cp's OEIS Frontend

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.

%I A369861 #14 Apr 09 2024 02:25:50
%S A369861 586,587,588,589,590,591,592,584,585,586,587,588,589,590,591,592,593,
%T A369861 748673,748674,748675,748676,748677,748678,748679,748680,748681,
%U A369861 709030,709031,709032,709033,709034,709035,709036,709037,709038,709039,513,514,515,516,517,518,519,520
%N 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.
%C A369861 The iterated function can also be defined as x -> c(x)*(10^L(x-c(x))-1) + x, where c(x) = A048762(x) = floor(x^(1/3))^3 is the largest perfect cube <= x; A055400(x) = x-c(x) is the "cube excess" of x, and L(x) = A055642(x) = floor(log_10(max(x,1))+1) is the number of decimal digits of x.
%C A369861 Often a(n+1) = a(n) + 1, especially when c(n+1) = c(n), in which case it is probable that all elements of the orbit of n+1 are just one larger than the elements of the orbit of n.
%H A369861 Eric Angelini, <a href="https://cinquantesignes.blogspot.com/2024/04/pseudo-loops-with-cubes.html">Pseudo-loops with cubes</a>, and post to math-fun discussion list; April 1, 2024.
%e A369861 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.
%e A369861 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.
%e A369861 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.
%e A369861 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.
%e A369861 For n = 64 -> 640 (= 8^3 + 128) -> 512128 = 80^3 + 128, we have a(n) = 128.
%o A369861 (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))}
%o A369861 (Python)
%o A369861 import sympy
%o A369861 def A369861(n: int):
%o A369861     while True:
%o A369861         C = sympy.integer_nthroot(n, 3)[0]**3; L = A055642(n-C)
%o A369861         if sympy.multiplicity(10, C) == L: return n-C
%o A369861         n += C * (10**L - 1)
%Y A369861 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).
%Y A369861 Cf. A369860 (a-values)
%K A369861 nonn,base
%O A369861 1,1
%A A369861 _M. F. Hasler_, Apr 03 2024