A087207 -id:A087207 - cp's OEIS frontend

A285329 a(n) = A013928(A007947(n)).

0, 1, 2, 1, 3, 4, 5, 1, 2, 6, 7, 4, 8, 9, 10, 1, 11, 4, 12, 6, 13, 14, 15, 4, 3, 16, 2, 9, 17, 18, 19, 1, 20, 21, 22, 4, 23, 24, 25, 6, 26, 27, 28, 14, 10, 29, 30, 4, 5, 6, 31, 16, 32, 4, 33, 9, 34, 35, 36, 18, 37, 38, 13, 1, 39, 40, 41, 21, 42, 43, 44, 4, 45, 46, 10, 24, 47, 48, 49, 6, 2, 50, 51, 27, 52, 53, 54, 14, 55, 18, 56, 29, 57, 58, 59, 4, 60, 9, 20, 6

Offset: 1

Antti Karttunen, Apr 17 2017

nonn

For n > 1, a(n) gives the (one-based) index of the column where n is located in array A284311, or respectively, index of the row where n is in A284457. A008479 gives the other index.

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

Cf. A008479 (the other index).
Cf. A005117, A007947, A008683, A013928, A019565, A064273, A087207, A285328.
Cf. array A284311 (A284457).

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 a013928(n): return sum(1 for i in range(1, n) if core(i) == i)
print([a013928(a007947(n)) for n in range(1, 101)]) # Indranil Ghosh, Apr 18 2017

from math import prod, isqrt
from sympy import primefactors, mobius
def A285329(n):
    m=prod(primefactors(n))-1
    return sum(mobius(k)*(m//k**2) for k in range(1,isqrt(m)+1)) # Chai Wah Wu, May 12 2024

a(n) = A013928(A007947(n)).
Other identities. For all n >= 0:
If A008683(n) <> 0 [when n is squarefree, A005117], a(n) = A013928(n), otherwise a(n) = a(A285328(n)).
a(A019565(n)) = A064273(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.

A372888 Sum of binary ranks of all strict integer partitions of n, where the binary rank of a partition y is given by Sum_i 2^(y_i-1).

Original entry on oeis.org

0, 1, 2, 7, 13, 31, 66, 138, 279, 581, 1173, 2375, 4783, 9630, 19316, 38802, 77689, 155673, 311639, 623845, 1248179, 2497719, 4996387, 9995304, 19992908, 39990902, 79986136, 159983241, 319975073, 639971495, 1279962115, 2559966847, 5119970499, 10240030209

Offset: 0

Gus Wiseman, May 23 2024

nonn

			The strict partitions of 6 are (6), (5,1), (4,2), (3,2,1), with respective binary ranks 32, 17, 10, 7 with sum 66, so a(6) = 66.

Alois P. Heinz, Table of n, a(n) for n = 0..3321

Row sums of A118462 (binary ranks of strict partitions).
For Heinz number the non-strict version is A145519, row sums of A215366.
For Heinz number (not binary rank) we have A147655, row sums of A246867.
The non-strict version is A372890.
A000009 counts strict partitions, ranks A005117.
A048675 gives binary rank of prime indices, distinct A087207.
A277905 groups all positive integers by binary rank of prime indices.
Binary indices (A048793):
- length A000120, complement A023416
- min A001511, opposite A000012
- max A029837 or A070939, opposite A070940
- sum A029931, product A096111
- reverse A272020
- complement A368494, sum A359400
- opposite A371572, sum A230877
- opposite complement A371571, sum A359359
Cf. A000041, A005940, A015716, A018819, A019565, A118457, A225620, A344086.

b:= proc(n, i) option remember; `if`(i*(i+1)/2 [0, p[1]*2^(i-1)]
          +p)(b(n-i, min(n-i, i-1)))))
    end:
a:= n-> b(n$2)[2]:
seq(a(n), n=0..33);  # Alois P. Heinz, May 23 2024

Table[Total[Total[2^(#-1)]& /@ Select[IntegerPartitions[n],UnsameQ@@#&]],{n,0,10}]

a(n) = Sum_{k=1..n} 2^(k-1) * A015716(n,k). - Alois P. Heinz, May 24 2024

A293447 Fully additive with a(p^e) = e * A000225(PrimePi(p)), where PrimePi(n) = A000720(n) and A000225(n) = (2^n)-1.

