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A140250 a(n) is the largest cube <= A066525(n).

%I A140250 #21 May 22 2023 10:07:04
%S A140250 343,15625,34965783,106496424,3023464536,3659383421,7222633237,
%T A140250 10403062487,11179320256,11993263569,25881801912,36495256013,
%U A140250 40672093519,47516597848,49917330568,63616767488,84200449887,96323848704,573234910443,972947676429
%N A140250 a(n) is the largest cube <= A066525(n).
%C A140250 Suggested by Carlos Rivera's Prime Puzzles & Problems Connection, Puzzle 443 (which asks if a sum of consecutive cubes can be a cube or a prime cube).
%H A140250 Nathaniel Johnston, <a href="/A140250/b140250.txt">Table of n, a(n) for n = 1..1000</a>
%H A140250 Carlos Rivera, <a href="https://www.primepuzzles.net/puzzles/puzz_443.htm">Puzzle 443. Sum of cubes of consecutive primes</a>, The Prime Puzzles and Problems Connection.
%e A140250 In A066525 the first term is 503, the sum of cubes of the first four consecutive primes, 2 3 5 7. The cube just less than 503 is 343, a(1) in this sequence.
%p A140250 with(numtheory): P:=proc(n) add(ithprime(k)^3,k=1..n): end:
%p A140250 A098563 := proc(n)local m: option remember: if(n=0)then return 0: fi: m:=procname(n-1)+2: while true do if(isprime(P(m)))then return m:fi: m:=m+2:od: end:
%p A140250 A140250 := proc(n)return floor(surd(P(A098563(n)),3))^3: end:
%p A140250 seq(A140250(n),n=1..20); # _Nathaniel Johnston_, Apr 21 2011
%t A140250 Floor[CubeRoot[#]]^3&/@Select[Accumulate[Prime[Range[400]]^3],PrimeQ] (* _Harvey P. Dale_, May 22 2023 *)
%Y A140250 Cf. A066525, A098563, A140251.
%K A140250 nonn
%O A140250 1,1
%A A140250 _Enoch Haga_, May 15 2008
%E A140250 Edited by _N. J. A. Sloane_, Aug 25 2008
%E A140250 a(11)-a(20) from _Nathaniel Johnston_, Apr 21 2011