A369861 The orbit of n under iterations of x -> concatenate(A048762(x), A055400(x)) enters a pseudo-loop x(k) = a^3 * 10^((k-k0)*A055642(b)) + b for k > k0. This sequence lists the b-value.
586, 587, 588, 589, 590, 591, 592, 584, 585, 586, 587, 588, 589, 590, 591, 592, 593, 748673, 748674, 748675, 748676, 748677, 748678, 748679, 748680, 748681, 709030, 709031, 709032, 709033, 709034, 709035, 709036, 709037, 709038, 709039, 513, 514, 515, 516, 517, 518, 519, 520
Offset: 1
Examples
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, a(1) = 586, as does the cube root up to an additional factor 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) -> 5832000587 (since 180^3 is again at distance 587) -> ... We see that in this sequence each term is one more than that of the preceding sequence, whence also a(2) = 587 = a(1)+1. Starting with 8, we get 8 -> 80 (since the largest cube <= 8 is 8, at distance 0) -> 6416 (since the cube 64 is at distance 16, two less than in 1's orbit) -> 5832584 (since 5832 = 18^3 is at distance 584, again 2 less than in 1's orbit) -> 5832000584 (since 180^3 is again at distance 584) -> ... We see that in this sequence each term is 2 (resp. 8) less than the corresponding term of 1's (resp. 7's) orbit (with the initial term deleted). Hence also a(8) = 584 = a(7)-8 = a(1)-2. From here on subsequent terms will again increase by 1 up to n = 17. Starting with 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 roots get multiplied by 10 and the distance from the cubes remains the same, a(18) = 748673. For n = 64 -> 640 (= 8^3 + 128) -> 512128 = 80^3 + 128, we have a(n) = 128.
Links
- Eric Angelini, Pseudo-loops with cubes, and post to math-fun discussion list; April 1, 2024.
Crossrefs
Programs
-
PARI
A369861(n)={until(, my(c=sqrtnint(n,3), v=valuation(c,10), L=logint(max(n-c^3,1),10)+1); L==v*3 && return(n-c^3); n += c^3*(10^L-1))}
-
Python
import sympy def A369861(n: int): while True: C = sympy.integer_nthroot(n, 3)[0]**3; L = A055642(n-C) if sympy.multiplicity(10, C) == L: return n-C n += C * (10**L - 1)
Comments