Original entry on oeis.org

0, 1, 3, 2, 7, 4, 15, 3, 6, 8, 31, 5, 63, 16, 10, 4, 127, 7, 255, 9, 18, 32, 511, 6, 14, 64, 9, 17, 1023, 11, 2047, 5, 34, 128, 22, 8, 4095, 256, 66, 10, 8191, 19, 16383, 33, 13, 512, 32767, 7, 30, 15, 130, 65, 65535, 10, 38, 18, 258, 1024, 131071, 12, 262143, 2048, 21, 6, 70, 35, 524287, 129, 514, 23, 1048575, 9, 2097151, 4096, 17, 257, 46, 67, 4194303, 11, 12

Offset: 1

Antti Karttunen, Nov 09 2017

nonn

Original, equal definition: totally additive with a(p^e) = e * A005187(2^(PrimePi(p)-1)), where PrimePi(n) = A000720(n).

Antti Karttunen, Table of n, a(n) for n = 1..8192
Index entries for sequences computed from indices in prime factorization

Cf. A000225, A000720, A003557, A005187, A087207.

Cf. also A046645, A048675, A293442, A331591, A331740.

A005187(n) = { my(s=n); while(n>>=1, s+=n); s; }; \\ This function from Charles R Greathouse IV
A293447(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2] * A005187(2^(primepi(f[k, 1])-1))); }

(define (A293447 n) (cond ((= 1 n) 0) (else (+ (A005187 (A087207 n)) (A293447 (A003557 n))))))
;; Alternatively:
(define (A293447 n) (if (= 1 n) 0 (+ (A005187 (A000079 (+ -1 (A061395 n)))) (A293447 (/ n (A006530 n))))))

Totally additive with a(p^e) = e * A005187(2^(PrimePi(p)-1)), where PrimePi(n) = A000720(n).
a(1) = 0, and for n > 1, a(n) = A005187(A087207(n)) + a(A003557(n)).
Other identities:
For all n >= 1, a(A293442(n)) = A046645(n).
For all n >= 2 and all k >= 0, a(n^k) = k*a(n).
For all n >= 1, a(n) >= A048675(n) >= A331740(n) >= A331591(n).

Definition simplified by Antti Karttunen, Feb 05 2020

A218403 Bitwise OR of all proper divisors of n; a(1) = 0 by convention.

Original entry on oeis.org

0, 1, 1, 3, 1, 3, 1, 7, 3, 7, 1, 7, 1, 7, 7, 15, 1, 15, 1, 15, 7, 11, 1, 15, 5, 15, 11, 15, 1, 15, 1, 31, 11, 19, 7, 31, 1, 19, 15, 31, 1, 31, 1, 31, 15, 23, 1, 31, 7, 31, 19, 31, 1, 31, 15, 31, 19, 31, 1, 31, 1, 31, 31, 63, 13, 63, 1, 55, 23, 47, 1, 63, 1

Offset: 1

Reinhard Zumkeller, Oct 28 2012

			n=20: properDivisors(20) = {1, 2, 4, 5, 10}, 0001 OR 0010 OR 0100 OR 0101 OR 1010 = 1111 -> a(20) = 15;
n=21: properDivisors(21) = {1, 3, 7}, 001 OR 011 OR 111 = 111 -> a(21) = 7;
n=22: properDivisors(22) = {1, 2, 11}, 0001 OR 0010 OR 1011 = 1111 -> a(22) = 11;
n=23: properDivisors(23) = {1} -> a(23) = 23;
n=24: properDivisors(24) = {1, 2, 3, 4, 6, 8, 12}, 0001 OR 0010 OR 0011 OR 0100 OR 0110 OR 1000 OR 1100 = 1111 -> a(24) = 15;
n=25: properDivisors(25) = {1, 5}, 001 OR 101 = 101 -> a(25) = 5.

Reinhard Zumkeller, Table of n, a(n) for n = 1..8191
Eric Weisstein's World of Mathematics, OR
Wikipedia, Bitwise operation OR
Index entries for sequences related to binary expansion of n

Cf. A001065, A027751, A000225 (subsequence), A123345, A227320, A293214, A293215.

import Data.Bits ((.|.))
a218403 = foldl (.|.)  0 . a027751_row :: Integer -> Integer

Table[BitOr@@Most[Divisors[n]],{n,80}] (* Harvey P. Dale, Nov 09 2012 *)

A218403(n) = { my(s=0); fordiv(n,d,if(dAntti Karttunen, Oct 08 2017

a(n) <= A218388(n).
a(A000040(n)) = 1.
From Antti Karttunen, Oct 08 2017: (Start)
a(n) = A087207(A293214(n)).
A227320(n) <= a(n) <= A001065(n).
(End)

A273258 Write the distinct prime divisors p of n in the (PrimePi(p) - 1)-th place, ignoring multiplicity. Decode the resulting number after first reversing the code, ignoring any leading zeros.

