A065642 -id:A065642 - cp's OEIS frontend

A285332 a(0) = 1, a(1) = 2, a(2n) = A019565(a(n)), a(2n+1) = A065642(a(n)).

1, 2, 3, 4, 6, 9, 5, 8, 15, 12, 14, 27, 10, 25, 7, 16, 210, 45, 35, 18, 105, 28, 462, 81, 21, 20, 154, 125, 30, 49, 11, 32, 10659, 420, 910, 75, 78, 175, 33, 24, 3094, 315, 385, 56, 780045, 924, 374, 243, 110, 63, 55, 40, 4389, 308, 170170, 625, 1155, 60, 286, 343, 42, 121, 13, 64, 54230826, 31977, 28405, 630, 1330665, 1820, 714

Offset: 0

Antti Karttunen, Apr 17 2017

Note the indexing: the domain starts from 0, while the range excludes zero.
This sequence can be represented as a binary tree. Each left hand child is produced as A019565(n), and each right hand child as A065642(n), when the parent node contains n >= 2:
                                     1
                                     |
                  ...................2...................
                 3                                       4
       6......../ \........9                   5......../ \........8
      / \                 / \                 / \                 / \
     /   \               /   \               /   \               /   \
    /     \             /     \             /     \             /     \
  15       12         14       27         10       25          7       16
210 45   35  18    105  28  462  81     21  20  154  125     30 49   11  32
etc.
Where will 38 appear in this tree? It is a reasonable assumption that by iterating A087207 starting from 38, as A087207(38) = 129, A087207(129) = 8194, A087207(8194) = 1501199875790187, ..., we will eventually hit a prime A000040(k), most likely with a largish index k. This prime occurs at the penultimate edge at right, as a(A000918(k)) = a((2^k)-2), and thus 38 occurs somewhere below it as a(m) = 38, m > k. All the numbers that share prime factors with 38, namely 76, 152, 304, 608, 722, ..., occur similarly late in this tree, as they form the rightward branch starting from 38. Alternatively, by iterating A285330 (each iteration moves one step towards the root) starting from 38, we might instead first hit some power of 3, or say, one of the terms of A033845 (the rightward branch starting from 6), in which case the first prime encountered would be a(2)=3 and 38 would appear on the left-hand side instead of the right-hand side subtree.
As long as it remains conjecture that A019565 has no cycles, it is certainly also an open question whether this is a permutation of the natural numbers: If A019565 has any cycles, then neither any of the terms in those cycles nor any A065642-trajectories starting from those terms (that is, numbers sharing same prime factors) may occur in this tree.
Sequence exhibits some outrageous swings, for example, a(703) = 224, but a(704) is 1427 decimal digits (4739 binary digits) long, thus it no longer fits into a b-file.
However, the scatter plot of A286543 gives some flavor of the behavior of this sequence even after that point. - Antti Karttunen, Dec 25 2017

Antti Karttunen, Table of n, a(n) for n = 0..703
Michael De Vlieger, Diagram of the binary tree of a(n) showing 1 <= n <= 2^8.
Antti Karttunen and David J. Seal, Discussion on SeqFan mailing list
Index entries for sequences that are permutations of the natural numbers

Inverse: A285331.
Cf. A007947, A019565, A033845, A048675, A065642, A087207, A109162, A285319, A285320, A285328, A285330, A285333, A286542, A286543.
Compare also to permutation A285112 and array A285321.

Block[{a = {1, 2}}, Do[AppendTo[a, If[EvenQ[i], Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[a[[i/2 + 1]], 2], If[# == 1, 1, Function[{n, c}, SelectFirst[Range[n + 1, n^2], Times @@ FactorInteger[#][[All, 1]] == c &]] @@ {#, Times @@ FactorInteger[#][[All, 1]]}] &[a[[(i - 1)/2 + 1]] ] ]], {i, 2, 70}]; a] (* Michael De Vlieger, Mar 12 2021 *)

A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n,n,n = n+r; while(A007947(n) <> r, n = n+r); n); };
A285332(n) = { if(n<=1,n+1,if(!(n%2),A019565(A285332(n/2)),A065642(A285332((n-1)/2)))); };
for(n=0, 4095, write("b285332.txt", n, " ", A285332(n)));

from operator import mul
from sympy import prime, primefactors
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # This function from Chai Wah Wu
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n = n + r
    while a007947(n)!=r:
        n+=r
    return n
def a(n):
    if n<2: return n + 1
    if n%2==0: return a019565(a(n//2))
    else: return a065642(a((n - 1)//2))
print([a(n) for n in range(51)]) # Indranil Ghosh, Apr 18 2017

