cp's OEIS Frontend

A329126 a(n) is the least positive number which yields a multiple of n when its decimal digits (which are necessarily 0's and 1's) are read in any base.

%I A329126 #126 Aug 09 2025 14:45:12
%S A329126 1,110,101010,111100,100010001000100010,1111110,
%T A329126 10000010000010000010000010000010000010,11111111000,
%U A329126 1010101010101010100,1100110011001100110,100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000100000000010,11111111111100
%N A329126 a(n) is the least positive number which yields a multiple of n when its decimal digits (which are necessarily 0's and 1's) are read in any base.
%C A329126 a(n) might be called the "smallest trans-basic multiple of n."
%C A329126 In order to be a valid binary number, the terms may contain only 0's and 1's.
%C A329126 The number of 1's in a(n) is conjectured to be n; the number of 0's separating each one digit is usually A268336(n)-1 for small n. The number of trailing 0's is A051903(n).
%C A329126 The first 21 terms have been found and verified to be minimal via an advanced search; the 13th term (see b-file) contains 146 digits, and in general for every prime n the corresponding a(n) is conjectured to contain 2+(n-1)^2 0's and 1's.
%C A329126 A lower bound for a(n) is given by a(A032742(n)). Proof: If a(n) were smaller than a(A032742(n)), then a(A032742(n)) would not be the smallest trans-basic multiple of A032742(n); a(n) would be. By definition a(n) is the smallest trans-basic multiple of n, so we have a contradiction; QED.
%C A329126 To verify a trans-basic multiple of n for n > 2, one must only: A) make sure the string has some multiple of n of '1' digits; B) make sure the string ends with at least one '0' digit; and C) check that, for all prime bases below n, the resulting number is divisible by n. If these three conditions are met, the string is a trans-basic multiple of n.
%C A329126 While the formula given below is guaranteed to provide a trans-basic multiple of n, it does not always yield a(n) which by definition is the smallest such number. [Corrected by _M. F. Hasler_, Nov 14 2019]
%C A329126 From _N. J. A. Sloane_, Nov 12 2019: (Start)
%C A329126 For each n, the values of (string a(n) read in base b)/n for b = 1,2,3,... give a sequence of integers.
%C A329126 For n=1 this is the all-1's sequence A000012.
%C A329126 For n=2, a(2) = 110 which in base b is b+b^2. Divided by 2 we get (b+b^2)/2, which evaluated at b = 1,2,3,4,... is 1,3,6,10,..., the triangular numbers A000217.
%C A329126 For n=3, we get (b+b^3+b^5)/3, which is A220892.
%C A329126 For n=4, we get A328994. (End)
%C A329126 See A329000 = (1, 6, 42, 60, 139810, 126, ...) for a(n) converted from base 2 to base 10, i.e., the numbers which yield the terms here when written in base 2. - _M. F. Hasler_, Nov 09 2021
%H A329126 Alon Ran, <a href="/A329126/b329126.txt">Table of n, a(n) for n = 1..21</a>
%H A329126 Alon Ran, <a href="/A329126/a329126.txt">Comments on this sequence</a>
%F A329126 To generate an upper bound on a(n), start with n 1's (this is required to ensure that it is divisible by n in bases n+1, 2n+1, etc.)
%F A329126 Next, place A268336(n)-1 0's in between the 1's (this ensures that the powers that are added will always sum to 0 (mod n)).
%F A329126 Finally, add A051903(n) 0's on the right (this is to ensure that the number will be divisible by n in bases that are roots of factors of n).
%F A329126 Note that this formula does not always yield the minimal solution a(n). For instance, a(10) is obtained from the above result by grouping the 1's in pairs and separating the pairs by two 0's.
%F A329126 a(n) <= A329338(n), with equality except for n = 10, 14, 15, ... - _M. F. Hasler_, Nov 14 2019
%e A329126 a(3) = 101010:
%e A329126   101010_2  =     42 =    14*3;
%e A329126   101010_3  =    273 =    91*3;
%e A329126   101010_4  =   1092 =   364*3;
%e A329126   101010_5  =   3255 =  1085*3;
%e A329126   101010_6  =   7998 =  2666*3;
%e A329126   101010_7  =  17157 =  5719*3;
%e A329126   101010_8  =  33288 = 11096*3;
%e A329126   101010_9  =  59787 = 19929*3;
%e A329126   101010_10 = 101010 = 33670*3;
%e A329126   101010_11 = 162393 = 54131*3;
%e A329126   101010_12 = 250572 = 83524*3;
%e A329126 and so on. All the resulting values are multiples of 3.
%o A329126 (PARI) \\ See A329338 for an upper bound which equals a(n) in many cases, e.g., all n < 14 except for n = 10. - _M. F. Hasler_, Nov 10 2021
%Y A329126 Cf. A000012, A000217, A051903, A220892, A268336, A329338 (an upper bound), A328994.
%Y A329126 A329000 gives a(n) read in base 2 and converted to base 10.
%Y A329126 See also A329443.
%K A329126 nonn,base,nice
%O A329126 1,2
%A A329126 _Alon Ran_, Nov 05 2019
%E A329126 I have weakened some of the assertions in the Comments section, since they seemed to be unproved. See Alon Ran's comments (see Links). - _N. J. A. Sloane_, Dec 02 2019
%E A329126 Definition corrected, following a remark by _Don Reble_, by _M. F. Hasler_, Nov 09 2021
%E A329126 The present definition has been reworded by _Peter Munn_, Nov 17 2021, and by _N. J. A. Sloane_, Nov 29 2021