Original entry on oeis.org

1, 2, 2, 2, 2, 6, 2, 2, 2, 10, 2, 6, 2, 14, 6, 2, 2, 6, 2, 10, 10, 22, 2, 6, 2, 26, 2, 14, 2, 30, 2, 2, 14, 34, 6, 6, 2, 38, 22, 10, 2, 70, 2, 22, 6, 46, 2, 6, 2, 10, 26, 26, 2, 6, 10, 14, 34, 58, 2, 30, 2, 62, 10, 2, 14, 154, 2, 34, 38, 42, 2, 6, 2, 74, 6, 38, 6, 286, 2, 10, 2, 82, 2, 70, 22, 86

Offset: 1

Michael De Vlieger, Aug 28 2016

Encode n with the function f(n) = noting the distinct prime divisors p of n by writing "1" in the (PrimePi(n) - 1)-th place, e.g, f(6) = f(12) = "11". This function is akin to A054841(n) except we don't note the multiplicity e of p in n, rather merely note "1" if e > 0.
This sequence decodes f(n) by reversing the digits.
If we decode f(n) without reversal, we have A007947(n), since f(n) sets any multiplicity e > 1 of prime divisor p of n to 1.
All terms except a(1) are of the form 2x with x odd. a(1) = 1, since f(1) = "0" and stands unaffected in reversal and decoding, and any zeros to the right of all 1's are lost in reversal. Thus f(15) = "110" reversed becomes "011" -> "11" decoded equals 2 * 3 = 6. Because we lose leading zeros, we always have 1 in position 1, which decoded is interpreted as the factor 2.
a(p) for p prime = 2, since primes are written via f(p) as 1 in the (PrimePi(p)-1)-th place. There is only one 1 in this number (similar to a perfect power of ten decimally) and when it is reversed, the number loses all leading zeros to become "1" -> 2. This also applies to prime powers p^e, since e is rendered as 1 by f(p^e), i.e., f(p^e) = f(p).

			a(3) = 2 since f(3) = "10" reversed becomes "01", loses leading zeros to become "1" -> 2.
a(6) = a(12) = "11" reversed stays the same -> 2 * 3 = 6.
a(15) = "110" reversed becomes "011", loses leading zeros to become "11" -> 6.
a(42) = "1101" reversed becomes "1011" -> 70 (a(70) = 42).

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from indices in prime factorization

Cf. A007947, A019565, A030101, A054841 (analogous encoding algorithm), A069799, A087207, A137502, A276379, A293448 (a bijective variant of this sequence).

Table[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ If[# == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> 1 &, f]]@ FactorInteger@ #] &@ n, {n, 86}]

(define (A273258 n) (A019565 (A030101 (A087207 n)))) ;; Antti Karttunen, Jun 18 2017

a(n) = A019565(A030101(A087207(n))). - Antti Karttunen, Jun 18 2017

For all n, a(A039956(n)) = A293448(A039956(n)). - Antti Karttunen, Nov 21 2017

A276379 Write a "1" for each distinct prime divisor p of n in the (pi(p) - 1)-th place, ignoring multiplicity.

Original entry on oeis.org

0, 1, 10, 1, 100, 11, 1000, 1, 10, 101, 10000, 11, 100000, 1001, 110, 1, 1000000, 11, 10000000, 101, 1010, 10001, 100000000, 11, 100, 100001, 10, 1001, 1000000000, 111, 10000000000, 1, 10010, 1000001, 1100, 11, 100000000000, 10000001, 100010, 101, 1000000000000, 1011, 10000000000000, 10001, 110

Offset: 1

Michael De Vlieger, Sep 02 2016

a(n) notes the distinct prime divisors p of n by writing "1" in the (pi(n)-1)-th place. Zeros hold the places of primes q less than the greatest prime divisor p that do not divide n. Thus a(n) consists of 1's and 0's like a binary number where each bit value, instead of representing 2^k, represents prime(k + 1).
a(n) = A054841(n) with all nonzero digits converted to 1's.
a(n) = a(A007947(n)), that is, a number n shares a value of a(n) with the largest squarefree divisor A007947(n). Thus a(18) = a(6) = 11.
a(p) = 1 in the leftmost place followed by (pi(p)-1) zeros.
This function is akin to A054841(n) except we don't note the multiplicity e of p in n, rather merely note "1" if e > 0.
Unlike A054841(1024) = 10, there are no overflows in a(n) into the next place that encodes prime(p+1) due to "carry". 1024 = 2^10, thus a(1024) = a(2^e) = 1, with e >= 1 = 1.

