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Numbers (4p+1)/3 where p is a Sophie Germain prime p > 3.
2

%I #40 Sep 14 2018 04:34:10

%S 7,15,31,39,55,71,111,119,151,175,231,239,255,311,319,335,375,391,479,

%T 559,575,591,655,679,791,855,871,879,911,959,991,1015,1079,1215,1271,

%U 1351,1359,1375,1399,1471,1631,1639,1719,1879,1919,1935,1975,1999,2015,2079,2111,2135,2311,2415,2519,2535,2575,2631

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

%C It appears that this is a subsequence of A179882.

%C 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

%H Muniru A Asiru, <a href="/A317510/b317510.txt">Table of n, a(n) for n = 1..5000</a>

%t lst = {}; Do[If[PrimeQ[p] && PrimeQ[2 p + 1], AppendTo[lst, (4 p + 1)/3]], {p, 5, 2*10^3}]; lst

%t 4 (Select[Prime@Range[3, 300], PrimeQ[2 # + 1] &] + 1)/3 - 1 (* _Robert G. Wilson v_, Jul 30 2018 *)

%o (PARI) lista(nn) = {forprime (p=5, nn, if (isprime(2*p+1), print1((4*p+1)/3, ", ")););} \\ _Michel Marcus_, Aug 27 2018

%o (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

%Y Cf. A005384, A179882.

%Y Subsequence of A004767, and of A004771.

%K nonn

%O 1,1

%A _Hilko Koning_, Jul 30 2018