;; With memoization-macro definec.
(definec (A285332 n) (cond ((<= n 1) (+ n 1)) ((even? n) (A019565 (A285332 (/ n 2)))) (else (A065642 (A285332 (/ (- n 1) 2))))))

a(0) = 1, a(1) = 2, a(2n) = A019565(a(n)), a(2n+1) = A065642(a(n)).
For n >= 0, a(2^n) = A109162(2+n). [The left edge of the tree.]
For n >= 0, a(A000225(n)) = A000079(n). [Powers of 2 occur at the right edge of the tree.]
For n >= 2, a(A000918(n)) = A000040(n). [And the next vertices inwards contain primes.]
For n >= 2, a(A036563(1+n)) = A001248(n). [Whose right children are their squares.]
For n >= 0, a(A055010(n)) = A000244(n). [Powers of 3 are at the rightmost edge of the left subtree.]
For n >= 2, a(A129868(n-1)) = A062457(n).
A048675(a(n)) = A285333(n).
A046523(a(n)) = A286542(n).

A285321 Square array A(1,k) = A019565(k), A(n,k) = A065642(A(n-1,k)), read by descending antidiagonals.

Original entry on oeis.org

2, 3, 4, 6, 9, 8, 5, 12, 27, 16, 10, 25, 18, 81, 32, 15, 20, 125, 24, 243, 64, 30, 45, 40, 625, 36, 729, 128, 7, 60, 75, 50, 3125, 48, 2187, 256, 14, 49, 90, 135, 80, 15625, 54, 6561, 512, 21, 28, 343, 120, 225, 100, 78125, 72, 19683, 1024

Offset: 1

Antti Karttunen, Apr 17 2017

A permutation of the natural numbers > 1.

Otherwise like array A284311, but the columns come in different order.

			The top left 12x6 corner of the array:
   2,   3,  6,     5,  10,  15,  30,      7,  14,  21,  42,   35
   4,   9, 12,    25,  20,  45,  60,     49,  28,  63,  84,  175
   8,  27, 18,   125,  40,  75,  90,    343,  56, 147, 126,  245
  16,  81, 24,   625,  50, 135, 120,   2401,  98, 189, 168,  875
  32, 243, 36,  3125,  80, 225, 150,  16807, 112, 441, 252, 1225
  64, 729, 48, 15625, 100, 375, 180, 117649, 196, 567, 294, 1715

Antti Karttunen, Table of n, a(n) for n = 1..120; the first 15 antidiagonals of array

Transpose: A285322.
Cf. A019565, A065642.
Cf. A008479 (index of the row where n is located), A087207 (of the column).
Cf. arrays A284311, A285325, also A285332.

a065642[n_] := Module[{k}, If[n == 1, Return[1], k = n + 1; While[ EulerPhi[k]/k != EulerPhi[n]/n, k++]]; k];
A[1, k_] := Times @@ Prime[Flatten[Position[#, 1]]]&[Reverse[ IntegerDigits[k, 2]]];
A[n_ /; n > 1, k_] := A[n, k] = a065642[A[n - 1, k]];
Table[A[n - k + 1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 17 2019 *)

from operator import mul
from sympy import prime, primefactors
def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # This function from Chai Wah Wu
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n = n + r
    while a007947(n)!=r:
        n+=r
    return n
def A(n, k): return a019565(k) if n==1 else a065642(A(n - 1, k))
for n in range(1, 11): print([A(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Apr 18 2017

(define (A285321 n) (A285321bi (A002260 n) (A004736 n)))
(define (A285321bi row col) (if (= 1 row) (A019565 col) (A065642 (A285321bi (- row 1) col))))

A(1,k) = A019565(k), A(n,k) = A065642(A(n-1,k)).

For all n >= 2: A(A008479(n), A087207(n)) = n.

A285331 Inverse for A285332: a(1) = 0, a(2) = 1, a(A019565(n)) = 2a(n), a(A065642(n)) = 1 + 2a(n).

