A170818 -id:A170818 - cp's OEIS frontend

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)).

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)).

A324891 a(n) = sigma(A170818(n)), where A170818(n) is the part of n composed of prime factors of form 4k+1.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 14, 1, 6, 1, 18, 1, 1, 6, 1, 1, 1, 1, 31, 14, 1, 1, 30, 6, 1, 1, 1, 18, 6, 1, 38, 1, 14, 6, 42, 1, 1, 1, 6, 1, 1, 1, 1, 31, 18, 14, 54, 1, 6, 1, 1, 30, 1, 6, 62, 1, 1, 1, 84, 1, 1, 18, 1, 6, 1, 1, 74, 38, 31, 1, 1, 14, 1, 6, 1, 42, 1, 1, 108, 1, 30, 1, 90, 6, 14, 1, 1, 1, 6, 1, 98, 1, 1, 31, 102, 18, 1, 14, 6

Offset: 1

Antti Karttunen, Mar 27 2019

Antti Karttunen, Table of n, a(n) for n = 1..20000
Index entries for sequences related to sigma(n)

Cf. A000203, A000593, A170818, A324893.

A324891(n) = { my(f=factor(n)); prod(i=1, #f~, if(f[i, 1]%4>1, 1, ((f[i,1]^(1+f[i,2]))-1)/(f[i,1]-1))); };

A170818(n) = { my(f=factor(n)); prod(i=1, #f~, if(f[i, 1]%4>1, 1, f[i, 1])^f[i, 2]); }; \\ From A170818
A324891(n) = sigma(A170818(n));

Multiplicative with a(p^e) = (p^(e+1) - 1)/(p-1) if p == 1 (mod 4), otherwise a(p^e) = 1.
a(n) = A000203(A170818(n)).
a(n) = A000593(n) / A324893(n).

A097706 Part of n composed of prime factors of form 4k+3.

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 7, 1, 9, 1, 11, 3, 1, 7, 3, 1, 1, 9, 19, 1, 21, 11, 23, 3, 1, 1, 27, 7, 1, 3, 31, 1, 33, 1, 7, 9, 1, 19, 3, 1, 1, 21, 43, 11, 9, 23, 47, 3, 49, 1, 3, 1, 1, 27, 11, 7, 57, 1, 59, 3, 1, 31, 63, 1, 1, 33, 67, 1, 69, 7, 71, 9, 1, 1, 3, 19, 77, 3, 79, 1, 81, 1, 83, 21

Offset: 1

Ralf Stephan, Aug 30 2004

Largest term of A004614 that divides n. - Peter Munn, Apr 15 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Equivalent sequence for distinct prime factors: A170819.
Equivalent sequences for prime factors of other forms: A000265 (2k+1), A170818 (4k+1), A072436 (not 4k+3), A248909 (6k+1), A343431 (6k+5).
Range of values: A004614.
Positions of 1's: A072437.
Cf. also A065338, A065339, A260728, A286363.

a:= n-> mul(`if`(irem(i[1], 4)=3, i[1]^i[2], 1), i=ifactors(n)[2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Jun 09 2014

a[n_] := Product[{p, e} = pe; If[Mod[p, 4] == 3, p^e, 1], {pe, FactorInteger[n]}]; Array[a, 100] (* Jean-François Alcover, Jun 16 2015, updated May 29 2019 *)

a(n)=local(f); f=factor(n); prod(k=1, matsize(f)[1], if(f[k, 1]%4<>3, 1, f[k, 1]^f[k, 2]))

from sympy import factorint
from operator import mul
def a072436(n):
    f=factorint(n)
    return 1 if n == 1 else reduce(mul, [1 if i%4==3 else i**f[i] for i in f])
def a(n): return n/a072436(n) # Indranil Ghosh, May 08 2017

from math import prod
from sympy import factorint
def A097706(n): return prod(p**e for p, e in factorint(n).items() if p & 3 == 3) # Chai Wah Wu, Jun 28 2022

a(n) = n/A072436(n).
a(A004614(n)) = A004614(n).
a(A072437(n)) = 1.
a(n) = A000265(n)/A170818(n). - Peter Munn, Apr 15 2021

A343430 Part of n composed of prime factors of the form 3k-1.

