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A101274
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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.
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5
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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
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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}];
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PROG
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(Python)
from itertools import count, islice
def A101274_gen(): # generator of terms
aset1, aset2, alist, n = {1}, set(), [1], 1
for k in count(2):
bset2 = {k<<1}
if (k<<1) not in aset2:
for d in aset1:
if (m:=d+k) in aset2:
break
bset2.add(m)
else:
yield k-n
n = k
alist.append(k)
aset1.add(k)
aset2.update(bset2)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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