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

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

Cf. A000578, A009421.

Sequence in context: A155785 A077525 A083765 * A088279 A225898 A028109

Adjacent sequences: A061338 A061339 A061340 * A061342 A061343 A061344

Keyword

base,nonn

Author

Erich Friedman, Jun 06 2001

Status

approved