Original entry on oeis.org

0, 1, 2, 3, 6, 4, 14, 7, 5, 12, 30, 9, 62, 10, 8, 15, 126, 19, 254, 25, 24, 252, 510, 39, 13, 76, 11, 21, 1022, 28, 2046, 31, 38, 316, 18, 79, 4094

Offset: 1

Antti Karttunen, Apr 17 2017, comments edited Apr 19 2017

Note the indexing: the domain starts from 1, while the range includes also zero.
For the question whether this sequence and A285332 are permutations of natural numbers, see comments in A285332 and the conjecture stated in A019565.
As a practical problem, it seems next-to-impossible to compute even the value of a(38). Even though we know that 38 certainly is not in a finite cycle of A019565, because A048675(38) = 129, A048675(129) = 8194 and A048675(8194) = 4503599627370561 which factorizes as 3^2 * 37 * 71 * 190483425427 (thus is not squarefree and A285320(38) = 3), the value of a(38) is most likely so huge that it will not fit into the data section or even into a b-file. The same problem applies to all numbers that share prime factors with 38, namely 76, 152, 304, 608, 722, ...
Terms a(39) .. a(61) are [632, 51, 8190, 60, 16382, 505, 17, 72057594037927932, 32766, 159, 29, 103, 1016, 153, 65534, 319, 50, 43, 16376, 131014, 131070, 57, 262142].
The name is slightly misleading. The given definition of a(n) is not always very helpful to compute the terms (cf. example of n = 38), it is actually not clear whether the sequence is well defined. - M. F. Hasler, Mar 01 2018

			a(1) = 0 and a(2) = 1 by definition.
a(3) = a(prime(2)) = a(A019565(2^1)) = 2*a(2) = 2.
a(4) = a(2^2) = a(A065642(2)) = 1 + 2*a(2) = 3.
a(5) = a(prime(3)) = a(A019565(2^2)) = 2*a(4) = 6.
a(9) = a(3^2) = a(A065642(3)) = 1 + 2*a(3) = 5.
a(10) = a(2*5) = a(prime(1)*prime(3)) = a(A019565(2^0+2^2)) = 2*a(1+4) = 12.
To compute a(38), write 38 = prime(1)*prime(8) = A019565(2^7+2^0), so a(38) = 2*a(129). To compute this, use 129 = prime(2)*prime(14) = A019565(2^13+2^1), so a(129) = 2*a(8194). But 8194 = prime(1)*prime(7)*prime(53) = A019565(2^0+2^6+2^52), so a(8194) = 2*a(4503599627370561)...

Inverse: A285332.
Cf. A005117, A008683, A019565, A048675, A065642, A087207, A285319, A285320, A285328.
Compare also to permutation A285111.

A285331(n)={ if(n<=2,n-1,if(moebius(n)<>0, 2*A285331(A048675(n)), 1+2*A285331(A285328(n))))} \\ See A048675 & A285328 for respective PARI code. (We avoid content duplication leading to obsolete code.)

;; With memoization-macro definec.
(definec (A285331 n) (cond ((<= n 2) (- n 1)) ((not (zero? (A008683 n))) (* 2 (A285331 (A048675 n)))) (else (+ 1 (* 2 (A285331 (A285328 n)))))))

a(1) = 0, a(2) = 1, and for n > 2, if A008683(n) <> 0 [when n is squarefree], a(n) = 2*a(A048675(n)), otherwise a(n) = 1 + 2*a(A285328(n)).

For all n >= 0, a(A285332(n)) = n.

A284342 Numbers n such that A065642(n) < n*lpf(n), where lpf = least prime factor (A020639).

Original entry on oeis.org

12, 18, 24, 36, 40, 45, 48, 50, 54, 56, 60, 63, 72, 75, 80, 84, 90, 96, 98, 100, 108, 112, 120, 126, 132, 135, 144, 147, 150, 156, 160, 162, 168, 175, 176, 180, 189, 192, 196, 198, 200, 204, 208, 216, 224, 225, 228, 234, 240, 242, 245, 250, 252, 264, 270, 275, 276, 280, 288, 294, 297, 300

Offset: 1

Gionata Neri, Mar 25 2017

nonn

Numbers n for which A065642(n) < A285109(n). Positions of terms > 1 in A285337. - Antti Karttunen, Apr 19 2017

For any n in this sequence, k*n is also in this sequence. No term is squarefree. For any distinct primes p and q with p > q, p^2*q and p*q^(ceiling(log_q(p))) are in this sequence. - Charlie Neder, Oct 29 2018

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A007947, A020639, A065642, A285100 (complement), A285109, A285337.

