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

%I #4 Jun 05 2015 12:59:27

%S 1,3,5,8,10,14,15,17,20,23,26,29,31,33,35,36,38,39,41,43,45,46,50,52,

%T 55,57,60,61,65,67,71,73,76,78,79,81,83,86,90,93,96,98,100,102,105,

%U 107,109,110,113,114,117,118,120,124,126,129,131,134,136,138,140

%N Numbers 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="/A258028/b258028.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, so a(1) = 1;

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

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

%e D(prime(4), 3) = 17 - 3*13 + 3*11 - 7 > 0.

%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, A258029, A258030, A258031.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jun 05 2015