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A340695 a(n) is the next perfect power after the earliest occurrence of n consecutive perfect powers, all of which are squares with exponents equal to 2.

%I A340695 #19 Jan 19 2021 21:25:52
%S A340695 8,64,216,512,1728,4913,12167,19683,32768,74088,148877,175616,328509,
%T A340695 493039,753571,1259712,1860867,2406104,3375000,4330747,5929741,
%U A340695 8365427,11089567,13824000,18191447,23639903,28934443,36264691,43614208,53582633,67917312,81182737,97972181
%N A340695 a(n) is the next perfect power after the earliest occurrence of n consecutive perfect powers, all of which are squares with exponents equal to 2.
%C A340695 The exponent of a(n) is > 2 thus terminating the progression of n consecutive preceding squares with exponents = 2 (A111245).
%C A340695 Is this sequence strictly increasing? - _David A. Corneth_, Jan 19 2021
%H A340695 David A. Corneth, <a href="/A340695/b340695.txt">Table of n, a(n) for n = 1..6963</a> (terms <= 10^22)
%H A340695 David A. Corneth, <a href="/A340695/a340695.gp.txt">PARI program</a>
%e A340695 See A340661.
%e A340695 From _David A. Corneth_, Jan 19 2021: (Start)
%e A340695 a(3) = 216 as in the perfect powers we see ..., 128 = 2^7, 144 = 12^2, 169 = 13^2, 196 = 14^2, 216 = 6^3, ... . We write them as powers of m^k where k is chosen as large as possible such that m and k are integers.
%e A340695 Then between two perfect powers with k > 2 (being 128 = 2^7 and 216 = 6^3) we have three consecutive perfect powers with k = 2. As 216 closes this earliest streak of 3, a(3) = 216. (End)
%o A340695 (PARI) \\ See Corneth link
%Y A340695 Cf. A000290, A001597, A111245, A340661, A340662, A340663, A340664.
%K A340695 nonn
%O A340695 1,1
%A A340695 _Hugo Pfoertner_, Jan 18 2021