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A143822
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Primes p such that sigma_0((p*p + 1)/2) = 4.
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1
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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MAPLE
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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
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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