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Search: id:A026272
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| A026272 |
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a(n) = a(m) if a(m) has already occurred exactly once and n = a(m) + m + 1, else a(n) = least positive integer that has not yet occurred. |
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+0
8
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| 1, 2, 1, 3, 2, 4, 5, 3, 6, 7, 4, 8, 5, 9, 10, 6, 11, 7, 12, 13, 8, 14, 15, 9, 16, 10, 17, 18, 11, 19, 20, 12, 21, 13, 22, 23, 14, 24, 15, 25, 26, 16, 27, 28, 17, 29, 18, 30, 31, 19, 32, 20, 33, 34, 21, 35, 36, 22, 37, 23, 38, 39, 24, 40, 41, 25
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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This sequence displays every positive integer exactly twice, and the gap between the two occurrences of n contains exactly n other values. The first occurrence of n precedes the first occurrence of n+1. (cont.)
Also related to the Wythoff array (A035513) and the Para-Fibonacci sequence (A035513) where every positive integer is displayed exactly once in the whole array. Take any integer n in A026272 and let C = number of terms from the beginning of the sequence to the second occurrence of n. Then C = (2nd term after n in the applicable sequence for n in A035513). (cont.)
Also in the second occurrence of n in A026272, let N=n ( - one term) = (first term value after n in the applicable sequence for n in A035513). In this format the second occurrence of n in A026272 will produce in A035513, n itself and two of the succeeding terms of n in the Wythoff array where every positive integer can only be displayed once. (cont.)
In A026272 if |a(n)-a(n+1)| > 10 then phi ~ a(n)/|a(n)-a(n+1)|. When n -> infinity it will converge to phi. - Dan Joyce (danj_536(AT)msn.com), Apr 13 2001
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CROSSREFS
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Cf. A035513.
a(n) = A026242(n+2) - 1 = A026350(n+3) - 2 = A026354(n+4) - 3.
Adjacent sequences: A026269 A026270 A026271 this_sequence A026273 A026274 A026275
Sequence in context: A116928 A034391 A094173 this_sequence A022447 A117194 A024467
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KEYWORD
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nonn,easy,nice
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AUTHOR
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Clark Kimberling (ck6(AT)evansville.edu)
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