A236255 Prime numbers s for which there exist primes p, q, r such that phi(pqs)=phi(rs^3), sigma(pqs)=sigma(rs^3).

2, 3, 5, 7, 31, 43, 139, 157, 191, 269, 293, 337, 463, 557, 593, 683, 709, 757, 769, 983, 1021, 1567, 1583, 2293, 2309, 2689, 2707, 2801, 2917, 3319, 3323, 3583, 3823, 4271, 5507, 5557, 6037, 6043, 6079, 6151, 6469, 6779, 6959, 6977, 7207, 7963, 8419, 8429, 8521, 8627, 8663, 8861, 8887, 9677, 9769, 10163, 10613, 10847, 11003

Offset: 1

Author

Vladimir Letsko, Jan 21 2014

Keywords

• nonn

Comments

Obviously tau(pqs) = tau(rs^3). So we have pairs of terms of A134922.

s = 593 is the least number such that there are just two matching pairs: (593*381187517*703949, 593^3*763079633) and (593*3911429*780389, 593^3*8680337). And for s = 853693 there are as many as 3 matching pairs.

Examples

			2 is in the sequence because for p = 11, q = 29, r = 71 we have phi(pqs)=phi(rs^3) and sigma(pqs)=sigma(rs^3).
		

Links

• Vladimir Letsko, Math Marathon, problem 163 (in Russian)

Crossrefs

cf. A134922, A000010, A000203

Programs

Maple

is_A236255:=proc(s::prime) local f,Q, c,d,cc,p,q,r;
f:=false:c:=2*s^2+1:
cc:=(c^2-1)/2;
Q:=numtheory[divisors](cc):
for d in Q do q:=d+c:
if isprime(q) then
p:=c+cc/(q-c): if pA236255(s) then print(s) fi od: