A317510 Numbers (4p+1)/3 where p is a Sophie Germain prime p > 3.

7, 15, 31, 39, 55, 71, 111, 119, 151, 175, 231, 239, 255, 311, 319, 335, 375, 391, 479, 559, 575, 591, 655, 679, 791, 855, 871, 879, 911, 959, 991, 1015, 1079, 1215, 1271, 1351, 1359, 1375, 1399, 1471, 1631, 1639, 1719, 1879, 1919, 1935, 1975, 1999, 2015, 2079, 2111, 2135, 2311, 2415, 2519, 2535, 2575, 2631

Offset: 1

Author

Hilko Koning, Jul 30 2018

Keywords

• nonn

Comments

It appears that this is a subsequence of A179882.

Define a set of consecutive positive odd numbers {1,......, (A077065(n)-1)} with n >= 3 and skip the number A077065(n)/2. Then the contraharmonic mean of that set gives the sequence. For example: ContraharmonicMean[{1, 3, 7, 9}] = 7, ContraharmonicMean[{1, 3, 5, 7, 9, 13, 15, 17, 19, 21}] = 15, ContraharmonicMean[{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 25, 27, 29, 31, 33, 35, 37,39, 41, 43, 45}] = 31. - Hilko Koning, Aug 28 2018

Let p be a Sophie Germain prime and define h = 2p + 1 a safe prime. Then the contraharmonic mean of the totatives of h is given by: CHM(h) = (Sum_{1 <= k < h, gcd(k, h) = 1} k^2) / (Sum_{1 <= k < h, gcd(k, h) = 1} k). Since h is prime, all integers k = 1, 2, ... , h - 1 are coprime to h. Then, CHM(h) = ((h - 1) * h * (2h-1) / 6) / ((h - 1) * h / 2). Thus CHM = (2h-1) / 3 = (4p+1) / 3. These values are integers precisely when p == 2 mod 3, which holds for all Sophie Germain primes, p >= 5. The resulting values for the sequence A317510, which is therefore a subsequence of A179882. - Hilko Koning, Jun 17 2025

Links

• Muniru A Asiru, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A005384, A179882.

Subsequence of A004767, and of A004771.

Programs

GAP

a:=[];; for p in [3..2000] do if IsPrime(p) and IsPrime(2*p+1) then Add(a,(4*p+1)/3); fi; od; a; # Muniru A Asiru, Aug 28 2018

Mathematica

lst = {}; Do[If[PrimeQ[p] && PrimeQ[2 p + 1], AppendTo[lst, (4 p + 1)/3]], {p, 5, 2*10^3}]; lst
4 (Select[Prime@Range[3, 300], PrimeQ[2 # + 1] &] + 1)/3 - 1 (* Robert G. Wilson v, Jul 30 2018 *)

PARI

lista(nn) = {forprime (p=5, nn, if (isprime(2*p+1), print1((4*p+1)/3, ", ")););} \\ Michel Marcus, Aug 27 2018