Select[Range[2, 300], Function[{n, c, lpf}, SelectFirst[Range[n + 1, n^2], Times @@ FactorInteger[#][[All, 1]] == c &] < n lpf] @@ {#1, Times @@ #2, #2[[1]]} & @@ {#, FactorInteger[#][[All, 1]]} &] (* Michael De Vlieger, Oct 31 2018 *)

for(n=1,300,for(k=1,n^2-n,a=factorback(factorint(n)[,1]); b=factorback(factorint(n+k)[,1]); c=vecmin(factor(n)[,1]); if(a==b&&n+k

A020639(n) = if(1==n,n,vecmin(factor(n)[, 1]));
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n,n,n = n+r; while(A007947(n) <> r, n = n+r); n); };
isA284342(n) = (A065642(n) < n*A020639(n));
n=0; k=1; while(k <= 10000, n=n+1; if(isA284342(n),write("b284342.txt", k, " ", n);k=k+1));
\\ Antti Karttunen, Apr 19 2017

from operator import mul
from sympy import primefactors
from functools import reduce
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n = n + r
    while a007947(n)!=r:
        n+=r
    return n
print([n for n in range(10, 301) if a065642(n)Indranil Ghosh, Apr 20 2017

A285111 Permutation of nonnegative integers: a(1) = 0, a(2) = 1, a(A005117(1+n)) = 2a(n), a(A065642(n)) = 1 + 2a(n).

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 7, 5, 12, 16, 13, 14, 10, 24, 15, 32, 27, 26, 25, 28, 20, 48, 55, 9, 30, 11, 21, 64, 54, 52, 31, 50, 56, 40, 111, 96, 110, 18, 51, 60, 22, 42, 41, 49, 128, 108, 223, 17, 103, 104, 61, 62, 447, 100, 43, 112, 80, 222, 109, 192, 220, 57, 63, 36, 102, 120, 113, 44, 84, 82, 895, 98, 256, 99, 221, 216, 446, 34, 207, 23

Offset: 1

Antti Karttunen, Apr 17 2017

nonn

Note the indexing: the domain starts from 1, while the range includes also zero.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers

Inverse: A285112.
Cf. A005117, A008683, A013928, A065642, A285328.
Similar or related permutations: A243343, A243345, A277695, A284571.

from operator import mul
from sympy import primefactors
from sympy.ntheory.factor_ import core
from functools import reduce
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a285328(n):
    if core(n) == n: return 1
    k=n - 1
    while k>0:
        if a007947(k) == a007947(n): return k
        else: k-=1
def a013928(n): return sum([1 for i in range(1, n) if core(i) == i])
def a(n):
    if n<3: return n - 1
    if core(n)==n: return 2*a(a013928(n))
    else: return 1 + 2*a(a285328(n))
print([a(n) for n in range(1, 121)]) # Indranil Ghosh, Apr 17 2017

a(1) = 0, a(2) = 1, and for n > 2, if A008683(n) <> 0 [when n is squarefree], a(n) = 2*a(A013928(n)), otherwise a(n) = 1 + 2*a(A285328(n)).

A285112 Permutation of natural numbers: a(0) = 1, a(1) = 2, a(2n) = A005117(1+a(n)), a(2n+1) = A065642(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 6, 8, 7, 25, 14, 27, 10, 12, 13, 16, 11, 49, 39, 125, 22, 28, 42, 81, 15, 20, 19, 18, 21, 169, 26, 32, 17, 121, 79, 343, 65, 117, 205, 625, 35, 44, 43, 56, 69, 84, 133, 243, 23, 45, 33, 40, 31, 361, 30, 24, 34, 63, 277, 2197, 41, 52, 53, 64, 29, 289, 199, 1331, 130, 6241, 563, 2401, 106, 325, 193, 351, 335, 1025, 1030, 3125, 58

Offset: 0

Antti Karttunen, Apr 17 2017

Note the indexing: the domain starts from 0, while the range excludes zero.
This sequence can be represented as a binary tree. Each left hand child is produced as A005117(1+n), and each right hand child as A065642(n), when the parent node contains n >= 2:
                                    1
                                    |
                 ...................2...................
                3                                       4
      5......../ \........9                   6......../ \........8
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
  7       25         14       27         10       12         13       16
11 49   39  125    22  28   42  81     15  20   19  18     21  169  26  32
etc.

