A267113 -id:A267113 - cp's OEIS frontend

A267116 Bitwise-OR of the exponents of primes in the prime factorization of n.

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 3, 1, 1, 1, 4, 1, 3, 1, 3, 1, 1, 1, 3, 2, 1, 3, 3, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 3, 3, 1, 1, 5, 2, 3, 1, 3, 1, 3, 1, 3, 1, 1, 1, 3, 1, 1, 3, 6, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 5, 4, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 5, 1, 3, 3, 2, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 1, 3, 3, 1, 1, 3

Offset: 1

Antti Karttunen, Feb 03 2016

nonn

			For n = 4 = 2^2, bitwise-OR of 2 alone is 2, thus a(4) = 2.
For n = 6 = 2^1 * 3^1, when we take a bitwise-or of 1 and 1, we get 1, thus a(6) = 1.
For n = 24 = 2^3 * 3^1, bitwise-or of 3 and 1 ("11" and "01" in binary) gives "11", thus a(24) = 3.
For n = 210 = 2^1 * 3^1 * 5^1 * 7^1, bitwise-or of 1, 1, 1 and 1 gives 1, thus a(210) = 1.
For n = 720 = 2^4 * 3^2 * 5^1, bitwise-or of 4, 2 and 1 ("100", "10" and "1" in binary) gives 7 ("111" in binary), thus a(720) = 7.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from exponents in factorization of n

Cf. A001222, A003986, A007814, A028234, A067029.
Cf. A000290 (indices of even numbers).
Cf. A000037 (indices of odd numbers).
Nonunit terms of A005117, A062503, A113849 give the positions of ones, twos, fours respectively in this sequence.
Sequences with similar definitions: A260728, A267113, A267115 (bitwise-AND) and A268387 (bitwise-XOR of exponents).
Sequences with related analysis: A267114, A268374, A268375, A268376.
Sequences A088529, A136565 and A181591 coincide with a(n) for n: 2 <= n < 24.
A003961, A059896 are used to express relationship between terms of this sequence.
Related to A087207 via A225546.

read("transforms"):
A267116 := proc(n)
    local a,e ;
    a := 0 ;
    for e in ifactors(n)[2] do
        a := ORnos(a,op(2,e)) ;
    end do:
    a ;
end proc: # R. J. Mathar, Feb 16 2021

{0}~Join~Rest@ Array[BitOr @@ Map[Last, FactorInteger@ #] &, 120] (* Michael De Vlieger, Feb 04 2016 *)

a(n)=my(f = factor(n)); my(b = 0); for (k=1, #f~, b = bitor(b, f[k,2]);); b; \\ Michel Marcus, Feb 05 2016

a(n)=if(n>1, fold(bitor, factor(n)[,2]), 0) \\ Charles R Greathouse IV, Aug 04 2016

from functools import reduce
from operator import or_
from sympy import factorint
def A267116(n): return reduce(or_,factorint(n).values(),0) # Chai Wah Wu, Aug 31 2022

a(1) = 0; for n > 1: a(n) = A067029(n) OR a(A028234(n)). [Here OR stands for bitwise-or, A003986.]
Other identities and observations. For all n >= 1:
a(n) = A007814(n) OR A260728(n) OR A267113(n).
a(n) = A001222(n) - A268374(n).
A268387(n) <= a(n) <= A001222(n).
From Peter Munn, Jan 08 2020: (Start)
a(A059896(n,k)) = a(n) OR a(k).
a(A003961(n)) = a(n).
a(n^2) = 2*a(n).
a(n) = A087207(A225546(n)).
a(A225546(n)) = A087207(n).
(End)

A260728 Bitwise-OR of the exponents of all 4k+3 primes in the prime factorization of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 1, 1, 0, 0, 2, 1, 0, 1, 1, 1, 1, 0, 0, 3, 1, 0, 1, 1, 0, 1, 0, 1, 2, 0, 1, 1, 0, 0, 1, 1, 1, 2, 1, 1, 1, 2, 0, 1, 0, 0, 3, 1, 1, 1, 0, 1, 1, 0, 1, 3, 0, 0, 1, 1, 0, 1, 1, 1, 2, 0, 0, 1, 1, 1, 1, 1, 0, 4, 0, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 1, 0, 2, 3, 0, 0, 1, 1, 0, 1, 0, 1, 3, 0, 1, 1, 1, 0, 1, 1, 0, 2, 1, 1, 1

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

A001481 (numbers that are the sum of 2 squares) gives the positions of even terms in this sequence, while its complement A022544 (numbers that are not the sum of 2 squares) gives the positions of odd terms.

