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Numbers k such that p(k)^p(k + 1) < p(k + 2)^p(k), where p(k) = prime(k).
2

%I #5 Dec 15 2022 14:00:21

%S 1,2,3,10,33,41,45,52,60,98,113,120,262,294,296,318,343,377,408,428,

%T 444,475,477,486,572,601,673,700,774,837,870,913,934,936,944,984,1050,

%U 1115,1169,1182,1230,1232,1287,1391,1445,1456,1550,1584,1647,1653,1674

%N Numbers k such that p(k)^p(k + 1) < p(k + 2)^p(k), where p(k) = prime(k).

%e For k = 3, we have 78125 = p(3)^p(4) < p(5)^p(3) = 161051.

%t p[n_] := Prime[n];

%t u = Select[Range[3000], p[#]^p[# + 1] < p[# + 2]^p[#] &] (* A358895 *)

%t Prime[u] (* A358896 *)

%Y Cf. A000040, A053089, A358896.

%K nonn

%O 1,2

%A _Clark Kimberling_, Dec 06 2022