A292274 -id:A292274 - cp's OEIS frontend

A163511 a(0)=1. a(n) = p(A000120(n)) * Product_{m=1..A000120(n)} p(m)^A163510(n,m), where p(m) is the m-th prime.

1, 2, 4, 3, 8, 9, 6, 5, 16, 27, 18, 25, 12, 15, 10, 7, 32, 81, 54, 125, 36, 75, 50, 49, 24, 45, 30, 35, 20, 21, 14, 11, 64, 243, 162, 625, 108, 375, 250, 343, 72, 225, 150, 245, 100, 147, 98, 121, 48, 135, 90, 175, 60, 105, 70, 77, 40, 63, 42, 55, 28, 33, 22, 13, 128

Offset: 0

Leroy Quet, Jul 29 2009

This is a permutation of the positive integers.
From Antti Karttunen, Jun 20 2014: (Start)
Note the indexing: the domain starts from 0, while the range excludes zero, thus this is neither a bijection on the set of nonnegative integers nor on the set of positive natural numbers, but a bijection from the former set to the latter.
Apart from that discrepancy, this could be viewed as yet another "entanglement permutation" where the two complementary pairs to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with the complementary pair even numbers (taken straight) and odd numbers in the order they appear in A003961: (A005843/A003961). See also A246375 which has almost the same recurrence.
Note how the even bisection halved gives the same sequence back. (For a(0)=1, take ceiling of 1/2).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
This irregular table can be represented as a binary tree. Each child to the left is obtained by doubling the parent, and each child to the right is obtained by applying A003961 to the parent:
                                     1
                                     |
                  ...................2...................
                 4                                       3
       8......../ \........9                   6......../ \........5
      / \                 / \                 / \                 / \
     /   \               /   \               /   \               /   \
    /     \             /     \             /     \             /     \
  16       27         18       25         12       15         10       7
32  81   54  125    36  75   50  49     24  45   30  35     20  21   14 11
etc.
Sequence A005940 is obtained by scanning the same tree level by level in mirror image fashion. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees, and A252463 gives the parent of the node containing n.
A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 1 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is smaller than the right child, and A252744(n) is an indicator function for those nodes.
(End)
Note that the idea behind maps like this (and the mirror image A005940) admits also using alternative orderings of primes, not just standard magnitude-wise ordering (A000040). For example, A332214 is a similar sequence but with primes rearranged as in A332211, and A332817 is obtained when primes are rearranged as in A108546. - Antti Karttunen, Mar 11 2020
From Lorenzo Sauras Altuzarra, Nov 28 2020: (Start)
This sequence is generated from A228351 by applying the following procedure: 1) eliminate the compositions that end in one unless the first one, 2) subtract one unit from every component, 3) replace every tuple [t_1, ..., t_r] by Product_{k=1..r} A000040(k)^(t_k) (see the examples).
Is it true that a(n) = A337909(n+1) if and only if a(n+1) is not a term of A161992?
Does this permutation have any other cycle apart from (1), (2) and (6, 9, 16, 7)? (End)
From Antti Karttunen, Jul 25 2023: (Start)
(In the above question, it is assumed that the starting offset would be 1 instead of 0).
Questions:
Does a(n) = 1+A054429(n) hold only when n is of the form 2^k times 1, 3 or 7, i.e., one of the terms of A029748?
It seems that A007283 gives all fixed points of map n -> a(n), like A335431 seems to give all fixed points of map n -> A332214(n). Is there a general rule for mappings like these that the fixed points (if they exist) must be of the form 2^k times a certain kind of prime, i.e., that any odd composite (times 2^k) can certainly be excluded? See also note in A029747.
(End)
If the conjecture given in A364297 holds, then it implies the above conjecture about A007283. See also A364963. - Antti Karttunen, Sep 06 2023
Conjecture: a(n^k) is never of the form x^k, for any integers n > 0, k > 1, x >= 1. This holds at least for squares, cubes, seventh and eleventh powers (see A365808, A365801, A366287 and A366391). - Antti Karttunen, Sep 24 2023, Oct 10 2023.
See A365805 for why the above holds for any n^k, with k > 1. - Antti Karttunen, Nov 23 2023

