cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Views

Author

Ralf Stephan, Aug 30 2004

Keywords

Comments

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

Crossrefs

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.

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    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]))
    
  • Python
    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
    
  • Python
    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

Formula

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

A072437 Numbers with no prime factors of form 4*k+3.

Original entry on oeis.org

1, 2, 4, 5, 8, 10, 13, 16, 17, 20, 25, 26, 29, 32, 34, 37, 40, 41, 50, 52, 53, 58, 61, 64, 65, 68, 73, 74, 80, 82, 85, 89, 97, 100, 101, 104, 106, 109, 113, 116, 122, 125, 128, 130, 136, 137, 145, 146, 148, 149, 157, 160, 164, 169, 170, 173, 178, 181, 185, 193, 194
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 17 2002

Keywords

Comments

m is a term iff A072436(m) = m.
These numbers have density zero (Pollack).

Crossrefs

Cf. A004144, A002144, A002145, A004613 (odd terms).
A097706(a(n)) = 1.
Cf. A187811 (complement).

Programs

Formula

n>0 such that A001842(n)=0. - Benoit Cloitre, Apr 24 2003
A005091(a(n)) = 0. - Reinhard Zumkeller, Jan 07 2013
A065339(a(n)) = 0 . - R. J. Mathar, Jan 28 2025

A286363 Least number with the same prime signature as {the largest divisor of n with only prime factors of the form 4k+3} has: a(n) = A046523(A097706(n)).

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 1, 4, 1, 2, 2, 1, 2, 2, 1, 1, 4, 2, 1, 6, 2, 2, 2, 1, 1, 8, 2, 1, 2, 2, 1, 6, 1, 2, 4, 1, 2, 2, 1, 1, 6, 2, 2, 4, 2, 2, 2, 4, 1, 2, 1, 1, 8, 2, 2, 6, 1, 2, 2, 1, 2, 12, 1, 1, 6, 2, 1, 6, 2, 2, 4, 1, 1, 2, 2, 6, 2, 2, 1, 16, 1, 2, 6, 1, 2, 2, 2, 1, 4, 2, 2, 6, 2, 2, 2, 1, 4, 12, 1, 1, 2, 2, 1, 6, 1, 2, 8, 1, 2, 2, 2, 1, 6, 2, 1, 4, 2, 2, 2
Offset: 1

Views

Author

Antti Karttunen, May 08 2017

Keywords

Crossrefs

Programs

  • Python
    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 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 a046523(n/a072436(n)) # Indranil Ghosh, May 09 2017
  • Scheme
    (define (A286363 n) (A046523 (A097706 n)))
    

Formula

a(n) = A046523(A097706(n)).
a(n) = A286361(A267099(n)).

A072438 Remove prime factors of form 4*k+1.

Original entry on oeis.org

1, 2, 3, 4, 1, 6, 7, 8, 9, 2, 11, 12, 1, 14, 3, 16, 1, 18, 19, 4, 21, 22, 23, 24, 1, 2, 27, 28, 1, 6, 31, 32, 33, 2, 7, 36, 1, 38, 3, 8, 1, 42, 43, 44, 9, 46, 47, 48, 49, 2, 3, 4, 1, 54, 11, 56, 57, 2, 59, 12, 1, 62, 63, 64, 1, 66, 67, 4, 69, 14, 71, 72, 1, 2, 3, 76, 77, 6, 79, 16, 81, 2
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 17 2002

Keywords

Comments

a(n) <= n; a(a(n)) = a(n).
All factors p^m of a(n) are of the form p=2 or p=4*k+3.

Examples

			a(90) = a(2*3*3*5) = a(2*(4*0+3)^2*(4*1+1)^1) = 2*3^2*1 = 18.
		

Crossrefs

Programs

  • Maple
    a:= n-> mul(`if`(irem(i[1], 4)=1, 1, i[1]^i[2]), i=ifactors(n)[2]):
    seq(a(n), n=1..100);  # Alois P. Heinz, Jun 09 2014
  • Mathematica
    a[n_] := n/Product[{p, e} = pe; If[Mod[p, 4] == 1, p^e, 1], {pe, FactorInteger[n]}];
    Array[a, 100] (* Jean-François Alcover, May 29 2019 *)
  • PARI
    a(n) = my(f=factor(n)); for (i=1, #f~, if ((f[i,1] % 4) == 1, f[i,1] = 1)); factorback(f); \\ Michel Marcus, Jun 09 2014
    
  • Python
    from sympy import factorint, prod
    def a(n):
        f = factorint(n)
        return 1 if n == 1 else prod(i**f[i] for i in f if i%4 != 1) # Indranil Ghosh, May 08 2017

Formula

Multiplicative with a(p)=(if p==1 (mod 4) then 1 else p).
Showing 1-4 of 4 results.