cp's OEIS Frontend

A309151 Lexicographically earliest infinite sequence of distinct terms starting with a(1) = 1 such that, for n>1, a(n) doesn't share any digit with the cumulative sum a(1) + a(2) + a(3) + ... + a(n-1) + a(n).

%I A309151 #46 Jan 22 2022 23:39:27
%S A309151 1,2,3,5,4,6,7,8,9,10,11,12,14,22,20,23,24,30,13,15,16,18,40,19,17,25,
%T A309151 21,31,27,26,28,29,32,34,39,33,35,36,41,43,37,38,47,42,44,45,46,48,49,
%U A309151 53,50,54,52,55,60,57,58,59,56,62,63,68,51,61,65,64,66,69,67,70,71,73,83,80,74,75,72,76,77,78,79,81,87,82
%N A309151 Lexicographically earliest infinite sequence of distinct terms starting with a(1) = 1 such that, for n>1, a(n) doesn't share any digit with the cumulative sum a(1) + a(2) + a(3) + ... + a(n-1) + a(n).
%C A309151 Comments from _N. J. A. Sloane_, Jul 15 2019: (Start)
%C A309151 If we leave out the word "infinite", the "Lexicographically earliest sequence of distinct terms starting with a(1) = 1 such that, for n>1, a(n) doesn't share any digit with the cumulative sum a(1) + a(2) + a(3) + ... + a(n-1) + a(n)" is the single-term sequence 1.
%C A309151 By computer, _Eric Angelini_ and _Jean-Marc Falcoz_ have shown that there is no earlier start than the 84-term sequence shown in the DATA section, that is,
%C A309151 D = 1, 2, 3, 5, 4, 6, 7, 8, 9, 10, 11, 12, 14, 22, 20, 23, 24, 30, 13, 15, 16, 18, 40, 19, 17, 25, 21, 31, 27, 26, 28, 29, 32, 34, 39, 33, 35, 36, 41, 43, 37, 38, 47, 42, 44, 45, 46, 48, 49, 53, 50, 54, 52, 55, 60, 57, 58, 59, 56, 62, 63, 68, 51, 61, 65, 64, 66, 69, 67, 70, 71, 73, 83, 80, 74, 75, 72, 76, 77, 78, 79, 81, 87, 82
%C A309151 The partial sums to this point are
%C A309151 S = 1, 3, 6, 11, 15, 21, 28, 36, 45, 55, 66, 78, 92, 114, 134, 157, 181, 211, 224, 239, 255, 273, 313, 332, 349, 374, 395, 426, 453, 479, 507, 536, 568, 602, 641, 674, 709, 745, 786, 829, 866, 904, 951, 993, 1037, 1082, 1128, 1176, 1225, 1278, 1328, 1382, 1434, 1489, 1549, 1606, 1664, 1723, 1779, 1841, 1904, 1972, 2023, 2084, 2149, 2213, 2279, 2348, 2415, 2485, 2556, 2629, 2712, 2792, 2866, 2941, 3013, 3089, 3166, 3244, 3323, 3404, 3491, 3573
%C A309151 To show that the 84 initial terms are as claimed, we must show that there is at least one infinite continuation satisfying the condition. For this, extend D by the terms 4204, 3334, 66666, 33334, 666666, 333334, 6666666, ... (compare A308900). As a result, S is extended by 7777, 11111, 77777, 111111, 777777, 1111111, 7777777, ..., and corresponding terms of D and S have no digits in common.
%C A309151 So the first 84 terms are correct.
%C A309151 It should be straightforward to use this method to show that the 1000 terms in the a-file are correct. This a-file could then be changed to a b-file. (End)
%H A309151 Carole Dubois, <a href="/A309151/b309151.txt">Table of n, a(n) for n = 1..5001</a>
%H A309151 Carole Dubois, <a href="/A309151/a309151.jpg">graph</a>
%H A309151 Carole Dubois, <a href="/A309151/a309151_1.jpg">graph for list of successive unauthorized sums</a>
%H A309151 Jean-Marc Falcoz, <a href="/A309151/a309151.txt">Conjectured table of n, a(n) for n = 1..1000</a>
%e A309151 The sequence starts with 1, 2, 3, 5, 4, 6, 7, 8, 9, 10, 11, 12, 14,...
%e A309151 We can't assign 4 to a(4) as the cumulative sum at that stage would be 10 and a cumulative sum ending in 0 cannot be extended by any integer without infringing the rules.
%e A309151 Thus we assign 5 to a(4) and 4 to a(5). We cannot assign 13 to a(13) as the cumulative sum would then be 91, with the digit 1 of 91 colliding with the digit 1 of 13. Thus a(13) = 14. Etc.
%Y A309151 Partial sums form A324773, A316914 (one digit is shared by the cumulative sum instead of zero digit here), A316915 (two digits shared), A326638 (three digits shared), A326639 (four digits shared), A326640 (five digits shared).
%Y A309151 See also A308900.
%K A309151 base,nonn
%O A309151 1,2
%A A309151 _Eric Angelini_ and _Jean-Marc Falcoz_, Jul 14 2019
%E A309151 Added "infinite" and "for n>1" to definition. - _N. J. A. Sloane_, Jul 15 2019