OFFSET
1,1
COMMENTS
From François Marques, Jul 11 2021: (Start)
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.
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.
(End)
LINKS
François Marques, Table of n, a(n) for n = 1..100
EXAMPLE
2272^3 = 11728027648 = 1_1728_0_27_64_8 (where the underscores indicate concatenation).
PROG
(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));
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
(Python)
from sympy import integer_nthroot
def iscube(n): return integer_nthroot(n, 3)[1]
def ok3(n, c):
if n%10 == 9 or (c == 1 and n%10 == 0): return False
if c > 1 and iscube(n): return True
d = str(n)
for i in range(1, len(d)):
if iscube(int(d[:i])) and ok3(int(d[i:]), c+1): return True
return False
print([r for r in range(61000) if ok3(r**3, 1)]) # Michael S. Branicky, Jul 11 2021
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Erich Friedman, Jun 06 2001
STATUS
approved