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).

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

Offset: 1

Author

Eric Angelini and Jean-Marc Falcoz, Jul 14 2019

Keywords

• base
• nonn

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)

Examples

			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.
		

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

Crossrefs

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).

See also A308900.

Extensions

Added "infinite" and "for n>1" to definition. - N. J. A. Sloane, Jul 15 2019