			a(1) = 0 since 1 is the empty product. a(0) is undefined.
a(6) = a(12) = 11, since 6 and 12 are products of the 1st and 2nd primes (i.e., 2 and 3). Thus we write 1's in the corresponding places. Any number n that is the product only of powers e >= 1 of 2 and 3 (e.g., 24, 96, 144, etc.) has a(n) = 11.
a(42) = 1011, since the prime divisors of 42 are 2, 3 and 7. Any number n that is the product only of powers e >= 1 of all of 2, 3 and 7 has a(n) = 1011.
a(70) = 1101, since its prime divisors are 2, 5 and 7.

Cf. A027748, A054841 (write multiplicity instead of 1 in the (pi(p)-1)th place), A079067 (reverse 0's and 1's in a(n) and convert to decimal), A087207 (a(n) interpreted as a binary number), A273258 (a(n) reversed and converted to decimal).

Sequence A087207 shown in base-2.

a:= n-> add(10^numtheory[pi](i[1]), i=ifactors(n)[2])/10:
seq(a(n), n=1..53);  # Alois P. Heinz, Feb 10 2020

f[n_] := If[n == 1, {0}, Function[k, ReplacePart[Table[0, {PrimePi[k[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> 1 &, k]]@ FactorInteger@ n]; Table[FromDigits@ Reverse@ f@ n, {n, 45}] (* or *)
FromDigits[IntegerDigits[#, 2]] & /@ Table[Floor@ Total[2^(PrimePi /@ FactorInteger[n][[All, 1]] - 1)], {n, 45}] (* latter program after Jean-François Alcover at A087207 *)

a(n) = A054841(A007947(n)) = A007088(A087207(n)). - Antti Karttunen, Jun 18 2017

G.f.: Sum_{k>=1} 10^(k-1) * x^prime(k) / (1 - x^prime(k)). - Ilya Gutkovskiy, Feb 10 2020

A297404 A binary representation of the positive exponents that appear in the prime factorization of a number, shown in decimal.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 4, 2, 1, 1, 3, 1, 1, 1, 8, 1, 3, 1, 3, 1, 1, 1, 5, 2, 1, 4, 3, 1, 1, 1, 16, 1, 1, 1, 2, 1, 1, 1, 5, 1, 1, 1, 3, 3, 1, 1, 9, 2, 3, 1, 3, 1, 5, 1, 5, 1, 1, 1, 3, 1, 1, 3, 32, 1, 1, 1, 3, 1, 1, 1, 6, 1, 1, 3, 3, 1, 1, 1, 9, 8, 1, 1, 3, 1, 1

Offset: 1

Rémy Sigrist, Dec 29 2017

This sequence is similar to A087207; here we encode the exponents, there the prime numbers appearing in the prime factorization of a number.
The binary representation of a(n) shows which exponents appear in the prime factorization of n, but without multiplicities:
- for any prime number p and k > 0, if p^k divides n but p^(k+1) does not divide n, then a(n) AND 2^(k-1) = 2^(k-1) (where AND denotes the bitwise AND operator),
- conversely, if a(n) AND 2^(k-1) = 2^(k-1) for some k > 0, then there is prime number p such that p^k divides n but p^(k+1) does not divide n.

			For n = 90:
- 90 = 5^1 * 3^2 * 2^1,
- the exponents appearing in the prime factorization of 90 are 1 and 2,
- hence a(90) = 2^(1-1) + 2^(2-1) = 3.

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A000120, A000695, A001694, A004709, A005117, A038109, A052485, A059404, A087207, A328400.

Array[Total@ Map[2^(# - 1) &, Union[FactorInteger[#][[All, -1]] ]] - Boole[# == 1] &, 86] (* Michael De Vlieger, Dec 29 2017 *)

a(n) = my (x=Set(factor(n)[,2]~)); sum(i=1, #x, 2^(x[i]))/2

a(p^k) = 2^(k-1) for any prime number p and k > 0.
a(n^2) = A000695(2 * a(n)) / 2 for any n > 0.
a(n) <= 1 iff n is squarefree (A005117).
a(n) <= 3 iff n is cubefree (A004709).
a(n) is odd iff n belongs to A052485 (weak numbers).
a(n) is even iff n belongs to A001694 (powerful numbers).
a(n) AND 2 = 2 iff n belongs to A038109 (where AND denotes the bitwise AND operator).
A000120(a(n)) <= 1 iff n belongs to A072774 (powers of squarefree numbers).
A000120(a(n)) > 1 iff n belongs to A059404.
If gcd(m, n) = 1, then a(m * n) = a(m) OR a(n) (where OR denotes the bitwise OR operator).
a(n) = a(A328400(n)). - Peter Munn, Oct 02 2023

A332208 Numbers k such that the squarefree kernel of sigma(k) is equal to the squarefree kernel of 2*k.

