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.

A347771 Unitary nontotient numbers: values not in range of unitary totient function uphi(n).

Original entry on oeis.org

5, 9, 11, 13, 17, 19, 21, 23, 25, 27, 29, 33, 34, 35, 37, 38, 39, 41, 43, 45, 47, 49, 50, 51, 53, 55, 57, 59, 61, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 81, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 98, 99, 101, 103, 105, 107, 109, 110, 111, 113, 114, 115, 117, 118, 119, 121, 122, 123, 125, 129, 131, 133, 134, 135
Offset: 1

Views

Author

Eric Chen, Sep 13 2021

Keywords

Comments

Numbers not appearing in A047994.
Indices of -1 in A135347.
Unitary version of A007617.
This sequence to A047994 is A007617 to A000010.
This sequence to A135347 is A007617 to A049283 (for the case that no such numbers exist, A135347 uses -1 and A049283 uses 0).
All odd numbers not of the form 2^k-1 (i.e. not in A000225) are in this sequence, since uphi(n) = A047994(n) is an even number unless n is a power of 2 (A000079), in this case uphi(n) = n-1.
The intersection of this sequence and A049225 is empty, since for squarefree numbers, all divisors are unitary divisors, note that the intersection of this sequence and A002202 is not empty, the number 110 is in both sequences.

Crossrefs

Programs

  • Mathematica
    Select[Range[135], Length[invUPhi[#]] == 0 &] (* Amiram Eldar, Apr 01 2023, using the function invUPhi from A361966 *)
  • PARI
    A047994(n)=my(f=factor(n)~); prod(i=1, #f, f[1, i]^f[2, i]-1)
    is(n)=for(k=1,n^2,if(A047994(k)==n,return(0)));1 \\ after A047994

Formula

A361967(a(n)) = 0. - Amiram Eldar, Apr 01 2023

A340521 List of possible orders of automorphism groups of finite groups.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 12, 16, 18, 20, 22, 24, 28, 30, 32, 36, 40, 42, 44, 46, 48, 52, 54, 56, 58, 60
Offset: 1

Views

Author

Des MacHale, Feb 05 2021

Keywords

Comments

The terms shown here match the initial terms of all of A002174, A002202, A049225, but this is a strictly different sequence since it is known that it contains 3^7 = 2187 (which is the smallest odd term greater than 1), whereas for the other three sequences all terms greater than 1 are even.
This is a supersequence of A002202 since |Aut(Z/nZ)| = phi(n). - Jianing Song, Feb 05 2021
John Bray has produced a group G of order 3^2*19 = 171 such that |Aut G| = 1026 = 2.3^3.19. So 1026 is in the present sequence but is not in A002202. So the present sequence contains both odd and even terms not in A002202. - Des MacHale, Feb 10 2021
For more about this problem, see the references in A137315.

Crossrefs

A064234 The least k such that A063994(k) = Product_{primes p dividing k} gcd(p-1, k-1) = n, or 0 if there's no such k.

Original entry on oeis.org

1, 3, 28, 5, 66, 7, 232, 45, 190, 11, 276, 13, 1106, 0, 286, 17, 1854, 19, 3820, 891, 2752, 23, 1128, 595, 2046, 0, 532, 29, 1770, 31, 9952, 425, 1288, 0, 2486, 37, 8474, 0, 742, 41, 3486, 43, 7612, 5589, 2356, 47, 13584, 325, 9850, 0, 20554, 53, 5778, 0, 12926, 435, 13282, 59, 42540, 61
Offset: 1

Views

Author

Robert G. Wilson v, Sep 22 2001

Keywords

Comments

From Richard N. Smith, Jul 15 2019: (Start)
The comment in the "Mathematica" section is not true: A063994(65513) = 76 (thus a(76) = 65513 instead of 0), but 76 is an even nontotient (in the sequence A005277).
The first counterexample of the comment is A063994(1541) = 484, which is an even nontotient, for the counterexamples <= 2^20, see the link.
Also A063994(1072871) = 68. (thus a(68) = 1072871).
Conjecture: a(n) = 0 iff n == 2 mod 4 and n+1 is composite, if this conjecture is true, then a(54), a(110), a(294), etc. would be 0.
Another conjecture: If A063994(k) = n and n == 2 mod 4, then n+1 is prime and k is a power of n+1. (End)