Original entry on oeis.org

1, 2, 1, 4, 5, 2, 1, 8, 1, 10, 11, 4, 1, 2, 5, 16, 17, 2, 1, 20, 1, 22, 23, 8, 25, 2, 1, 4, 29, 10, 1, 32, 11, 34, 5, 4, 1, 2, 1, 40, 41, 2, 1, 44, 5, 46, 47, 16, 1, 50, 17, 4, 53, 2, 55, 8, 1, 58, 59, 20, 1, 2, 1, 64, 5, 22, 1, 68, 23, 10, 71, 8, 1, 2, 25, 4, 11, 2, 1, 80, 1, 82, 83, 4, 85

Offset: 1

Peter Munn, Jun 08 2021

Largest term of A004612 that divides n.
Modulo 6, the prime numbers are partitioned into 4 nonempty sets: {2}, {3}, primes of the form 6k-1 (A007528) and primes of the form 6k+1 (A002476). The modulo 3 partition is nearly the same, but unites the only even prime, 2, with primes of the form 6k-1 in the set of primes we use here.
A positive integer m is a Loeschian number (a term of A003136) if and only if a(A007913(m)) = 1, that is the squarefree part of m has no prime factors of the form 3k-1.

			n = 60 has prime factorization 60 = 2 * 2 * 3 * 5. Factors 2 = 3*1 - 1 and 5 = 3*2 - 1 have form 3k-1, whereas 3 does not (having form 3k). Multiplying the factors of form 3k-1, we get 2 * 2 * 5 = 20. So a(60) = 20.

Equivalent sequences for prime factors of other forms: A006519 (2 only), A000265 (2k+1), A038500 (3 only), A038502 (3k+/-1), A170818 (4k+1), A097706 (4k-1), A248909 (6k+1), A343431 (6k-1).
Range of terms: A004612 (closure under multiplication of A003627).
Cf. A002476, A007528, squarefree part (A007913) of terms of A003136.
First 28 terms are the same as A247503.

f[p_, e_] := If[Mod[p, 3] == 2, p^e, 1]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jun 11 2021 *)

a(n) = {my(f = factor(n)); for (i=1, #f~, if ((f[i, 1] + 1) % 3, f[i, 1] = 1); ); factorback(f); } \\ after Michel Marcus at A248909

from math import prod
from sympy import factorint
def A343430(n): return prod(p**e for p, e in factorint(n).items() if p%3==2) # Chai Wah Wu, Dec 23 2022

Completely multiplicative with a(p) = p if p is of the form 3k-1, otherwise a(p) = 1.
For k >= 1, a(n) = a(k*n) / gcd(k, a(k*n)).
a(n) = A006519(n) * A343431(n).
a(n) = (n / A038500(n)) / A248909(n) = A038502(n) / A248909(n).
A006519(a(n)) = a(A006519(n)) = A006519(n).
A343431(a(n)) = a(A343431(n)) = A343431(n).
A038500(a(n)) = a(A038500(n)) = 1.
A248909(a(n)) = a(A248909(n)) = 1.

A248909 Completely multiplicative with a(p) = p if p = 6k+1 and a(p) = 1 otherwise.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 13, 7, 1, 1, 1, 1, 19, 1, 7, 1, 1, 1, 1, 13, 1, 7, 1, 1, 31, 1, 1, 1, 7, 1, 37, 19, 13, 1, 1, 7, 43, 1, 1, 1, 1, 1, 49, 1, 1, 13, 1, 1, 1, 7, 19, 1, 1, 1, 61, 31, 7, 1, 13, 1, 67, 1, 1, 7, 1, 1, 73, 37, 1, 19, 7, 13, 79, 1, 1

Offset: 1

Tom Edgar, Mar 06 2015

To compute a(n) replace primes not of the form 6k+1 in the prime factorization of n by 1.
The first place this sequence differs from A170824 is at n = 49.
For p prime, a(p) = p if p is a term in A002476 and a(p) = 1 if p = 2, p = 3 or p is a term in A007528.
a(n) is the largest term of A004611 that divides n. - Peter Munn, Mar 06 2021

			a(49) = 49 because 49 = 7^2 and 7 = 6*1 + 1.
a(15) = 1 because 15 = 3*5 and neither of these primes is of the form 6k+1.
a(62) = 31 because 62 = 31*2, 31 = 6*5 + 1, and 2 is not of the form 6k+1.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Index to divisibility sequences

Sequences used in a definition of this sequence: A002476, A004611, A007528, A020639, A028234, A032742.
Equivalent sequence for distinct prime factors: A170824.
Equivalent sequences for prime factors of other forms: A000265 (2k+1), A343430 (3k-1), A170818 (4k+1), A097706 (4k-1), A343431 (6k-1), A065330 (6k+/-1), A065331 (<= 3).

