1,5
%N Initialization a(n)=0 for all n; if a(n)=0, a(n+kp)=n where p=prime(n) and k=0,1,2,...
%C Conjecture: there is an infinity of numbers n such that a(n)=n.
%H Michel Lagneau, <a href="/A256908/b256908.txt">Table of n, a(n) for n = 1..10000</a>
%e n=1 => prime(1)=2 and a(1+2k)=1 => a(1)=a(3)=a(5)=... =1;
%e n=2 => prime(2)=3 and a(2+3k)=2 => a(2)=a(8)=... =2 (the element a(5) is already in the sequence);
%e n=3 => prime(3)=5 and a(3+5k)=3 => a(18)=3 (the elements a(n) for n<18 are already in the sequence), a(28)=3, a(48)=3,...
%p with(numtheory):nn:=500:T:=array(1..nn):
%p for i from 1 to nn do:
%p T[i]:=0:
%p od:
%p for n from 1 to nn do:
%p p:=ithprime(n):
%p for m from n by p to nn do:
%p if T[m]=0 then T[m]:=n:
%p else
%p fi:
%p od:
%p od:
%p print(T):
%K nonn
%O 1,2
%A _Michel Lagneau_, Apr 12 2015
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