Crossrefs

Programs

  • Mathematica
    f[ n_ ] := If[ n == 1, 1, Apply[ Times, GCD[ n - 1, Transpose[ FactorInteger[ n ] ] [ [ 1 ] ] - 1 ] ] ]; a = Table[ 0, {100} ]; Do[ m = f[ n ]; If[ m < 101 && a[ [ m ] ] == 0, a[ [ m ] ] = n ], {n, 1, 10^7} ]; a a(54) > 2*10^7. The zeros appear at positions that are the values in the sequence A005277, the nontotients: even n such that phi(m) = n has no solution. [Warning: This is wrong, see the "comment" section]
  • PARI
    a063994(n)=my(f=factor(n)[, 1]); prod(i=1, #f, gcd(f[i]-1, n-1))
    a(n)=if(n%4==2 && !isprime(n+1), 0, for(k=1, 2^30, if(a063994(k)==n,return(k)))) \\ Richard N. Smith, Jul 15 2019, after Charles R Greathouse IV in A063994

Extensions

a(54) - a(60) from Richard N. Smith, Jul 15 2019

A308828 Number of sequences that include all residues modulo n starting with x_0 = 0 and then x_i given recursively by x_{i+1} = a * x_i + c (mod n) where a and c are integers in the interval [0..n-1].

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 8, 18, 4, 10, 4, 12, 6, 8, 32, 16, 18, 18, 8, 12, 10, 22, 16, 100, 12, 162, 12, 28, 8, 30, 128, 20, 16, 24, 36, 36, 18, 24, 32, 40, 12, 42, 20, 72, 22, 46, 64, 294, 100, 32, 24, 52, 162, 40, 48, 36, 28, 58, 16, 60, 30, 108, 512, 48, 20, 66, 32, 44, 24
Offset: 1

Views

Author

Haris Ziko, Jun 27 2019

Keywords

Comments

All such sequences will be cyclic with period length n.
Different values of the integers a and c modulo n will always produce different sequences.
Only sequences starting with x_0 = 0 are included here. However using any other fixed number in the interval [0..n-1] for x_0 will result in the same cycles.
It appears that the set of distinct terms of this sequence gives A049225.
From Andrew Howroyd, Aug 21 2019: (Start)
Knuth calls these sequences linear congruential sequences. Theorem A in section 3.2.1.2 of the reference states that the period has length n if and only if:
1) c is relatively prime to n;
2) a - 1 is a multiple of p for every prime p dividing n;
3) a - 1 is a multiple of 4 if n is a multiple of 4.
(End)

Examples

			For n = 1:
  a = 0, c = 0: [0];
  #cycles = 1 -> a(1) = 1.
For n = 5:
  a = 1, c = 1: [0, 1, 2, 3, 4];
  a = 1, c = 2: [0, 2, 4, 1, 3];
  a = 1, c = 3: [0, 3, 1, 4, 2];
  a = 1, c = 4: [0, 4, 3, 2, 1];
  #cycles = 4 -> a(5) = 4.
For n = 8:
  a = 1, c = 1: [0, 1, 2, 3, 4, 5, 6, 7];
  a = 1, c = 3: [0, 3, 6, 1, 4, 7, 2, 5];
  a = 1, c = 5: [0, 5, 2, 7, 4, 1, 6, 3];
  a = 1, c = 7: [0, 7, 6, 5, 4, 3, 2, 1];
  a = 5, c = 1: [0, 1, 6, 7, 4, 5, 2, 3];
  a = 5, c = 3: [0, 3, 2, 5, 4, 7, 6, 1];
  a = 5, c = 5: [0, 5, 6, 3, 4, 1, 2, 7];
  a = 5, c = 7: [0, 7, 2, 1, 4, 3, 6, 5];
  #cycles = 8 -> a(8) = 8.
		

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 3, Random Numbers, Section 3.2.1.2, p. 16.

Crossrefs

Programs

  • PARI
    checkFullSet(v,n)={my(v2=vector(n), unique=1); for(i=1, n, my(j=v[i]+1); if(v2[j]==1, unique=0; break, v2[j]=1;);); unique;};
    doCycle(a,c,m)={my(v_=vector(m), x=c); v_[1]=c; for(i=1, m-1, v_[i+1]=(a*v_[i]+c)%m;); v_;};
    getCycles(m)={my(L=List()); for(a=0, m-1, for(c=0, m-1, my(v1=doCycle(a,c,m)); if(checkFullSet(v1,m), listput(L, v1)););); Mat(Col(L))};
    a(n)={my(M=getCycles(n)); matsize(M)[1]};

Formula

Conjecture: a(n) = A135616(n)/n.
Conjecture: a(n) = phi(n)*A003557(n / gcd(n,2)). - Andrew Howroyd, Aug 20 2019
Showing 1-4 of 4 results.