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.

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A244752 Square array read by antidiagonals in which rows are indexed by composite numbers w and row w gives n such that n^(w-1) == 1 (mod w^2).

Original entry on oeis.org

17, 33, 37, 49, 73, 65, 65, 109, 129, 80, 81, 145, 193, 82, 101, 97, 181, 257, 161, 201, 145, 113, 217, 321, 163, 301, 289, 197, 129, 253, 385, 242, 401, 433, 393, 26, 145, 289, 449, 244, 501, 577, 589, 199, 257, 161, 325, 513, 323, 601, 721, 785, 224, 513
Offset: 2

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Author

Felix Fröhlich, Jul 05 2014

Keywords

Comments

We can say that "w is a Wieferich pseudoprime to base n".
Any prime factor of w is a Wieferich prime to base n.

Examples

			Table starts
w=4: 17, 33, 49, 65, 81, 97, 113, ....
w=6: 37, 73, 109, 145, 181, 217, ....
w=8: 65, 129, 193, 257, 321, 385, ....
w=9: 80, 82, 161, 163, 242, 244, ....
w=10: 101, 201, 301, 401, 501, 601, ....
w=12: 145, 289, 433, 577, 721, 865, ....
w=14: 197, 393, 589, 785, 981, ....
....
		

Crossrefs

Programs

  • Mathematica
    T = {};
    For[w = 4, w <= 100, w++,
      If[PrimeQ[w], Continue[]];
      t = {};
      For [n = 2, n <= 10^5, n++,
       If[Mod[n^(w - 1), w^2] == 1, AppendTo[t, n]]];
      AppendTo[T, t]];
    Print[TableForm[T]];
    A244752 = {};
    For[c = 1, c <= 50, c++,
      For[r = 1, r <= c, r++, AppendTo[A244752, T[[r]][[c - r + 1]]]]];
    A244752 (* Robert Price, Sep 07 2019 *)
  • PARI
    forcomposite(w=2, 20, print1("w=", w, ": "); for(n=2, 10^3, if(Mod(n, w^2)^(w-1)==1, print1(n, ", "))); print(""))

Extensions

a(17)-a(55) from Robert Price, Sep 07 2019

A084653 Pseudoprimes whose prime factors do not divide any smaller pseudoprime.

Original entry on oeis.org

341, 1387, 2047, 8321, 13747, 18721, 19951, 31621, 60701, 83333, 88357, 219781, 275887, 422659, 435671, 513629, 514447, 587861, 604117, 653333, 680627, 710533, 722261, 741751, 769757, 916327, 1194649, 1252697, 1293337, 1433407, 1441091
Offset: 1

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Author

T. D. Noe, Jun 02 2003

Keywords

Comments

Here pseudoprime means a Fermat base-2 pseudoprime; sequence A001567, a composite number n such that n divides 2^(n-1) - 1. All numbers in this sequence seem to have only two prime factors - a conjecture that has been tested for all pseudoprimes < 10^15. The two prime factors are given in A084654 and A084655. The two prime factors are the same when the pseudoprime is the square of a Wieferich prime (A001220).

Examples

			a(2) = 1387 because 1387 = 19*73 and the smaller pseudoprimes (341, 561, 645, 1105) do not have the factors 19 or 73.
		

Crossrefs

A178871 2nd Wieferich prime base prime(n).

Original entry on oeis.org

3511, 1006003, 20771, 491531
Offset: 1

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Author

Jonathan Sondow, Jun 20 2010, Jun 24 2010

Keywords

Comments

2nd prime p such that p^2 divides prime(n)^(p-1) - 1.
2nd prime p such that p divides the Fermat quotient q_p(p_n) = ((p_n)^(p-1) - 1)/p, where p_n = prime(n).
a(5) is unknown: 71 is the only known prime p that divides q_p(11).
If a(5) is found, the sequence continues a(6) = 863, a(7) = 3, a(8) = 7, a(9) = 2481757.
See additional comments, references, links, and cross-refs in A039951 and A174422.

Examples

			a(1) = 3511 is the 2nd Wieferich prime A001220(2).
a(2) = 1006003 is the 2nd Mirimanoff prime A014127(2).
		

Crossrefs

Cf. A039951 = smallest prime p such that p^2 divides n^(p-1) - 1, A174422 = first Wieferich prime base prime(n).

Programs

  • PARI
    {default(primelimit, 10^7); for(n=1, 9, a=prime(n); c=0; forprime(p=2, 10^7, if(Mod(a, p^2)^(p-1)==1, c++; if(c==2, print1(p, ", "); next(2)))); print1(">10^7, "))} \\ Jens Kruse Andersen, Jun 18 2014

A182297 Wieferich numbers (2): positive odd integers q such that q and (2^A002326((q-1)/2)-1)/q are not relatively prime.