Original entry on oeis.org

6, 28, 120, 135, 270, 496, 672, 891, 1080, 1638, 1782, 3780, 8128, 18600, 20580, 24948, 26208, 30240, 32640, 32760, 35640, 41850, 44226, 55860, 66960, 164640, 167400, 185220, 199584, 200655, 273000, 293760, 307125, 401310, 441936, 446880, 502740, 523776, 544635, 614250, 707616, 802620, 819000, 884520

Offset: 1

Antti Karttunen, Feb 07 2020

nonn

Numbers k such that sigma(k) has the same set of distinct prime factors as 2*k.
Numbers k such that A007947(sigma(k)) is equal to A007947(2*k), or equally, that A087207(sigma(k)) is equal to A087207(2*k).
Of the first 256 terms 44 are odd, and none occurs in A228058. Compare also to A331752.

Antti Karttunen, Table of n, a(n) for n = 1..256
Index entries for sequences where any odd perfect numbers must occur

Cf. A000203, A007947, A080398, A087207, A228058, A331751, A331752.

Subsequences: A000396, A027687.

Select[Range[10^6], SameQ @@ Map[Times @@ FactorInteger[#][[All, 1]] &, {DivisorSigma[1, #], 2 #}] &] (* Michael De Vlieger, Feb 08 2020 *)

A007947(n) = factorback(factorint(n)[, 1]);
isA332208(n) = (A007947(sigma(n)) == A007947(2*n));

{n: A080398(n) == A007947(2n)}.

A351558 a(n) = A048675(gcd(n, A019565(n))).

Original entry on oeis.org

0, 0, 0, 2, 0, 4, 2, 0, 0, 0, 0, 0, 0, 0, 8, 6, 0, 0, 2, 0, 4, 0, 16, 0, 0, 0, 0, 2, 8, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 34, 0, 0, 10, 0, 0, 4, 0, 0, 0, 0, 0, 2, 32, 0, 2, 20, 8, 0, 0, 0, 4, 0, 0, 10, 0, 0, 2, 0, 64, 0, 4, 0, 0, 0, 0, 2, 0, 8, 2, 0, 0, 0, 0, 0, 0, 68, 0, 2, 16, 0, 2, 8, 0, 0, 0, 4, 0, 0, 0, 2, 4, 0, 66

Offset: 0

Antti Karttunen, Feb 19 2022

Cf. A004198, A007947, A019565, A087207, A351556.

Cf. also A324198, A351559.

Table[If[# == 1, 0, Total[#2*2^PrimePi[#1] & @@@ FactorInteger[#]]/2] &@ GCD[n, Times @@ Prime@ Flatten@ Position[Reverse@ IntegerDigits[n, 2], 1]], {n, 102}] (* Michael De Vlieger, Feb 20 2022 *)

A019565(n) = { my(m=1, p=1); while(n>0, p = nextprime(1+p); if(n%2, m *= p); n >>= 1); (m); };
A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
A351558(n) = A048675(gcd(n, A019565(n)));

a(n) = A048675(A351556(n)) = A048675(gcd(n, A019565(n))).

a(n) = n AND A087207(n), where AND is bitwise-and, A004198.

cp's OEIS Frontend

A285329 a(n) = A013928(A007947(n)).

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Python

Python

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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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PARI

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A372888 Sum of binary ranks of all strict integer partitions of n, where the binary rank of a partition y is given by Sum_i 2^(y_i-1).

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Maple

Mathematica

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A293447 Fully additive with a(p^e) = e * A000225(PrimePi(p)), where PrimePi(n) = A000720(n) and A000225(n) = (2^n)-1.

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PARI

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Extensions

A218403 Bitwise OR of all proper divisors of n; a(1) = 0 by convention.

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Haskell

Mathematica

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A273258 Write the distinct prime divisors p of n in the (PrimePi(p) - 1)-th place, ignoring multiplicity. Decode the resulting number after first reversing the code, ignoring any leading zeros.

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A276379 Write a "1" for each distinct prime divisor p of n in the (pi(p) - 1)-th place, ignoring multiplicity.

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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).