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A316918
For any k, the cumulative sum a(1) + a(2) + a(3) + ... + a(k) starts with a(k). Lexicographic first sequence of positive integers without duplicate terms having this property and extending itself forever.
3
61, 6, 7, 8, 9, 10, 1, 11, 12, 13, 15, 16, 18, 20, 2, 23, 25, 28, 31, 3, 35, 39, 43, 4, 48, 54, 5, 60, 67, 74, 83, 92, 102, 113, 126, 140, 14, 157, 174, 17, 196, 19, 220, 22, 246, 24, 276, 27, 310, 345, 34, 387, 430, 477, 530, 53, 595, 661, 66, 742, 75, 833, 84, 85, 944, 95, 96, 97, 98, 99, 100, 101, 1125, 1250, 1389, 1544
OFFSET
1,1
COMMENTS
The authors conjecture that it is impossible to extend the sequence forever if a(1) < 61.
With a(1) = 60, for instance, we will be stuck with the integer 5:
60,6,7,8,9,10,1,11,12,13,15,16,18,20,2,23,25,28,31,3,35,39,43,4,48,54,5 STOP (no unused integer can extend the sequence).
See here the sequence starting with a(1) = 60 and the cumulative sums:
60,6, 7, 8, 9, 10, 1, 11, 12, 13, 15, 16, 18, 20, 2, 23, 25, 28, 31, 3, 35,
60,66,73,81,90,100,101,112,124,137,152,167,185,205,207,230,255,283,314,317,352,
(the two lines above continue here):
39, 43, 4, 48, 54, 5, STOP
391,434,438,486,540,545.
LINKS
EXAMPLE
Here are the first terms of the sequence:
61,6,7,8,9,10,1,11,12,13,15,16,18,20,2,23,25,...
and here are the cumulative sums:
61,67,74,82,91,101,102,113,125,138,153,169,187,207,209, 232,257,...
If we align the a(n)s and the cumulative sums, we see that those always begin with the integer above it:
61,6, 7, 8, 9, 10, 1, 11, 12, 13, 15, 16, 18, 20, 2, 23, 25,...
61,67,74,82,91,101,102,113,125,138,153,169,187,207,209,232,257,...
CROSSREFS
Sequence in context: A345224 A198189 A102600 * A077331 A106426 A106416
KEYWORD
base,nonn
AUTHOR
STATUS
approved