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Composite squarefree numbers n such that the ratio (n + 1/3)/(p(i) + 1/3) is an integer, where p(i) are the prime factors of n.
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%I #14 Jun 04 2013 00:12:07

%S 133653,1280533,193638133,514276565,1421486733,1567953933,3857178453,

%T 3973200933,5411272533,7694639213,8021152533,8469827669,9820706133,

%U 15832804533,18238619373,22356801133,23037766613,25136796813,27315827733,32434329685,39817016633

%N Composite squarefree numbers n such that the ratio (n + 1/3)/(p(i) + 1/3) is an integer, where p(i) are the prime factors of n.

%C Also composite squarefree numbers n such that (3*p(i)+1) | (3*n+1).

%H Giovanni Resta, <a href="/A226112/b226112.txt">Table of n, a(n) for n = 1..67</a> (terms < 2*10^12)

%e The prime factors of 133653 are 3, 13, 23 and 149. We see that (133653 + 1/3)/(3 + 1/3) = 40096, (133653 + 1/3)/(13 + 1/3) = 10024, (133653 + 1/3)/(23 + 1/3) = 5728 and (133653 + 1/3)/(149 + 1/3) = 895. Hence 133653 is in the sequence.

%e The prime factors of 1127749 are 7, 31 and 5197. We see that

%e (1127749 + 1/3)/(7 + 1/3) = 153784, (1127749 + 1/3)/(31 + 1/3) = 35992 but (1127749 + 1/3)/(5197 + 1/3) = 422906/1949. Hence 1127749 is not in the sequence.

%p with(numtheory); A226112:=proc(i, j) local c, d, n, ok, p;

%p for n from 2 to i do if not isprime(n) then p:=ifactors(n)[2]; ok:=1;

%p for d from 1 to nops(p) do if p[d][2]>1 or not type((n+j)/(p[d][1]+j),integer) then ok:=0; break; fi; od;

%p if ok=1 then print(n); fi; fi; od; end: A226112(10^9,1/3);

%Y Cf. A208728, A225702-A225720, A226020, A226111, A226113, A226114.

%K nonn,hard

%O 1,1

%A _Paolo P. Lava_, May 29 2013

%E a(4)-a(21) from _Giovanni Resta_, Jun 02 2013