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 1-3 of 3 results.

A228870 Numbers n such that 2 * (1^n + 2^n + 3^n + ... + n^n) is not 0 (mod n).

Original entry on oeis.org

6, 12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 60, 66, 72, 78, 80, 84, 90, 96, 100, 102, 108, 110, 114, 120, 126, 132, 138, 140, 144, 150, 156, 160, 162, 168, 174, 180, 186, 192, 198, 200, 204, 210, 216, 220, 222, 228, 234, 240, 246, 252, 258, 260, 264, 270, 272
Offset: 1

Views

Author

T. D. Noe, Sep 06 2013

Keywords

Comments

These are the numbers not appearing in A228869; the even numbers not in A226872.
Also, positive integers n such that there exists an odd prime divisor p of n such that (p-1) also divides n (cf. A124240). - Max Alekseyev, Sep 07 2013
This sequence agrees with A088723 for many terms, but they are different.
If n is in the sequence, then so are the multiples of n. See A280187 for primitive members of this sequence. - Charles R Greathouse IV, Dec 28 2016

Crossrefs

Programs

  • Mathematica
    Select[Range[100], Mod[2*Sum[PowerMod[k, #, #], {k, #}], #] > 0 &]
  • PARI
    is(n)=my(f=factor(n)[,1]); for(i=1,#f, if(n%(f[i]-1)==0 && f[i]>2, return(1))); 0 \\ Charles R Greathouse IV, Dec 28 2016

A228869 Numbers n such that 2 * (1^n + 2^n + 3^n + ... + n^n) == 0 (mod n).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 79, 81, 82, 83
Offset: 1

Views

Author

T. D. Noe, Sep 06 2013

Keywords

Comments

See A228870 for the numbers not in this sequence.
Union of A226872 and the positive odd integers (A005408).
Also, positive integers n such that (p-1) does not divide n for every odd prime p dividing n (cf. A124240). - Max Alekseyev, Sep 07 2013

Crossrefs

Programs

  • Mathematica
    Select[Range[100], Mod[2*Sum[PowerMod[k, #, #], {k, #}], #] == 0 &]

A225821 a(n) = Product_{p | p is prime and p, p-1 both divide n}.

Original entry on oeis.org

1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 6, 1, 10, 1, 2, 1, 6, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 6, 1, 2, 1, 10, 1, 42, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 30, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 6, 1, 10, 1, 2, 1, 42
Offset: 1

Views

Author

Keywords

Comments

a(n) = 2 iff n is even and is a term of A226872. - Daniel Suteu, Jul 28 2019
From Bernard Schott, Jul 30 2019: (Start)
a(n) = n if n = 1, 2, 6, 42, 1806.
a(n) = 6 if n is of the form 2^i*3^j, i and j >= 1, so if n is a term of A033845.
a(n) = 10 if n is of the form 2^i*5^j, i >= 2 and j >= 1.
a(n) = 30 if n is of the form 2^i*3^j*5^k, i >=2, j >= 1 and k >= 1. (End)

Crossrefs

Programs

  • Mathematica
    fa=FactorInteger; d[m_]:= Product[If[IntegerQ[m/(fa[m][[i, 1]]-1)],fa[m][[i, 1]], 1], {i, Length@fa@m}]; Table[d[n], {n, 1, 333}]
  • PARI
    a(n)=my(f=factor(n)[,1]); prod(i=1,#f,if(n%(f[i]-1)==0,f[i],1)) \\ Charles R Greathouse IV, Nov 13 2013
  • Sage
    def A225821(n) : return prod(p for (p,m) in factor(n) if n%(p-1)==0) # Eric M. Schmidt, Jul 31 2013
    

Formula

a(n) = denominator(A031971(n)/n) = gcd(n, A027642(n)). - Daniel Suteu, Jul 28 2019
Showing 1-3 of 3 results.