If instead of bitwise-oring (A003986) we added in ordinary way the exponents of 4k+3 primes together, we would get the sequence A065339. For the positions where these two sequences differ see A260730.

			For n = 21 = 3^1 * 7^1 we compute A003986(1,1) = 1, thus a(21) = 1.
For n = 63 = 3^2 * 7^1 we compute A003986(2,1) = A003986(1,2) = 3, thus a(63) = 3.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A000035, A001481, A022544, A003986, A020639, A028234, A032742, A067029, A097706, A229062, A260730.
Cf. also A267113, A267116, A267099.
Differs from A065339 for the first time at n=21, where a(21) = 1, while A065339(21)=2.

Table[BitOr @@ (Map[Last, FactorInteger@ n /. {p_, } /; MemberQ[{0, 1, 2}, Mod[p, 4]] -> Nothing]), {n, 0, 120}] (* _Michael De Vlieger, Feb 07 2016 *)

from functools import reduce
from operator import or_
from sympy import factorint
def A260728(n): return reduce(or_,(e for p, e in factorint(n).items() if p & 3 == 3),0) # Chai Wah Wu, Jun 28 2022

(define (A260728 n) (cond ((< n 3) 0) ((even? n) (A260728 (/ n 2))) ((= 1 (modulo (A020639 n) 4)) (A260728 (A032742 n))) (else (A003986bi (A067029 n) (A260728 (A028234 n)))))) ;; A003986bi implements bitwise-or (see A003986).

If n < 3, a(n) = 0; thereafter, for any even n: a(n) = a(n/2), for any n with its smallest prime factor (A020639) of the form 4k+1: a(n) = a(A032742(n)), otherwise [when A020639(n) is of the form 4k+3] a(n) = A003986(A067029(n),a(A028234(n))).
Other identities. For all n >= 0:
A229062(n) = 1 - A000035(a(n)). [Reduced modulo 2 and complemented, the sequence gives the characteristic function of A001481.]
a(n) = a(A097706(n)). [The result depends only on the prime factors of the form 4k+3.]
a(n) = A267116(A097706(n)).
a(n) = A267113(A267099(n)).

A286361 Least number with the same prime signature as {the largest divisor of n with only prime factors of the form 4k+1} has: a(n) = A046523(A170818(n)).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 4, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 4, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 6, 1, 1, 2, 1, 2, 1, 1, 2, 2, 4, 1, 1, 2, 1, 2, 1, 2, 1, 1, 6, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 4, 2, 2, 1, 2, 2, 2, 1, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2

Offset: 1

Antti Karttunen, May 08 2017

nonn

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

Cf. A046523, A170818, A267099, A267113, A286363, A286364, A286365.

Differs from A063014 for the first time at n=25, where a(25) = 4, while A063014(25) = 3.

from sympy import factorint
from operator import mul
def P(n):
    f = factorint(n)
    return sorted([f[i] for i in f])
def a046523(n):
    x=1
    while True:
        if P(n) == P(x): return x
        else: x+=1
def a072438(n):
    f = factorint(n)
    return 1 if n == 1 else reduce(mul, [1 if i%4==1 else i**f[i] for i in f])
def a(n): return a046523(n/a072438(n)) # Indranil Ghosh, May 09 2017

(define (A286361 n) (A046523 (A170818 n)))

a(n) = A046523(A170818(n)).

a(n) = A286363(A267099(n)).