A248909 := proc(n)
    local a,pf;
    a := 1 ;
    for pf in ifactors(n)[2] do
        if modp(op(1,pf),6) = 1 then
            a := a*op(1,pf)^op(2,pf) ;
        end if;
    end do:
    a ;
end proc: # R. J. Mathar, Mar 14 2015

f[p_, e_] := If[Mod[p, 6] == 1, p^e, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 19 2020 *)

a(n) = {my(f = factor(n)); for (i=1, #f~, if ((f[i,1] - 1) % 6, f[i, 1] = 1);); factorback(f);} \\ Michel Marcus, Mar 11 2015

from sympy import factorint
def A248909(n):
    y = 1
    for p,e in factorint(n).items():
        y *= (1 if (p-1) % 6 else p)**e
    return y # Chai Wah Wu, Mar 15 2015

n=100; sixnplus1Primes=[x for x in primes_first_n(100) if (x-1)%6==0]
[prod([(x^(x in sixnplus1Primes))^y for x,y in factor(n)]) for n in [1..n]]

(define (A248909 n) (if (= 1 n) n (* (if (= 1 (modulo (A020639 n) 6)) (A020639 n) 1) (A248909 (A032742 n))))) ;; Antti Karttunen, Jul 09 2017

a(1) = 1; for n > 1, if A020639(n) = 1 (mod 6), a(n) = A020639(n) * a(A032742(n)), otherwise a(n) = a(A028234(n)). - Antti Karttunen, Jul 09 2017

a(n) = a(A065330(n)). - Peter Munn, Mar 06 2021

A170819 a(n) = product of distinct primes of the form 4k-1 that divide n.

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 7, 1, 3, 1, 11, 3, 1, 7, 3, 1, 1, 3, 19, 1, 21, 11, 23, 3, 1, 1, 3, 7, 1, 3, 31, 1, 33, 1, 7, 3, 1, 19, 3, 1, 1, 21, 43, 11, 3, 23, 47, 3, 7, 1, 3, 1, 1, 3, 11, 7, 57, 1, 59, 3, 1, 31, 21, 1, 1, 33, 67, 1, 69, 7, 71, 3, 1, 1, 3, 19, 77, 3, 79, 1, 3, 1, 83, 21, 1

Offset: 1

N. J. A. Sloane, Dec 23 2009

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A007947, A011765, A065339, A097706, A170817, A170818, A363340.

A170819 := proc(n) a := 1 ; for p in numtheory[factorset](n) do if p mod 4 = 3 then a := a*p ; end if; end do: a ; end proc:
seq(A170819(n),n=1..20) ; # R. J. Mathar, Jun 07 2011

Array[Times @@ Select[FactorInteger[#][[All, 1]], Mod[#, 4] == 3 &] &, 85] (* Michael De Vlieger, Feb 19 2019 *)

for(n=1,99, t=select(x->x%4==3, factor(n)[,1]); print1(prod(i=1,#t,t[i])","))

Multiplicative with a(p^e) = p^A011765(p+1), e > 0. - R. J. Mathar, Jun 07 2011

a(n) = A007947(A097706(n)) = A097706(A007947(n)). - Peter Munn, Jul 06 2023

Extended with PARI program by M. F. Hasler, Dec 23 2009

A267113 Bitwise-OR of the exponents of all 4k+1 primes in the prime factorization of n.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 2, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 2, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1

Offset: 1

Antti Karttunen, Feb 03 2016

nonn

			For n = 65 = 5 * 13 = (4+1)^1 * ((3*4)+1)^1, bitwise-or of 1 and 1 is 1, thus a(65) = 1.