Antti Karttunen, Table of n, a(n) for n = 0..116
Index entries for sequences that are permutations of the natural numbers

Inverse: A285111.
Cf. A005117, A065642.
Similar or related permutations: A243344, A243346, A252753, A277696, A284572.
Cf. also arrays A284457 & A284311.

a(0) = 1, a(1) = 2, a(2n) = A005117(1+a(n)), a(2n+1) = A065642(a(n)).

A284571 Permutation of natural numbers: a(1) = 1, a(A005117(1+n)) = 2a(n), a(A065642(1+n)) = 1 + 2a(n).

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 9, 5, 12, 32, 17, 18, 10, 24, 33, 64, 65, 34, 11, 36, 20, 48, 129, 7, 66, 19, 37, 128, 130, 68, 49, 22, 72, 40, 97, 96, 258, 14, 69, 132, 38, 74, 73, 21, 256, 260, 81, 13, 29, 136, 15, 98, 521, 44, 39, 144, 80, 194, 257, 192, 516, 23, 137, 28, 138, 264, 45, 76, 148, 146, 197, 42, 512, 147, 193, 520, 162, 26, 27

Offset: 1

Antti Karttunen, Apr 17 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers

Inverse: A284572.
Cf. A005117, A008683, A013928, A065642, A285328.
Similar or related permutations: A243343, A243345, A277695, A285111.

from operator import mul
from sympy import primefactors
from sympy.ntheory.factor_ import core
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a285328(n):
    if core(n) == n: return 1
    k=n - 1
    while k>0:
        if a007947(k) == a007947(n): return k
        else: k-=1
def a013928(n): return sum(1 for i in range(1, n) if core(i) == i)
def a(n):
    if n==1: return 1
    if core(n)==n: return 2*a(a013928(n))
    else: return 1 + 2*a(a285328(n) - 1)
[a(n) for n in range(1, 121)] # Indranil Ghosh, Apr 17 2017

a(1) = 1, for n > 1, if A008683(n) <> 0 [when n is squarefree], a(n) = 2*a(A013928(n)), otherwise a(n) = 1 + 2*a(A285328(n)-1).

A284572 Permutation of natural numbers: a(1) = 1, a(2n) = A005117(1+a(n)), a(2n+1) = A065642(1+a(n)).

Original entry on oeis.org

1, 2, 4, 3, 9, 6, 25, 5, 8, 14, 20, 10, 49, 39, 52, 7, 12, 13, 27, 22, 45, 33, 63, 15, 121, 79, 80, 65, 50, 85, 2809, 11, 16, 19, 169, 21, 28, 42, 56, 35, 529, 73, 92, 55, 68, 103, 128, 23, 32, 199, 244, 130, 100, 131, 243, 106, 132, 82, 153, 139, 172, 4619, 5620, 17, 18, 26, 289, 31, 40, 277, 340, 34, 44, 43, 841, 69, 1849, 91, 171, 58, 48

Offset: 1

Antti Karttunen, Apr 17 2017

This sequence can be represented as a binary tree. Each left hand child is produced as A005117(1+n), and each right hand child as A065642(1+n), when the parent node contains n:
                                    1
                 ................../ \..................
                2                                       4
      3......../ \........9                   6......../ \........25
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
  5       8          14       20         10       49         39       52
7  12   13 27      22  45   33  63     15  121  79  80     65  50   85 2809
etc.
Compare to A285112.

Antti Karttunen, Table of n, a(n) for n = 1..166
Index entries for sequences that are permutations of the natural numbers

Inverse: A284571.
Cf. A005117, A065642.
Similar or related permutations: A243344, A243346, A277696, A285112.

a(1) = 1, a(2n) = A005117(1+a(n)), a(2n+1) = A065642(1+a(n)).

