%I #26 Aug 08 2021 01:49:11
%S 22,303,1006,2272,3003,6001,6006,10006,30003,50015,50024,60001,60006,
%T 60025,100006,120015,121925,150005,150012,240005,300003,466085,500015,
%U 500024,600001,600006,600025,600075,1000006,1200015,1500005,1500012,1500015,2400005,2500006
%N Numbers not ending in 0 whose cubes are concatenations of other cubes.
%C From _François Marques_, Jul 11 2021: (Start)
%C If a and b are integers such that 3*a^2*b and 3*a*b^2 can be obtained as concatenations of cubes then a*10^k+b is a term in the list for k greater than the maximum of the number of digits in b^3, 3*a^2*b and 3*a*b^2.
%C Conjecture: the last digit of a(n) is in {1, 2, 3, 4, 5, 6, 8} and if it's 3 then there is k>=2 such that a(n) = 3*10^k+3.
%C (End)
%H François Marques, <a href="/A061341/b061341.txt">Table of n, a(n) for n = 1..100</a>
%e 2272^3 = 11728027648 = 1_1728_0_27_64_8 (where the underscores indicate concatenation).
%o (PARI) can_split(v,b=10,p=3)=if(#v==0,1,for(k=1,#v,if(ispower(fromdigits(v[1..k],b),p) && can_split(v[k+1..#v],b,p), return(1))); return(0));
%o isok(n,b=10,p=3)=if(n%b==0,return(0)); my(v=digits(n^p,b)); for(k=1,#v-1,if(ispower(fromdigits(v[1..k],b),p) && can_split(v[k+1..#v],b,p), return(1))); return(0); \\ _François Marques_, Jul 12 2021
%o (Python)
%o from sympy import integer_nthroot
%o def iscube(n): return integer_nthroot(n, 3)[1]
%o def ok3(n, c):
%o if n%10 == 9 or (c == 1 and n%10 == 0): return False
%o if c > 1 and iscube(n): return True
%o d = str(n)
%o for i in range(1, len(d)):
%o if iscube(int(d[:i])) and ok3(int(d[i:]), c+1): return True
%o return False
%o print([r for r in range(61000) if ok3(r**3, 1)]) # _Michael S. Branicky_, Jul 11 2021
%Y Cf. A000578, A009421.
%K base,nonn
%O 1,1
%A _Erich Friedman_, Jun 06 2001