OFFSET
1,2
COMMENTS
Comments from N. J. A. Sloane, Jul 15 2019: (Start)
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.
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,
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
The partial sums to this point are
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
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.
So the first 84 terms are correct.
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)
LINKS
Carole Dubois, Table of n, a(n) for n = 1..5001
Carole Dubois, graph
Carole Dubois, graph for list of successive unauthorized sums
Jean-Marc Falcoz, Conjectured table of n, a(n) for n = 1..1000
EXAMPLE
The sequence starts with 1, 2, 3, 5, 4, 6, 7, 8, 9, 10, 11, 12, 14,...
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.
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.
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Eric Angelini and Jean-Marc Falcoz, Jul 14 2019
EXTENSIONS
Added "infinite" and "for n>1" to definition. - N. J. A. Sloane, Jul 15 2019
STATUS
approved