			For n=3, whose binary representation is "11", we have A000120(3)=2, with A163510(3,1) = A163510(3,2) = 0, thus a(3) = p(2) * p(1)^0 * p(2)^0 = 3*1*1 = 3.
For n=9, "1001" in binary, we have A000120(9)=2, with A163510(9,1) = 0 and A163510(9,2) = 2, thus a(9) = p(2) * p(1)^0 * p(2)^2 = 3*1*9 = 27.
For n=10, "1010" in binary, we have A000120(10)=2, with A163510(10,1) = 1 and A163510(10,2) = 1, thus a(10) = p(2) * p(1)^1 * p(2)^1 = 3*2*3 = 18.
For n=15, "1111" in binary, we have A000120(15)=4, with A163510(15,1) = A163510(15,2) = A163510(15,3) = A163510(15,4) = 0, thus a(15) = p(4) * p(1)^0 * p(2)^0 * p(3)^0 * p(4)^0 = 7*1*1*1*1 = 7.
[1], [2], [1,1], [3], [1,2], [2,1] ... -> [1], [2], [3], [1,2], ... -> [0], [1], [2], [0,1], ... -> 2^0, 2^1, 2^2, 2^0*3^1, ... = 1, 2, 4, 3, ... - _Lorenzo Sauras Altuzarra_, Nov 28 2020

Inverse: A243071.
Cf. A000040, A000120, A000225, A000788, A003961, A007814, A054429, A055396, A064216, A135523, A161992, A163510, A245605, A245612, A246375, A246378, A246681, A161511, A228351, A243499, A243503, A243504, A269854, A280873, A285727, A290251, A293437, A337909.
Cf. A007283 (known positions where a(n)=n), A029747, A029748, A364255 [= gcd(n,a(n))], A364258 [= a(n)-n], A364287 (where a(n) < n), A364292 (where a(n) <= n), A364494 (where n|a(n)), A364496 (where a(n)|n), A364963, A364297.
Cf. A365808 (positions of squares), A365801 (of cubes), A365802 (of fifth powers), A365805 [= A052409(a(n))], A366287, A366391.
Cf. A005940, A332214, A332817, A366275 (variants).

f[n_] := Reverse@ Map[Ceiling[(Length@ # - 1)/2] &, DeleteCases[Split@ Join[Riffle[IntegerDigits[n, 2], 0], {0}], {k__} /; k == 1]]; {1}~Join~
Table[Function[t, Prime[t] Product[Prime[m]^(f[n][[m]]), {m, t}]][DigitCount[n, 2, 1]], {n, 120}] (* Michael De Vlieger, Jul 25 2016 *)

from sympy import prime
def A163511(n):
    if n:
        k, c, m = n, 0, 1
        while k:
            c += 1
            m *= prime(c)**(s:=(~k&k-1).bit_length())
            k >>= s+1
        return m*prime(c)
    return 1 # Chai Wah Wu, Jul 17 2023

