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A061341

Numbers not ending in 0 whose cubes are concatenations of other cubes.

22, 303, 1006, 2272, 3003, 6001, 6006, 10006, 30003, 50015, 50024, 60001, 60006, 60025, 100006, 120015, 121925, 150005, 150012, 240005, 300003, 466085, 500015, 500024, 600001, 600006, 600025, 600075, 1000006, 1200015, 1500005, 1500012, 1500015, 2400005, 2500006

(list; graph; refs; listen; history; text; internal format)

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