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A346789 a(n) is the smallest integer m, such that for every sufficiently large integer k, A165370(729*k+n) can be written as m followed by zero or more 9's.

%I A346789 #22 Oct 25 2024 16:11:51
%S A346789 0,1,11,111,1111,2666888,26688,268,2,12,112,3377,13377,358888,3788888,
%T A346789 2268,22,122,1122,4567,46778,477,1477,11477,222,36688,368,3,13,113,
%U A346789 1113,5566888,55688,558,5788,23,123,44788,144788,57777,777778,2558,25788,223,5666
%N A346789 a(n) is the smallest integer m, such that for every sufficiently large integer k, A165370(729*k+n) can be written as m followed by zero or more 9's.
%C A346789 This sequence was motivated by the problem of finding large terms in A346630.
%C A346789 The sequence has 9^3 = 729 terms. As it turns out, "sufficiently large" is >= 4609; 4609 mod 729 = 235, and a(235) = 1888888888 is both the largest term and the term having the largest sum of cubes of digits.
%H A346789 Ondrej Kutal, <a href="/A346789/b346789.txt">Table of n, a(n) for n = 0..728</a>
%F A346789 a(n) = A165370(729*7+n) minus the trailing 9's. - _Ondrej Kutal_, Oct 06 2024
%e A346789 Let f(j) be the sum of the cubes of the digits of j, and let g(k) be the smallest number such that f(g(k)) = k.
%e A346789 g(4609) is the concatenation of a(4609 mod 729) and 0 9's: a(4609 mod 729) = a(235) = 1888888888, and f(1888888888) = 4609.
%e A346789 This sequence can be applied to the problem of extending A346630. With a suitable computer program, it can be determined in a small fraction of a second that the four smallest numbers requiring 14 steps of summing the cubes of digits to reach the narcissistic number 153 are 12558 (which is A346630(14)) and its three smallest anagrams, 12585, 12855, and 15258. Then, given the terms of this sequence, it can immediately be shown that --
%e A346789 -- g(12558) is the concatenation of a(12558 mod 729) and 13 9's: a(12558 mod 729) = a(165) = 12888888, and 12558 - f(12888888) = 12558 - 3081 = 9477 = 13 * 9^3, so g(12558) is the concatenation 12888888//9999999999999 (a 21-digit number)
%e A346789 -- g(12585) is the concatenation of a(12585 mod 729) and 17 9's; 12585 mod 729 = 192, a(192) = 444, so g(12585) is the 20-digit number 44499999999999999999
%e A346789 -- g(12855) is the concatenation of a(12855 mod 729) = a(462) = 13347 and 17 9's (thus, a 22-digit number)
%e A346789 -- g(15258), since 15258/9^3 > 20, must have more than 20 digits, and so must g(k) for all k > 15258
%e A346789 -- consequently, the 20-digit number 44499999999999999999 is the smallest number j such that f(j) is one of {12558, 12585, 12855}, and thus the smallest number requiring 15 steps to reach 153; i.e., 44499999999999999999 = A346630(15).
%e A346789 Going one step further, since 44499999999999999999 mod 729 = 327, a(327) = 3777, and f(3777) = 1056, the concatenation of 3777 and (44499999999999999999 - 1056)/9^3 = 61042524005486967 9's, i.e., the 61042524005486971-digit number 3777999...9 (let's call this huge number "H"), is the smallest j such that f(j) = 44499999999999999999. And since the 2nd-smallest number requiring 15 steps to reach 153 is 44949999999999999999 (the smallest anagram of A346630(15)), and since dividing it by 9^3 yields 61659807956104252.39..., which clearly exceeds the number of digits in H, it follows that H = A346630(16).
%o A346789 (Python)
%o A346789 # code from A165370 is needed
%o A346789 def A346789(n):
%o A346789     g = A165370(729*7+n)
%o A346789     while g % 10 == 9:
%o A346789         g //= 10
%o A346789     return g # _Ondrej Kutal_, Oct 06 2024
%Y A346789 Cf. A055012, A165370, A346630.
%K A346789 nonn,base,fini
%O A346789 0,3
%A A346789 _Jon E. Schoenfield_, Aug 03 2021
%E A346789 Name revised by _Ondrej Kutal_, Oct 25 2024