For n >= 1, a(2n) is even, a(2n+1) is odd. a(2^k) = 2^(k+1), for all k >= 0.
From Antti Karttunen, Jun 20 2014: (Start)
a(0) = 1, a(1) = 2, a(2n) = 2*a(n), a(2n+1) = A003961(a(n)).
As a more general observation about the parity, we have:
For n >= 1, A007814(a(n)) = A135523(n) = A007814(n) + A209229(n). [This permutation preserves the 2-adic valuation of n, except when n is a power of two, in which cases that value is incremented by one.]
For n >= 1, A055396(a(n)) = A091090(n) = A007814(n+1) + 1 - A036987(n).
For n >= 1, a(A000225(n)) = A000040(n).
(End)
From Antti Karttunen, Oct 11 2014: (Start)
As a composition of related permutations:
a(n) = A005940(1+A054429(n)).
a(n) = A064216(A245612(n))
a(n) = A246681(A246378(n)).
Also, for all n >= 0, it holds that:
A161511(n) = A243503(a(n)).
A243499(n) = A243504(a(n)).
(End)
More linking identities from Antti Karttunen, Dec 30 2017: (Start)
A046523(a(n)) = A278531(n). [See also A286531.]
A278224(a(n)) = A285713(n). [Another filter-sequence.]
A048675(a(n)) = A135529(n) seems to hold for n >= 1.
A250245(a(n)) = A252755(n).
A252742(a(n)) = A252744(n).
A245611(a(n)) = A253891(n).
A249824(a(n)) = A275716(n).
A292263(a(n)) = A292264(n). [A292944(n) + A292264(n) = n.]
--
A292383(a(n)) = A292274(n).
A292385(a(n)) = A292271(n). [A292271(n) +  A292274(n) = n.]
--
A292941(a(n)) = A292942(n).
A292943(a(n)) = A292944(n).
A292945(a(n)) = A292946(n). [A292942(n) + A292944(n) + A292946(n) = n.]
--
A292253(a(n)) = A292254(n).
A292255(a(n)) = A292256(n). [A292944(n) + A292254(n) + A292256(n) = n.]
--
A279339(a(n)) = A279342(n).
a(A071574(n)) = A269847(n).
a(A279341(n)) = A279338(n).
a(A252756(n)) = A250246(n).
(1+A008836(a(n)))/2 = A059448(n).
(End)
From Antti Karttunen, Jul 26 2023: (Start)
For all n >= 0, a(A007283(n)) = A007283(n).
A001222(a(n)) = A290251(n).
(End)

More terms computed and examples added by Antti Karttunen, Jun 20 2014

A292383 Base-2 expansion of a(n) encodes the steps where numbers of the form 4k+3 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n.

Original entry on oeis.org

0, 0, 1, 0, 2, 2, 5, 0, 0, 4, 11, 4, 22, 10, 5, 0, 44, 0, 89, 8, 8, 22, 179, 8, 0, 44, 1, 20, 358, 10, 717, 0, 20, 88, 11, 0, 1434, 178, 45, 16, 2868, 16, 5737, 44, 8, 358, 11475, 16, 0, 0, 89, 88, 22950, 2, 17, 40, 176, 716, 45901, 20, 91802, 1434, 17, 0, 40, 40, 183605, 176, 356, 22, 367211, 0, 734422, 2868, 1, 356, 22, 90, 1468845, 32, 0, 5736, 2937691, 32

Offset: 1

Antti Karttunen, Sep 15 2017

			For n = 3, the starting value is of the form 4k+3, after which follows A252463(3) = 2, and A252463(2) = 1, the end point of iteration, and neither 2 nor 1 is of the form 4k+3, thus a(3) = 1*(2^0) + 0*(2^1) + 0*(2^2) = 1.
For n = 5, the starting value is not of the form 4k+3, after which follows A252463(5) = 3 (which is), continuing as before as 3 -> 2 -> 1, thus a(5) = 0*(2^0) + 1*(2^1) + 0*(2^2) + 0*(2^3) = 2.
For n = 10, the starting value is not of the form 4k+3, after which follows A252463(10) = 5 (also not 4k+3), and then A252463(5) = 3 (which is), continuing as before as 3 -> 2 -> 1, thus a(10) = 0*(2^0) + + 0*(2^1) + 1*(2^2) + 0*(2^3) + 0*(2^4) = 4.

Cf. A163511, A243071, A292274, A292373, A292377, A292380, A292381, A292382, A292384, A292385.

Table[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]], n, # > 1 &] /. k_ /; IntegerQ@ k :> If[Mod[k, 4] == 3, 1, 0], 2], {n, 84}] (* Michael De Vlieger, Sep 21 2017 *)

A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A252463(n) = if(!(n%2),n/2,A064989(n));
A292383(n) = if(1==n,0,(if(3==(n%4),1,0)+(2*A292383(A252463(n)))));

(define (A292383 n) (A292373 (A292384 n)))

a(1) = 0; for n > 1, a(n) = 2*a(A252463(n)) + [n ≡ 3 (mod 4)], where the last part of the formula is Iverson bracket, giving 1 only if n is of the form 4k+3, and 0 otherwise.
a(n) = A292373(A292384(n)).
a(n) = A292274(A243071(n)).
Other identities. For n >= 1:
a(2n) = 2*a(n).
a(n) + A292385(n) = A243071(n).
a(A163511(n)) = A292274(n).
A000120(a(n)) = A292377(n).

