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A232750
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a(0)=1, after which a(n) = Number of terms of A005228 which occur between each consecutive terms of A232739, in range A232739(n)..A232739(n+1).
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6
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1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1
(list;
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refs;
listen;
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OFFSET
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0
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COMMENTS
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Does a term larger than 1 ever appear?
Positions of zeros: 2, 5, 8, 12, 16, 20, 25, 30, 36, 42, 48, 54, 61, 68, 76, 84, 92, 101, 110, 119, 128, ...
and their first differences: 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 9, 9, 10, 10, 11, 11, 11, 12, 11, 13, 12, 13, ...
might be also interesting.
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LINKS
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FORMULA
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EXAMPLE
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The two sequences begin as:
A232739: 2, 4,6, 9, 13,17, 22, 28,34, 41, 49, 58,67, 77, ...
A005228: 1, 3, 7, 12, 18, 26, 35, 45, 56, 69, 83, ...
We let a(0)=1 stand for the number of terms of A005228 that are before the first term of A232739, namely that 1 which is less than 2, and thereafter we count the terms of A005228 that occur between each two consecutive terms of A232739, noting that 3 in the latter occurs between 2 and 4 in the former, thus a(1)=1, no terms in the latter occur between 4 and 6 in the former, thus a(2)=0, 7 in the latter occurs between 6 and 9 in the former, thus a(3)=1, and so on. See also the example in A232740.
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PROG
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(Scheme)
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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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