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A369860
The orbit of n under iterations of x -> c(x)*(10^L(x-c(x))-1) + x, where c(x) = floor(x^(1/3))^3, L(x) = floor(log_10(max(x,1))+1), enters a pseudo-loop x(k) = a^3 * 10^((k-k0)*L(b)) + b beyond some k0. This sequence lists the a-values.
2
18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 883, 883, 883, 883, 883, 883, 883, 883, 883, 581, 581, 581, 581, 581, 581, 581, 581, 581, 581, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 8, 8, 8, 8
OFFSET
1,1
COMMENTS
The iterated function can also be defined as x -> concatenate(c(x), A055400(x)), where c = A048762 gives the largest perfect cube <= x and x-c(x) = A055400(x) is the "cube excess" of x. L = A055642 gives the number of decimal digits.
The corresponding b-values are listed in A369861.
LINKS
Eric Angelini, Pseudo-loops with cubes, and post to math-fun discussion list; April 1, 2024
EXAMPLE
Starting with 1, we get 1 -> 10 -> 82 (since 8 is the largest cube <= 10, at distance 2) -> 6418 (since the cube 64 is at distance 18) -> 5832586 (since 5832 = 18^3 is at distance 586) -> 5832000586 (since 180^3 is again at distance 586) -> ...: Each time 3 '0's will be inserted in front of the remainder which remains always the same, as does the cube root a(1) = 18, up to factors of 10.
Starting with 2, we get 2 -> 11 (since the largest cube <= 2 is 1, at distance 1) -> 83 (since largest cube <= 11 is 8, at distance 2) -> 6419 (since the cube 64 is at distance 19) -> 5832587 (since 5832 = 18^3 is at distance 587). We see that in this sequence each term is just one more than that of the preceding sequence, so the cube root remains the same, a(2) = a(1) = 18.
For n = 18, we get 18 -> 810 (since the largest cube <= 18 is 8, at distance 10) -> 72981 (since the cube 729 is at distance 81) -> 689214060 (since 68921 = 41^3 is at distance 4060) -> 688465387748673 (since 688465387 = 883^3 is at distance 748673), from where on the cube root a(18) = 883 gets an additional factor 10 at each step, but the cube excess A055400 remains the same, A369861(18) = 748673.
See A369861 for more examples.
PROG
(PARI) A369860(n)={until(, my(c=sqrtnint(n, 3), v=valuation(c, 10), L=logint(max(n-c^3, 1), 10)+1); L==v*3 && return(c/10^v); n += c^3*(10^L-1))}
(Python)
import sympy # for integer_nthroot (A048766), multiplicity (A122840)
def A369860(n: int):
while True:
C = sympy.integer_nthroot(n, 3)[0]; L = A055642(n-C**3)
if sympy.multiplicity(10, C)*3 == L: return C//10**(L//3)
n += C**3 * (10**L - 1)
CROSSREFS
Cf. A000578 (cubes), A048766 (cube root), A048762 (largest cube <= n), A055400 (cube excess), A055642 (length of n in base 10), A122840 (10-valuation of n).
Cf. A369861 (b-values).
Sequence in context: A278716 A282814 A010857 * A158910 A243097 A040307
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Apr 05 2024
STATUS
approved