

A101274


a(1)=1; for n>1, a(n) is the smallest positive integer such that the set of all sums of adjacent elements up to and including a(n) contains no number more than once.


3



1, 2, 4, 5, 8, 10, 14, 21, 15, 16, 26, 25, 34, 22, 48, 38, 71, 40, 74, 90, 28, 69, 113, 47, 94, 54, 46, 143, 153, 83, 128, 49, 249, 75, 133, 225, 125, 131, 270, 145, 230, 199, 237, 206, 201, 299, 136, 346, 72, 272, 120, 55, 453, 247, 376, 427, 124, 535, 87, 242, 431, 283, 227, 212, 940, 318, 387, 311, 391, 325
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OFFSET

1,2


COMMENTS

Does the sequence together with the sums of adjacent elements include all positive integers? Choosing starting values other than a(1)=1 gives other sequences. We could ask, for a given n, which such sequences have the smallest sum of a(k) from k=1 to n.
The first differences of A005282. [Zak Seidov, Nov 06 2010]


LINKS

JeanMarc Falcoz, Table of n, a(n) for n = 1..483
E. Giaquinta and S. Grabowski, New algorithms for binary jumbled pattern matching, arXiv preprint arXiv:1210.6176 [cs.DS], 2012.  From N. J. A. Sloane, Jan 01 2013


EXAMPLE

a(8)=21 because the set of sums of adjacent elements to this point, call it s(7) is {1,2,3,4,5,6,7,8,9,10,11,12,13,14,17,18,19,20,23,24,27,29,30,32,37,41,43,44}.
The first number missing from this list is 15, but a(8) cannot equal 15 because 15+14=29 and 29 is already in s(7). Similarly a(8) cannot be 16 because 16+14=30.


MATHEMATICA

t = {1}; sms = {2}; k = 1;
Do[k++; While[Intersection[sms, k + t] != {}, k++]; sms = Join[sms, t + k, {2 k}]; AppendTo[t, k], {41}];
Differences[t] (* JeanFrançois Alcover, Feb 13 2019, after T. D. Noe in A005282 *)


CROSSREFS

Cf. A005282.
Sequence in context: A262937 A249508 A163295 * A080222 A050539 A277101
Adjacent sequences: A101271 A101272 A101273 * A101275 A101276 A101277


KEYWORD

nonn


AUTHOR

David S. Newman, Dec 20 2004


STATUS

approved



