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%I
%S 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,
%T 54,5,60,67,74,83,92,102,113,126,140,14,157,174,17,196,19,220,22,246,
%U 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
%N 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.
%C The authors conjecture that it is impossible to extend the sequence forever if a(1) < 61.
%C With a(1) = 60, for instance, we will be stuck with the integer 5:
%C 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).
%C See here the sequence starting with a(1) = 60 and the cumulative sums:
%C 60,6, 7, 8, 9, 10, 1, 11, 12, 13, 15, 16, 18, 20, 2, 23, 25, 28, 31, 3, 35,
%C 60,66,73,81,90,100,101,112,124,137,152,167,185,205,207,230,255,283,314,317,352,
%C (the two lines above continue here):
%C 39, 43, 4, 48, 54, 5, STOP
%C 391,434,438,486,540,545.
%H Jean-Marc Falcoz, <a href="/A316918/b316918.txt">Table of n, a(n) for n = 1..5004</a>
%e Here are the first terms of the sequence:
%e 61,6,7,8,9,10,1,11,12,13,15,16,18,20,2,23,25,...
%e and here are the cumulative sums:
%e 61,67,74,82,91,101,102,113,125,138,153,169,187,207,209, 232,257,...
%e If we align the a(n)s and the cumulative sums, we see that those always begin with the integer above it:
%e 61,6, 7, 8, 9, 10, 1, 11, 12, 13, 15, 16, 18, 20, 2, 23, 25,...
%e 61,67,74,82,91,101,102,113,125,138,153,169,187,207,209,232,257,...
%K base,nonn
%O 1,1
%A _Eric Angelini_ and _Jean-Marc Falcoz_, Jul 16 2018
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