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1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 3, 4, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5
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OFFSET
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1,3
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COMMENTS
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Can be generated recursively by first setting R_1 = (1), after which each R_n is obtained by replacing in R_{n-1} each term k with terms 1 .. k, followed by final n. This sequence is then obtained by concatenating all levels R_1, R_2, ..., R_inf together. See page 230 in Kubo-Vakil paper (page 6 in PDF).
Deleting all 1's and decrementing the remaining terms by one gives the sequence back.
The following simple Pascal-like triangle produces the same sequence. Construct a triangle T(n,k) of strings (with 0 <= k <= n), where T(0,0) = {1}, T(n,n) = {n+1}, and otherwise T(n,k) is the concatenation of T(n-1,k-1) and T(n-1,k). The first few rows of the triangle (where the strings T(n,k) are shown without spaces for legibility) are:
1
1,2
1,12,3
1,112,123,4
1,1112,112123,1234,5
1,11112,1112112123,1121231234,12345,6
...
Now read the strings across the rows to get the sequence. T(n,k) has length binomial(n,k). (End)
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LINKS
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FORMULA
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As a recurrence: If A036987(n) = 1 [when n is of the form 2^k -1], a(n) = A070939(n), else if a(n+1) = 1, a(n) = a(2^A000523(n) - A266349(n)), otherwise a(n) = a(n+1)-1.
Other identities. For all n >= 1:
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EXAMPLE
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Illustration of the sequence as a tree:
1
/ \
1 2
/ /|\
1 1 2 3_________
/ / /| | \ \ \
1 1 1 2 1 2 3__ 4________
/ / / /| | / \ |\ \ \ \ \ \ \
1 1 1 1 2 1 1 2 1 2 3 1 2 3 4 5
etc.
Compare with the illustration in A265332.
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PROG
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(Scheme, two variants)
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CROSSREFS
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Cf. A000225 (positions of records, where n appears first time).
Cf. A266640 (obtained from the mirror image of the same tree).
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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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