login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Numbers prime(k) such that D(prime(k), 3) > 0, where D( * , 3) = 3rd difference.
5

%I #4 Jun 05 2015 12:59:53

%S 3,7,13,23,37,53,67,73,89,97,103,107,113,131,139,157,173,181,193,211,

%T 223,233,241,263,277,293,307,311,317,337,359,373,389,409,421,433,449,

%U 457,461,479,491,499,509,523,547,563,577,593,613,631,653,661,691,719

%N Numbers prime(k) such that D(prime(k), 3) > 0, where D( * , 3) = 3rd difference.

%C Partition of the positive integers: A064149, A258027, A258028;

%C Corresponding partition of the primes: A258029, A258030, A258031.

%H Clark Kimberling, <a href="/A258030/b258030.txt">Table of n, a(n) for n = 1..1000</a>

%F D(prime(k), 3) = P(k+3) - 3*P(k+2) + 3*P(k+1) - P(k), where P(m) = prime(m) for m >= 1.

%e D(prime(1), 3) = 7 - 3*5 + 3*3 - 2 < 0;

%e D(prime(2), 3) = 11 - 3*7 + 3*5 - 3 > 0, so a(1) = prime(2) = 3;

%e D(prime(3), 3) = 13 - 3*11 + 3*7 - 5 < 0;

%e D(prime(4), 3) = 17 - 3*13 + 3*11 - 7 > 0, so a(2) = prime(4) = 7;

%t d = Differences[Table[Prime[n], {n, 1, 400}], 3];

%t u1 = Flatten[Position[d, 0]] (* A064149 *)

%t u2 = Flatten[Position[Sign[d], 1]] (* A258027 *)

%t u3 = Flatten[Position[Sign[d], -1]] (* A258028 *)

%t p1 = Prime[u1] (* A258029 *)

%t p2 = Prime[u2] (* A258030 *)

%t p3 = Prime[u3] (* A258031 *)

%Y Cf. A064149, A258027, A258028, A258029, A258031.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jun 05 2015