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

Cf. A170818, A267116, A260728.
Cf. A004144 (indices of zeros), A009003 (of nonzeros).
Differs from both A046080 and A083025 for the first time at n=65, which here a(65) = 1.

a(n) = A267116(A170818(n)).
Other identities. For all n >= 0:
a(n) = a(A170818(n)). [The result depends only on the prime factors of the form 4k+1.]
a(n) <= A083025(n).

A343431 Part of n composed of prime factors of the form 6k-1.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 1, 1, 5, 11, 1, 1, 1, 5, 1, 17, 1, 1, 5, 1, 11, 23, 1, 25, 1, 1, 1, 29, 5, 1, 1, 11, 17, 5, 1, 1, 1, 1, 5, 41, 1, 1, 11, 5, 23, 47, 1, 1, 25, 17, 1, 53, 1, 55, 1, 1, 29, 59, 5, 1, 1, 1, 1, 5, 11, 1, 17, 23, 5, 71, 1, 1, 1, 25, 1, 11, 1, 1, 5, 1, 41, 83, 1, 85, 1, 29, 11, 89, 5

Offset: 1

Peter Munn, Apr 15 2021

Completely multiplicative with a(p) = p if p is of the form 6k-1 and a(p) = 1 otherwise.

Largest term of A259548 that divides n.

Equivalent sequence for distinct prime factors: A170825.
Equivalent sequences for prime factors of other forms: A000265 (2k+1), A343430 (3k-1), A170818 (4k+1), A097706 (4k-1), A248909 (6k+1), A065330 (6k+/-1), A065331 (<= 3), A355582 (<= 5).
Range of terms: A259548.

f[p_, e_] := If[Mod[p, 6] == 5, p^e, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* after Amiram Eldar at A248909 *)

a(n) = {my(f = factor(n)); for (i=1, #f~, if ((f[i, 1] + 1) % 6, f[i, 1] = 1); ); factorback(f); } \\ after Michel Marcus at A248909

from math import prod
from sympy import factorint
def A343431(n): return prod(p**e for p, e in factorint(n).items() if not (p+1)%6) # Chai Wah Wu, Dec 26 2022

a(n) = n / A065331(n) / A248909(n) = A065330(n) / A248909(n).

A170817 a(n) = product of distinct primes of form 4k+1 that divide n.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 1, 1, 5, 1, 1, 13, 1, 5, 1, 17, 1, 1, 5, 1, 1, 1, 1, 5, 13, 1, 1, 29, 5, 1, 1, 1, 17, 5, 1, 37, 1, 13, 5, 41, 1, 1, 1, 5, 1, 1, 1, 1, 5, 17, 13, 53, 1, 5, 1, 1, 29, 1, 5, 61, 1, 1, 1, 65, 1, 1, 17, 1, 5, 1, 1, 73, 37, 5, 1, 1, 13, 1, 5, 1, 41, 1, 1, 85, 1, 29, 1

Offset: 1

N. J. A. Sloane, Dec 22 2009

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A170818, A170819, A097706, A083025, A170824, A170825.

a:= n-> mul (i, i=map (x-> x[1], select (x-> isprime (x[1]) and irem (x[1], 4)=1, ifactors(n)[2]))): seq (a(n), n=1..120);

Table[Times@@Select[Transpose[FactorInteger[n]][[1]],Mod[#,4]==1&], {n,90}] (* Harvey P. Dale, Dec 07 2012 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1] % 4 == 1, f[i,1], 1));} \\ Amiram Eldar, Jun 09 2025

Corrected and extended with Maple program by Alois P. Heinz, Dec 23 2009

cp's OEIS Frontend

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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A324891 a(n) = sigma(A170818(n)), where A170818(n) is the part of n composed of prime factors of form 4k+1.

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A097706 Part of n composed of prime factors of form 4k+3.

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A343430 Part of n composed of prime factors of the form 3k-1.

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A248909 Completely multiplicative with a(p) = p if p = 6k+1 and a(p) = 1 otherwise.

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A170819 a(n) = product of distinct primes of the form 4k-1 that divide n.

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A267113 Bitwise-OR of the exponents of all 4k+1 primes in the prime factorization of n.

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