A292271 a(n) = A292385(A163511(n)).

Original entry on oeis.org

0, 1, 2, 2, 4, 5, 4, 5, 8, 8, 10, 11, 8, 8, 10, 10, 16, 17, 16, 17, 20, 20, 22, 23, 16, 17, 16, 16, 20, 21, 20, 20, 32, 32, 34, 35, 32, 32, 34, 34, 40, 41, 40, 41, 44, 44, 46, 47, 32, 32, 34, 34, 32, 33, 32, 33, 40, 40, 42, 42, 40, 41, 40, 41, 64, 65, 64, 65, 68, 68, 70, 71, 64, 65, 64, 64, 68, 69, 68, 68, 80, 80, 82, 83, 80, 80, 82, 82, 88, 89, 88, 89, 92, 92

Offset: 0

Antti Karttunen, Sep 16 2017

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n

Cf. A163511, A292274, A292385.

a(n) = A292385(A163511(n)).

a(n) + A292274(n) = n.

A292592 a(n) = A292590(A245612(n)).

Original entry on oeis.org

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, 10, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 16, 17, 16, 17, 20, 20, 22, 23, 16, 17, 16, 16, 20, 21, 20, 21, 0, 0, 2, 2, 0, 1, 0, 0, 8, 8, 10, 11, 8, 8, 10, 10, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 32, 32, 34, 35, 32, 32, 34, 34, 40, 41, 40, 41

Offset: 0

Antti Karttunen, Sep 20 2017

Because A292590(n) = a(A245611(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 multiples of 3 in binary tree A245612 on that trajectory which leads from the root of the tree to the node containing A245612(n). See the examples.

			A245612(18) = 188, that is, at "node address" 18 in binary tree A245612 sits number 188. 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 the tree when starting from the root, to land in node containing number 188. Here, after 1 and 2, turn left from 2, turn left from 5, turn right from 14 and then turn left from 63 and then indeed one lands in 188. These turns correspond with the four lowermost bits of the code, "0010". When one selects multiples of 3 from this path 1 -> 2 -> 5 -> 14 -> 63 -> 188, the only one is 63, 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.
A245612(15) = 6, that is, at "node address" 15 in binary tree A245612 sits number 6. 15 in binary is "1111", which tells that 6 can be located in tree A245612 by going (after the initial root 1 and 2) three steps towards right from 2: 1 -> 2 -> 3 -> 4 -> 6. Of these numbers, only 3 and 6 are multiples of 3, thus the mask to obtain the corresponding bits from "1111" is "00101" (5 in binary), thus a(15) = 5.
A245612(31) = 7, that is, at "node address" 31 in binary tree A245612 sits number 7. 31 in binary is "11111", which tells that 7 can be located in tree A245612 by going (after the initial root 1 and 2) four steps towards right from 2: 1 -> 2 -> 3 -> 4 -> 6 -> 7. Of these numbers, only 3 and 6 are multiples of 3, thus the mask to obtain the corresponding bits from "11111" is "001010" (ten in binary), thus a(31) = 10.

Cf. A004198, A292590, A292593.
Cf. also A292377.
Differs from related A292274 for the first time at n=31, where a(31) = 10, while A292274(31) = 11. Compare also the scatter plots.

f[n_] := f[n] = Which[n == 1, 0, Mod[n, 3] == 2, Ceiling[n/3], True, (Times @@ Power[If[# == 1, 1, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger[2 n - 1] + 1)/2]; g[n_] := g[n] = If[n == 1, 0, 2 g[f@ n] + Boole[Divisible[n, 3]]]; h[n_] := (Times @@ Power[If[# == 1, 1, NextPrime@ #] & /@ First@ #, Last@ #] + 1)/2 &@ Transpose@ FactorInteger@If[n == 0, 1, Prime[#] Product[ Prime[m]^(Map[ Ceiling[(Length@ # - 1)/2] &, DeleteCases[ Split@ Join[ Riffle[ IntegerDigits[n, 2], 0], {0}], {k__} /;k == 1]][[-m]]), {m, #}] &[DigitCount[n, 2, 1]]]; Array[g@ h@ # &, 108, 0] (* Michael De Vlieger, Sep 22 2017 *)

a(n) + A292593(n) = n, a(n) AND A292593(n) = 0.

a(n) AND n = a(n), where AND is bitwise-AND (A004198).

