



0, 0, 0, 1, 0, 0, 2, 2, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 8, 8, 10, 11, 8, 8, 10, 11, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 16, 17, 16, 17, 20, 20, 22, 22, 16, 17, 16, 17, 20, 20, 22, 22, 0, 0, 2, 2, 0, 1, 0, 0, 8, 8, 10, 11, 8, 8, 10, 11, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 32, 32, 34, 35, 32, 32, 34, 35, 40, 41, 40, 40, 44
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OFFSET

0,7


COMMENTS

Because A292383(n) = a(A243071(n)), the sequence works as a "masking function" where the 1bits in a(n) (always a subset of the 1bits in binary expansion of n) indicate which numbers are of the form 4k+3 in binary tree A005940 on that trajectory which leads from the root of the tree to the node containing A163511(n). This works because A243071(n) = A054429(A156552(n)), a bitflipped variant of Leonid Broukhis's unarybinary encoded compressed factorization of natural numbers, A156552(n) being an inverse of Doudna map f(n) = A005940(1+n).


LINKS

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n
Index entries for sequences computed from indices in prime factorization


FORMULA

a(n) = A292383(A163511(n)).
a(n) + A292271(n) = n, a(n) AND A292271(n) = 0.
a(n) AND n = a(n), where AND is bitwiseAND (A004198).


EXAMPLE

A163511(18) = 54, that is, at "node address" 18 in binary tree A163511 (which is the mirror image of A005940) sits number 54. 18 in binary is "10010", which when read from left to right (after the most significant bit which is always 1) gives the directions to follow in either tree when starting from the root, so that we land in number 54. (E.g. in A005940tree, turn right from 2, turn right from 4, turn left from 8 and then turn right from 27 and one lands in 54, this corresponds with the four lowermost bits of the code, "0010". In A163511 the sense of direction is just reversed). When one selects the numbers of the form 4k+3 from this path 1 > 2 > 4 > 8 > 27 > 54, one sees that only one is 27, which corresponds with the second rightmost bit (which also is the only 1bit) in the code, which can be masked with 2 (binary "10"), thus a(18) = 2.
A163511(15) = 7, that is, at "node address" 15 in binary tree A163511 sits number 7. 15 in binary is "1111", which tells that 7 can be located in mirrorimage tree A005940 by going (after the initial root 1 and 2) three steps towards left from 2: 1 > 2 > 3 > 5 > 7. Of these numbers, only 3 and 7 are of the form 4k+3, thus the mask with which to obtain the corresponding bits from "1111" is "00101" (5 in binary), thus a(15) = 5.
A163511(31) = 11, that is, at "node address" 31 in binary tree A163511 sits number 11. 31 in binary is "11111", which tells that 11 can be located in mirrorimage tree A005940 by going (after the initial root 1 and 2) four steps towards left from 2: 1 > 2 > 3 > 5 > 7 > 11. Of these numbers, only 3, 7 and 11 are of the form 4k+3, thus the mask with which to obtain the corresponding bits from "11111" is "001011" (11 in binary), thus a(31) = 11.


MATHEMATICA

f[n_] := Reverse@ Map[Ceiling[(Length@ #  1)/2] &, DeleteCases[Split@ Join[Riffle[IntegerDigits[n, 2], 0], {0}], {k__} /; k == 1]]; Map[FromDigits[Reverse@ NestWhileList[Function[k, Which[k == 1, 1, EvenQ@ k, k/2, True, Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, 1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ k]], #, # > 1 &] /. k_ /; IntegerQ@ k :> If[Mod[k, 4] == 3, 1, 0], 2] &, {1}~Join~Table[Function[t, Prime[t] Product[Prime[m]^(f[n][[m]]), {m, t}]][DigitCount[n, 2, 1]], {n, 120}]] (* Michael De Vlieger, Sep 22 2017 *)


CROSSREFS

Cf. A004198, A163511, A243071, A292271, A292383.
Differs from related A292592 for the first time at n=31, where a(31) = 11, while A292592(31) = 10. Compare also the scatter plots.
Sequence in context: A242284 A306595 A292592 * A317675 A204423 A112170
Adjacent sequences: A292271 A292272 A292273 * A292275 A292276 A292277


KEYWORD

nonn,look,base


AUTHOR

Antti Karttunen, Sep 16 2017


EXTENSIONS

Comments and examples from Antti Karttunen, Sep 22 2017


STATUS

approved



