A143822 Primes p such that sigma_0((p*p + 1)/2) = 4.

13, 17, 23, 31, 37, 53, 67, 89, 97, 103, 109, 113, 127, 137, 149, 151, 163, 167, 179, 197, 211, 223, 227, 229, 241, 263, 269, 277, 281, 283, 311, 331, 347, 359, 367, 373, 383, 389, 397, 419, 431, 433, 439, 479, 491, 503, 509, 541, 547, 587, 601, 617, 619, 653

Offset: 1

Author

Ctibor O. Zizka, Sep 02 2008

Keywords

• easy
• nonn

Comments

A048161 are primes p such that sigma_0((p*p+1)/2)= 2. Primes p such that sigma_0((p*p+1)/2)= 3 gives all RMS numbers (A140480) with 2 divisors (prime RMS numbers, prime NSW numbers (A088165)) and all RMS numbers with 4 divisors as those are a multiple of two nonequal RMS prime numbers. In general we look after primes p such that sigma_0((p*p+1)/2) equals some given integer k. RMS numbers n=p_1*...*p_t have k=2^t divisors (p_i prime, t integer >=1) and sigma_2(p_1*...*p_t)=(2^t)* (q_1^r_1 *...* q_t^r_t), q_j prime, r_t integer >=1.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005 (sigma_0), A048161, A088165, A140480, A002315.

Programs

Maple

A066885 := proc(n) local p; p :=ithprime(n) ; (p^2+1)/2 ; end: A000005 := proc(n) numtheory[tau](n) ; end: for n from 2 to 300 do if A000005(A066885(n)) = 4 then printf("%d,",ithprime(n)) ; fi; od: # R. J. Mathar, Sep 04 2008

Mathematica

Select[Range[650], PrimeQ[#] && DivisorSigma[0, (#^2 + 1)/2] == 4 &] (* Amiram Eldar, Mar 11 2020 *)
Select[Prime[Range[150]],DivisorSigma[0,(#^2+1)/2]==4&] (* Harvey P. Dale, Sep 22 2022 *)

Extensions

97 inserted and extended by R. J. Mathar, Sep 04 2008