A292944 a(n) = A292272(A004754(n)) - 2*A053644(n).

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 2, 2, 0, 1, 2, 2, 4, 5, 4, 4, 0, 1, 2, 2, 4, 5, 4, 4, 8, 9, 10, 10, 8, 9, 8, 8, 0, 1, 2, 2, 4, 5, 4, 4, 8, 9, 10, 10, 8, 9, 8, 8, 16, 17, 18, 18, 20, 21, 20, 20, 16, 17, 18, 18, 16, 17, 16, 16, 0, 1, 2, 2, 4, 5, 4, 4, 8, 9, 10, 10, 8, 9, 8, 8, 16, 17, 18, 18, 20, 21, 20, 20, 16, 17, 18, 18, 16, 17, 16, 16, 32, 33, 34, 34, 36, 37, 36, 36

Offset: 0

Antti Karttunen, Sep 28 2017

In binary expansion (A007088) of n, clear the most significant bit and all those 1-bits that have another 1-bit at their left side, except for the second most significant 1-bit, even in cases where the binary expansion begins as "11...".

Because A292943(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 6k+3 (odd multiples of three) in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).

			For n = 23, 10111 in binary, when we clear (change to zero) the most significant bit (always 1) and also all 1-bits that have 1's at their left side, we are left with 100, which in binary stands for 4, thus a(23) = 4.
For n = 27, 11011 in binary, when we clear the most significant bit, and also all 1-bits that have 1's at their left side except the second most significant, we are left with 1010, which in binary stands for ten, thus a(27) = 10.

Antti Karttunen, Table of n, a(n) for n = 0..16383
Index entries for sequences related to binary expansion of n

Cf. A004754, A005940, A048735, A163511, A292272, A292943.

Cf. also A292247, A292248, A292254, A292256, A292264, A292271, A292274, A292592, A292593, A292942, A292946.

(define (A292944 n) (let ((x (+ n (A053644 n)))) (- (A292272 x) (A053644 x))))
(define (A292944 n) (- (A292272 (A004754 n)) (* 2 (A053644 n))))
(define (A292944 n) (A292943 (A163511 n)))

a(n) = A292272(A004754(n)) - 2*A053644(n).
a(n) = A292943(A163511(n)).
Other identities. For all n >= 0:
a(n) + A292264(n) = A292942(n) + a(n) + A292946(n) = a(n) + A292254(n) + A292256(n) = n.
a(n) = a(n) AND n; a(n) AND A292264(n) = 0, where AND is bitwise-and (A004198).

A292254 a(n) = A292253(A163511(n)).

Original entry on oeis.org

0, 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 9, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 18, 19, 16, 16, 16, 17, 16, 16, 16, 17, 32, 32, 32, 33, 32, 32, 32, 32, 32, 32, 32, 32, 36, 36, 38, 39, 32, 32, 32, 32, 32, 32, 34, 34, 32, 32, 32, 32, 32, 32, 34, 35, 64, 64, 64, 64, 64, 64, 66, 67, 64, 64, 64, 65, 64, 64, 64, 65, 64, 64, 64, 65, 64, 64, 64, 64, 72, 72, 72, 72, 76, 76

Offset: 0

Antti Karttunen, Sep 28 2017

nonn

Because A292253(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 the numbers that are either of the form 12k+1 or of the form 12k+11 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).

The AND - XOR formula just restates the fact that J(3|n) = J(-1|n)*J(-3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Cf. A005940, A163511, A292253.

Cf also A292247, A292248, A292256, A292264, A292271, A292274, A292592, A292593, A292942, A292944, A292946 (for similarly constructed sequences).