A285100 Numbers k for which A065642(k) = A285109(k).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 41, 42, 43, 44, 46, 47, 49, 51, 52, 53, 55, 57, 58, 59, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 76, 77, 78, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 97, 99, 101

Offset: 1

Antti Karttunen, Apr 19 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Complement: A284342.
Positions of ones in A285337.
Subsequences: A000961, A005117.
Cf. A065642, A285109.

A020639(n) = if(1==n,n,vecmin(factor(n)[, 1]));
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n,n,n = n+r; while(A007947(n) <> r, n = n+r); n); };
isA285100(n) = (A065642(n) == n*A020639(n));
n=0; k=1; while(k <= 10000, n=n+1; if(isA285100(n),write("b285100.txt", k, " ", n);k=k+1));

from operator import mul
from sympy import primefactors
from functools import reduce
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a020639(n): return 1 if n==1 else primefactors(n)[0]
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n += r
    while a007947(n)!=r:
        n+=r
    return n
print([n for n in range(1, 102) if a065642(n) == n*a020639(n)]) # Indranil Ghosh, May 24 2017

;; With Antti Karttunen's IntSeq-library.
(define A285100 (MATCHING-POS 1 1 (lambda (n) (= (A065642 n) (A285109 n)))))

A285327 Row 3 of A285325: a(n) = A048675(A065642(A065642(A019565(n)))).

Original entry on oeis.org

0, 3, 6, 5, 12, 7, 10, 9, 24, 11, 18, 13, 20, 15, 18, 17, 48, 19, 22, 21, 36, 23, 26, 25, 40, 27, 34, 29, 36, 31, 34, 33, 96, 35, 38, 37, 68, 39, 42, 41, 72, 43, 50, 45, 52, 47, 50, 49, 80, 51, 54, 53, 68, 55, 58, 57, 72, 59, 66, 61, 68, 63, 66, 65, 192, 67, 70, 69, 132, 71, 74, 73, 136, 75, 82, 77, 84, 79, 82, 81, 144, 83, 86, 85, 100, 87, 90, 89

Offset: 0

Antti Karttunen, Apr 19 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 0..8191

Row 3 of A285325 (after the initial zero).

Cf. A007947, A019565, A048675, A065642, A285323, A285324.

A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
A048675(n) = my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; \\ Michel Marcus, Oct 10 2016
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n,n,n = n+r; while(A007947(n) <> r, n = n+r); n); };
A285327(n) = A048675(A065642(A065642(A019565(n))));

(define (A285327 n) (A048675 (A065642 (A065642 (A019565 n)))))

a(n) = A048675(A065642(A065642(A019565(n)))).

cp's OEIS Frontend

A285332 a(0) = 1, a(1) = 2, a(2n) = A019565(a(n)), a(2n+1) = A065642(a(n)).

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A285321 Square array A(1,k) = A019565(k), A(n,k) = A065642(A(n-1,k)), read by descending antidiagonals.

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A285331 Inverse for A285332: a(1) = 0, a(2) = 1, a(A019565(n)) = 2*a(n), a(A065642(n)) = 1 + 2*a(n).

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A284342 Numbers n such that A065642(n) < n*lpf(n), where lpf = least prime factor (A020639).

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A285111 Permutation of nonnegative integers: a(1) = 0, a(2) = 1, a(A005117(1+n)) = 2*a(n), a(A065642(n)) = 1 + 2*a(n).

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A285112 Permutation of natural numbers: a(0) = 1, a(1) = 2, a(2n) = A005117(1+a(n)), a(2n+1) = A065642(a(n)).

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A284571 Permutation of natural numbers: a(1) = 1, a(A005117(1+n)) = 2*a(n), a(A065642(1+n)) = 1 + 2*a(n).

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A284572 Permutation of natural numbers: a(1) = 1, a(2n) = A005117(1+a(n)), a(2n+1) = A065642(1+a(n)).

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A285331 Inverse for A285332: a(1) = 0, a(2) = 1, a(A019565(n)) = 2a(n), a(A065642(n)) = 1 + 2a(n).

A285111 Permutation of nonnegative integers: a(1) = 0, a(2) = 1, a(A005117(1+n)) = 2a(n), a(A065642(n)) = 1 + 2a(n).

A284571 Permutation of natural numbers: a(1) = 1, a(A005117(1+n)) = 2a(n), a(A065642(1+n)) = 1 + 2a(n).