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Cubes, not ending in 0, formed by concatenating other cubes.
2

%I #20 Jul 21 2021 09:37:50

%S 10648,27818127,1018108216,11728027648,27081081027,216108018001,

%T 216648648216,1001801080216,27008100810027,125112533753375,

%U 125180086413824,216010800180001,216064806480216,216270112515625,1000180010800216,1728648081003375,1812501158328125

%N Cubes, not ending in 0, formed by concatenating other cubes.

%H Sean A. Irvine, <a href="/A009421/b009421.txt">Table of n, a(n) for n = 1..75</a>

%F a(n) = A061341(n)^3. - _Michel Marcus_, Apr 23 2018

%e 2272^3 = 1_1728_0_27_64_8.

%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**3 for r in range(122000) if ok3(r**3, 1)]) # _Michael S. Branicky_, Jul 11 2021

%Y Cf. A000578, A019548, A061341.

%K nonn,base

%O 1,1

%A R. Muller

%E Name clarified by _Tanya Khovanova_, Jul 10 2021