Search: a315615 -id:a315615
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A269501
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Subsequence immediately following the instances of n in the sequence is n, n-1, ..., 1, n+1, n+2, ....
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+0
4
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1, 1, 2, 2, 1, 3, 3, 2, 3, 1, 4, 4, 3, 4, 2, 4, 1, 5, 5, 4, 5, 3, 5, 2, 5, 1, 6, 6, 5, 6, 4, 6, 3, 6, 2, 6, 1, 7, 7, 6, 7, 5, 7, 4, 7, 3, 7, 2, 7, 1, 8, 8, 7, 8, 6, 8, 5, 8, 4, 8, 3, 8, 2, 8, 1, 9, 9, 8, 9, 7, 9, 6, 9, 5, 9, 4, 9, 3, 9, 2, 9, 1, 10, 10, 9, 10, 8, 10, 7, 10, 6, 10, 5, 10, 4, 10, 3, 10, 2, 10, 1
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OFFSET
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0,3
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COMMENTS
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The sequence includes every ordered pair of positive integers exactly once as consecutive terms of the sequence. Through n = k^2, it has every pair i,j with 0 < i,j <= k.
Can be regarded as an irregular triangle where row k contains 1, k, k, k-1, k, k-2, ..., 2, k, with 2n-1 terms.
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LINKS
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FORMULA
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Let r = ceiling(sqrt(n)) = A003059(n). If n and r have the same parity, a(n) = (r^2-n)/2 + 1; otherwise a(n) = r.
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EXAMPLE
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The first 3 occurs as a(5), so a(6) = 3, the first term of 3, 2, 1, 4, 5, 6, .... The second 3 is thus a(6), so a(7) = 2. The third 3 is a(8), so a(9) = 1. The fourth 3 is a(12), now we start incrementing, and a(13) = 4.
The triangle starts:
1
1, 2, 2
1, 3, 3, 2, 3
1, 4, 4, 3, 4, 2, 4
1, 5, 5, 4, 5, 3, 5, 2, 5
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PROG
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(PARI) a(n) = my(r = if(n<=0, 0, sqrtint(n-1)+1); if((n-r)%2, r, (r^2-n)/2 + 1)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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