
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(n1) + a(n)" is the singleterm sequence 1.
By computer, Eric Angelini and JeanMarc Falcoz have shown that there is no earlier start than the 84term 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 afile are correct. This afile could then be changed to a bfile. (End)