A268377 Numbers n such that any prime factor of the form 4k+1 has even multiplicity.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 14, 16, 18, 19, 21, 22, 23, 24, 25, 27, 28, 31, 32, 33, 36, 38, 42, 43, 44, 46, 47, 48, 49, 50, 54, 56, 57, 59, 62, 63, 64, 66, 67, 69, 71, 72, 75, 76, 77, 79, 81, 83, 84, 86, 88, 92, 93, 94, 96, 98, 99, 100, 103, 107, 108, 112, 114, 118, 121, 124, 126, 127, 128, 129, 131, 132, 133

Offset: 1

Antti Karttunen, Feb 03 2016

nonn

Closed under multiplication.

			Neither 5 or 10 (= 2*5) are included, because the prime factor 5 (of the form 4k+1) occurs just once.
6 = 2*3 is present, as there are no prime factors of 4k+1 present at all, and zero is an even number.
Also 25 (5*5) and 50 (2*5*5) and 75 (3*5*5) and 625 (5*5*5*5) are included, because in all of them, the prime factor 5 (of the form 4k+1) occurs an even number of times.

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

Cf. A267113.
Cf. A268378 (a subsequence).
Cf. A001481, A267099.

{1}~Join~Select[Range@ 140, NoneTrue[FactorInteger@ #, And[Mod[First@ #, 4] == 1, OddQ@ Last@ #] &] &] (* Michael De Vlieger, Feb 04 2016, Version 10 *)

isok(n) = {my(f = factor(n)); for (i=1, #f~, if (((f[i,1] % 4) == 1) && (f[i,2] % 2), return (0));); return (1);} \\ Michel Marcus, Feb 04 2016

(define A268377 (MATCHING-POS 1 1 (COMPOSE even? A267113)))

A268378 Numbers whose prime factorization includes at least one prime factor of form 4k+3 and any prime factor of the form 4k+1 has even multiplicity.

Original entry on oeis.org

3, 6, 7, 9, 11, 12, 14, 18, 19, 21, 22, 23, 24, 27, 28, 31, 33, 36, 38, 42, 43, 44, 46, 47, 48, 49, 54, 56, 57, 59, 62, 63, 66, 67, 69, 71, 72, 75, 76, 77, 79, 81, 83, 84, 86, 88, 92, 93, 94, 96, 98, 99, 103, 107, 108, 112, 114, 118, 121, 124, 126, 127, 129, 131, 132, 133, 134, 138, 139, 141, 142, 144, 147, 150

Offset: 1

Antti Karttunen, Feb 03 2016

nonn

Closed under multiplication.

			6 = 2*3 is included, as there is a prime factor of the form 4k+3 present.
75 = 3 * 5 * 5 is included, as there is a prime factor of the form 4k+3 present and the prime factor of the form 4k+1 (5) is present twice.

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

Cf. A065339, A267113.
Cf. A004431, A267099.
Subsequence of A268377.
Differs from A221264 for the first time at n=38, which here is a(38) = 75, a value missing from A221264.

Select[Range@ 150, AnyTrue[#, Mod[First@ #, 4] == 3 &] && NoneTrue[#, And[Mod[First@ #, 4] == 1, OddQ@ Last@ #] &] &@ FactorInteger@ # &] (* Michael De Vlieger, Feb 04 2016, Version 10 *)

isok(n) = {my(f = factor(n), nb3 = 0); for (i=1, #f~, if (((f[i,1] % 4) == 1) && (f[i,2] % 2), return (0)); if ((f[i,1] % 4) == 3, nb3++);); return (nb3);} \\ Michel Marcus, Feb 04 2016

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A267116 Bitwise-OR of the exponents of primes in the prime factorization of n.

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A260728 Bitwise-OR of the exponents of all 4k+3 primes in the prime factorization of n.

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A286361 Least number with the same prime signature as {the largest divisor of n with only prime factors of the form 4k+1} has: a(n) = A046523(A170818(n)).

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A268377 Numbers n such that any prime factor of the form 4k+1 has even multiplicity.

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Crossrefs

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Mathematica

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Scheme

A268378 Numbers whose prime factorization includes at least one prime factor of form 4k+3 and any prime factor of the form 4k+1 has even multiplicity.

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Examples

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Programs

Mathematica

PARI