(define (A292254 n) (A292253 (A163511 n)))

a(n) = A292253(A163511(n)).
a(n) = A292264(n) AND (A292274(n) XOR A292942(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987). [See comments.]
For all n >= 0, a(n) + A292944(n) + A292256(n) = n.

A292256 a(n) = A292255(A163511(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 6, 6, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 8, 8, 8, 9, 12, 12, 12, 12, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 6, 6, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 2, 16, 16, 16, 16, 16, 16

Offset: 0

Antti Karttunen, Sep 28 2017

nonn

Because A292255(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 the numbers that are either of the form 12k+5 or of the form 12k+7 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).

The AND - XOR formulas just restate the fact that J(3|n) = J(-1|n)*J(-3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Cf. A005940, A163511, A292255.

Cf. also A292247, A292248, A292254, A292264, A292271, A292274, A292592, A292593, A292942, A292944, A292946 (for similarly constructed sequences).

(define (A292256 n) (A292255 (A163511 n)))

a(n) = A292255(A163511(n)).
a(n) = A292264(n) AND (A292274(n) XOR A292946(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).
a(n) = A292264(n) AND (A292271(n) XOR A292942(n)). [See comments].
For all n >= 0, a(n) + A292944(n) + A292254(n) = n.

A292942 a(n) = A292941(A163511(n)).

Original entry on oeis.org

0, 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 9, 8, 8, 8, 9, 16, 16, 16, 16, 16, 16, 18, 19, 16, 16, 16, 16, 16, 16, 18, 18, 32, 32, 32, 33, 32, 32, 32, 33, 32, 32, 32, 32, 36, 36, 38, 39, 32, 32, 32, 33, 32, 32, 32, 32, 32, 32, 32, 33, 36, 36, 36, 37, 64, 64, 64, 64, 64, 64, 66, 67, 64, 64, 64, 64, 64, 64, 66, 66, 64, 64, 64, 65, 64, 64, 64, 65, 72, 72, 72, 72, 76, 76

Offset: 0

Antti Karttunen, Sep 28 2017

nonn

Because A292941(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 6k+1 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).

The AND - XOR formula is just a restatement of the fact that J(-3|n) = J(-1|n)*J(3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Cf. A005940, A163511, A292941.

Cf. also A292247, A292248, A292254, A292256, A292264, A292271, A292274, A292592, A292593, A292944, A292946 (for similarly constructed sequences).

(define (A292942 n) (A292941 (A163511 n)))

a(n) = A292941(A163511(n)).
a(n) = A292264(n) AND (A292254(n) XOR A292274(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987). [See comments.]
For all n >= 0, a(n) + A292944(n) + A292946(n) = n.

A292946 a(n) = A292945(A163511(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 4, 4, 4, 5, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 8, 8, 8, 8, 8, 8, 10, 10, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 4, 4, 4, 5, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0

Offset: 0

Antti Karttunen, Sep 28 2017

nonn

Because A292945(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 6k+5 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).

The AND - XOR formulas just restate the fact that J(-3|n) = J(-1|n)*J(3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Cf. A005940, A163511, A292945.

Cf. also A292247, A292248, A292254, A292256, A292264, A292271, A292274, A292592, A292593, A292942, A292944 (for similarly constructed sequences).

(define (A292946 n) (A292945 (A163511 n)))

a(n) = A292945(A163511(n)).
a(n) = A292264(n) AND (A292256(n) XOR A292274(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).
a(n) = A292264(n) AND (A292254(n) XOR A292271(n)). [See comments.]
For all n >= 0, A292942(n) + A292944(n) + a(n) = n.

cp's OEIS Frontend

A163511 a(0)=1. a(n) = p(A000120(n)) * Product_{m=1..A000120(n)} p(m)^A163510(n,m), where p(m) is the m-th prime.

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A292383 Base-2 expansion of a(n) encodes the steps where numbers of the form 4k+3 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n.

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A292271 a(n) = A292385(A163511(n)).

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A292592 a(n) = A292590(A245612(n)).

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A292944 a(n) = A292272(A004754(n)) - 2*A053644(n).

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A292254 a(n) = A292253(A163511(n)).

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A292256 a(n) = A292255(A163511(n)).

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A292942 a(n) = A292941(A163511(n)).

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