|
%I #31 Sep 23 2017 19:13:40
%S 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,
%T 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,
%U 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
%N a(n) = A292383(A163511(n)).
%C Because A292383(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits 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 bit-flipped variant of _Leonid Broukhis_'s unary-binary encoded compressed factorization of natural numbers, A156552(n) being an inverse of Doudna map f(n) = A005940(1+n).
%H Antti Karttunen, <a href="/A292274/b292274.txt">Table of n, a(n) for n = 0..8192</a>
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F a(n) = A292383(A163511(n)).
%F a(n) + A292271(n) = n, a(n) AND A292271(n) = 0.
%F a(n) AND n = a(n), where AND is bitwise-AND (A004198).
%e 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 A005940-tree, 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 1-bit) in the code, which can be masked with 2 (binary "10"), thus a(18) = 2.
%e 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 mirror-image 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.
%e 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 mirror-image 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.
%t 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 *)
%Y Cf. A004198, A163511, A243071, A292271, A292383.
%Y Differs from related A292592 for the first time at n=31, where a(31) = 11, while A292592(31) = 10. Compare also the scatter plots.
%K nonn,look,base
%O 0,7
%A _Antti Karttunen_, Sep 16 2017
%E Comments and examples from _Antti Karttunen_, Sep 22 2017
|
