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.

A253016 Numbers k such that 11^phi(k) == 1 (mod k^2), where phi(k) = A000010(k).

Original entry on oeis.org

71, 142, 284, 355, 497, 710, 994, 1420, 1491, 1988, 2485, 2840, 2982, 3976, 4970, 5680, 5964, 7455, 9940, 11928, 14910, 19880, 23856, 29820, 39760, 59640, 79520, 119280, 238560, 477120
Offset: 1

Views

Author

Felix Fröhlich, Dec 26 2014

Keywords

Comments

No further terms up to 10^9.
No more terms less than 10^10. - Robert G. Wilson v, Jan 18 2015
The first 30 terms are divisible by 71. Are there any terms not divisible by 71? - Robert Israel, Dec 30 2014
By Corollary 5.9 in Agoh, Dilcher, Skula (1997), if there are no further Wieferich primes to base 11 apart from 71, then the answer is no. - Felix Fröhlich, Dec 30 2014

Crossrefs

Programs

  • Maple
    select(t -> 11 &^ numtheory:-phi(t) mod t^2 = 1, [$1..10^6]); # Robert Israel, Dec 30 2014
  • Mathematica
    a253016[n_] := Select[Range[n], PowerMod[11,EulerPhi[#], #^2] == 1 &]; a253016[500000] (* Michael De Vlieger, Dec 29 2014; modified by Robert G. Wilson v, Jan 18 2015 *)
  • PARI
    for(n=2, 1e9, if(Mod(11, n^2)^(eulerphi(n))==1, print1(n, ", ")))

A247154 a(n) = smallest composite c such that n^(A000010(c)) == 1 (mod c^2), i.e., smallest composite Wieferich number to base n.

Original entry on oeis.org

4, 3279, 22, 3279, 41542, 330805, 4, 3279, 4, 1461, 142, 1812389, 1726, 3883, 4, 3279, 4, 35, 6, 1967
Offset: 1

Views

Author

Felix Fröhlich, Nov 21 2014

Keywords

Comments

a(21) > 10^9
a(22)-a(28): 39, 4, 128165, 4, 9, 22, 9
a(29) > 10^9
a(30)-a(33): 1123787, 4, 3279, 4
a(34) > 10^9

Crossrefs

Programs

  • PARI
    for(n=1, 20, forcomposite(c=1, 1e9, if(Mod(n, c^2)^(eulerphi(c))==1, print1(c, ", "); next({2}))); print1("--, "))

A250206 Least base b > 1 such that b^A000010(n) = 1 (mod n^2).

Original entry on oeis.org

2, 5, 8, 7, 7, 17, 18, 15, 26, 7, 3, 17, 19, 19, 26, 31, 38, 53, 28, 7, 19, 3, 28, 17, 57, 19, 80, 19, 14, 107, 115, 63, 118, 65, 18, 53, 18, 69, 19, 7, 51, 19, 19, 3, 26, 63, 53, 17, 18, 57, 134, 19, 338, 161, 3, 31, 28, 41, 53, 107, 264, 115, 19, 127, 99, 161, 143, 65, 28, 99, 11, 55
Offset: 1

Views

Author

Eric Chen, Feb 21 2015

Keywords

Comments

a(n) = least base b > 1 such that n is a Wieferich number (see A077816).
At least, b = n^2+1 can satisfy this equation, so a(n) is defined for all n.
Least Wieferich number (>1) to base n: 2, 1093, 11, 1093, 2, 66161, 4, 3, 2, 3, 71, 2693, 2, 29, 4, 1093, 2, 5, 3, 281, 2, 13, 4, 5, 2, ...; each is a prime or 4. It is 4 if and only if n mod 72 is in the set {7, 15, 23, 31, 39, 47, 63}.
Does every natural number (>1) appear in this sequence? If yes, do they appear infinitely many times?
For prime n, a(n) = A185103(n), does there exist any composite n such that a(n) = A185103(n)?

Examples

			a(30) = 107 since A000010(30) = 8, 30^2 = 900, and 107 is the least base b > 1 such that b^8 = 1 (mod 900).
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{b = 2, m = EulerPhi[n]}, While[ PowerMod[b, m, n^2] != 1, b++]; b]; f[1] = 2; Array[f, 72] (* Robert G. Wilson v, Feb 28 2015 *)
  • PARI
    a(n)=for(k=2,2^24,if((k^eulerphi(n))%(n^2)==1, return(k)))

Formula

a(prime(n)) = A039678(n) = A185103(prime(n)).
a(A077816(n)) = 2.
a(A242958(n)) <= 3.

A257660 Numbers n such that 13^phi(n) == 1 (mod n^2), where phi(n) = A000010(n).

Original entry on oeis.org

2, 863, 1726, 3452, 371953, 743906, 1487812, 1747591, 1859765, 2975624, 3495182, 3719530, 5242773, 6990364, 7439060, 8737955, 10485546, 14878120, 15993979, 17475910, 20971092, 26213865, 29756240, 31987958, 34951820, 41942184, 47981937, 52427730, 59512480
Offset: 1

Views

Author

Felix Fröhlich, Jul 26 2015

Keywords

Comments

The subsequence of primes in this sequence is A128667.

Crossrefs

Programs

  • Mathematica
    Select[Range@ 1000000, Mod[13^EulerPhi[#], #^2] == 1 &] (* Michael De Vlieger, Jul 27 2015 *)
  • PARI
    for(n=2, 1e9, if(Mod(13, n^2)^(eulerphi(n))==1, print1(n, ", ")))
Showing 1-4 of 4 results.