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A269175
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a(n) = number of distinct k for which A269174(k) = n; number of finite predecessors for pattern encoded in the binary expansion of n in Wolfram's Rule 124 cellular automaton.
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6
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1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 2, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 2, 1, 0, 2, 2, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 2, 0, 0, 2, 1, 1
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OFFSET
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0,32
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COMMENTS
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At positions A000225 seems to occur the record values of this sequence: 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, ... which seem to match with A000931 (Padovan sequence), or more exactly, with A182097 (Number of compositions (ordered partitions) into parts 2 and 3). Note that these values give also the number of predecessors for each "repunit-pattern" (2^n)-1 in Rule 110 cellular automaton, as rules 110 and 124 are mirror images of each other.
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LINKS
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PROG
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(Scheme)
(definec (A269175 n) (let loop ((p 0) (s 0)) (cond ((> p n) s) (else (loop (+ 1 p) (+ s (if (= n (A269174 p)) 1 0))))))) ;; Very straightforward and very slow.
;; Somewhat optimized version:
(definec (A269175 n) (if (zero? n) 1 (let ((nwid-1 (- (A000523 n) 1))) (let loop ((p (if (< n 2) 0 (A000079 nwid-1))) (s 0)) (cond ((> (A000523 p) nwid-1) s) (else (loop (+ 1 p) (+ s (if (= n (A269174 p)) 1 0)))))))))
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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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