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.

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

Jean-Marc Falcoz, Table of n, a(n) for n = 1..5004

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

Adjacent sequences: A316915 A316916 A316917 * A316919 A316920 A316921

Keyword

base,nonn

Author

Eric Angelini and Jean-Marc Falcoz, Jul 16 2018

Status

approved