

A218615


a(n) = binary code (shown here in decimal) of the position of natural number n in the beanstalktree A218776.


4



1, 3, 2, 6, 4, 14, 10, 26, 18, 58, 42, 122, 90, 106, 74, 202, 138, 458, 330, 970, 714, 842, 586, 1866, 1354, 1610, 1098, 3402, 2378, 3658, 2634, 6730, 4682, 14922, 10826, 31306, 23114, 27210, 19018, 59978, 43594, 51786, 35402, 109130, 76362, 117322, 84554, 248394
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OFFSET

1,2


COMMENTS

The binary code is the same as used by function generalcarcdr of MIT/GNU Scheme: a zero bit represents a cdr operation (taking the right hand side branch in the binary tree), and a one bit represents a car (taking the left hand side branch in the binary tree). The bits are interpreted from LSB to MSB, and the most significant one bit, rather than being interpreted as an operation, signals the end of the binary code.


LINKS

A. Karttunen, Table of n, a(n) for n = 1..1024
MIT/GNU Scheme 9.1. documentation, Function generalcarcdr


FORMULA

a(1)=1, for odd n, a(n) = A004754(a(A011371(n))), for even n, a(n) = A004755(a(A011371(n))).


EXAMPLE

As we can traverse to 4 in A218776tree (see the example there) by taking first the right branch (cdr) from the root, resulting bit 0 as the least significant bit of the code, then by taking the left branch (car) from 3 to get to 4, resulting bit 1 as the second rightmost bit of the code, which when capped with an extra terminationone, results binary code 110, 6 in decimal, thus a(4)=6.


PROG

(Scheme with memoization macro definec): (definec (A218615 n) (cond ((< n 2) n) ((odd? n) (A004754 (A218615 (A011371 n)))) (else (A004755 (A218615 (A011371 n))))))


CROSSREFS

a(n) = A054429(A218614(n)). Superset of A218791. Used to construct A218776, A218777. Cf. also A179016, A218787, A218788
Sequence in context: A162255 A164073 A286595 * A177828 A248479 A304533
Adjacent sequences: A218612 A218613 A218614 * A218616 A218617 A218618


KEYWORD

nonn,base


AUTHOR

Antti Karttunen, Nov 16 2012


STATUS

approved



