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Harvey P. Dale, <a href="/A104492/b104492.txt">Table of n, a(n) for n = 1..1000</a>
#-Floor[Surd[#, 3]]^3&/@Prime[Range[80]] (* Harvey P. Dale, Feb 15 2018 *)
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a(n) = prime(n) - floor(prime(n)^(1/3))^3. - Jon E. Schoenfield, Jan 17 2015
a(48) = 7 because the 48th prime is 223 and 223 - 6^3 = 7, while 223 - 7^3 = -120.
while 223-7^3 = -120.
lst={}; Do[p=Prime[n]; s=p^(1/3); f=Floor[s]; a=f^3; d=p-a; AppendTo[lst, d], {n, 6!}]; lst [From _(* _Vladimir Joseph Stephan Orlovsky_, Mar 11 2009] *)
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lst={}; Do[p=Prime[n]; s=p^(1/3); f=Floor[s]; a=f^3; d=p-a; AppendTo[lst, d], {n, 6!}]; lst [From _Vladimir Joseph Stephan Orlovsky (4vladimir(AT)gmail.com), _, Mar 11 2009]
_Jonathan Vos Post (jvospost3(AT)gmail.com), _, Mar 10 2005
lst={}; Do[p=Prime[n]; s=p^(1/3); f=Floor[s]; a=f^3; d=p-a; AppendTo[lst, d], {n, 6!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Mar 11 2009]
easy,nonn,new
a(48) = 7 because the 48th prime is 223, and 223 - 6^3 = 7,
easy,nonn,new
easy,nonn,new
Jonathan Vos Post (jvospost2jvospost3(AT)yahoogmail.com), Mar 10 2005