Original entry on oeis.org

21, 39, 55, 57, 105, 111, 147, 155, 165, 171, 183, 195, 201, 203, 205, 219, 231, 237, 253, 273, 285, 291, 301, 305, 309, 327, 333, 355, 357, 385, 399, 417, 429, 453, 465, 483, 489, 495, 497, 505, 507, 525, 543, 555, 579, 597, 605, 609, 615, 627, 633, 651, 655
Offset: 1

Views

Author

Felix Fröhlich, Apr 23 2012

Keywords

Comments

The primes in this sequence are A001220, the Wieferich primes. - Charles R Greathouse IV, Feb 02 2014
Odd prime p is a Wieferich prime if and only if A002326((p^2-1)/2) = A002326((p-1)/2). See the sixth comment to A001220 and my formula below. - Thomas Ordowski, Feb 03 2014

Examples

			21 is in the sequence because the multiplicative order of 2 mod 21 is 6, and (2^6-1)/21 = 3, which is not coprime to 21.
		

Crossrefs

For another definition of Wieferich numbers, see A077816.
Cf. A002326.

Programs

  • Maple
    with(numtheory):
    a:= proc(n) option remember; local q;
          for q from 2 +`if`(n=1, 1, a(n-1)) by 2
            while igcd((2^order(2, q)-1)/q, q)=1 do od; q
        end:
    seq (a(n), n=1..60);  # Alois P. Heinz, Apr 23 2012
  • Mathematica
    Select[Range[1, 799, 2], GCD[#, (2^MultiplicativeOrder[2, #] - 1)/#] > 1 &] (* Alonso del Arte, Apr 23 2012 *)
  • PARI
    is(n)=n%2 && gcd(lift(Mod(2,n^2)^znorder(Mod(2,n))-1)/n,n)>1 \\ Charles R Greathouse IV, Feb 02 2014

Formula

Odd numbers q such that A002326((q^2-1)/2) < q * A002326((q-1)/2). Other positive odd integers satisfy the equality. - Thomas Ordowski, Feb 03 2014
Odd numbers q such that gcd(A165781((q-1)/2), q) > 1. - Thomas Ordowski, Feb 12 2014

A238736 Balancing Wieferich primes: primes p that divide their Pell quotients, where the Pell quotient of p is A000129(p - (2/p))/p and (2/p) is a Jacobi symbol.

Original entry on oeis.org

13, 31, 1546463
Offset: 1

Views

Author

John Blythe Dobson, Mar 04 2014

Keywords

Comments

Williams 1982 (p. 86), notes that p = 13, 31 and 1546463 are the only primes less than 10^8 for which the Pell quotient vanishes mod p. Elsenhans and Jahnel, "The Fibonacci sequence modulo p^2," p. 5, report in effect that there are no more such primes p < 10^9.
Williams 1991 (p. 440), and Sun 1995 pt. 3, Theorem 3.3, together prove a set of formulas connecting the Pell quotient with the Fermat quotient (base 2) (A007663) and harmonic numbers like H(floor(p/8)) (see example in the Formula section below). As is well known, the vanishing of the Fermat quotient (base 2) is a necessary condition for the failure of the first case of Fermat's Last Theorem (see discussion under A001220); and in light of a corresponding result of Dilcher and Skula concerning this type of harmonic number, the vanishing of the Pell quotient mod p is also a necessary condition for the failure of the first case of Fermat's Last Theorem.
There are no more terms up to 10^10.
Using the PARI script by Charles R Greathouse IV, I have extended the search from 10^10 to 10^12 without finding a further solution. - John Blythe Dobson, Mar 30 2015
Also primes p such that p^2 divides A001109((p - (2/p))/2). - Jianing Song, Oct 08 2018
From Felix Fröhlich, May 18 2019: (Start)
The term "balancing Wieferich prime" comes from Rout, 2016.
Primes p that satisfy the congruence B_{p-(8/p)} == 0 (mod p^2), where B_i denotes the i-th balancing number A001109(i) and (a/b) denotes the Jacobi symbol (cf. Rout, 2016, (1.6)).
Primes p such that the period of the balancing sequence (A001109) modulo p is equal to the period of the balancing sequence modulo p^2 (cf. Panda, Rout, 2014, p. 275).
Under the abc conjecture for the number field Q(sqrt(2)) there exist at least (log(x)/log(log(x)))*(log(log(log(x))))^m balancing non-Wieferich primes <= x such that p == 1 (mod k) for any integers k > 2, m > 0 (cf. Dutta, Patel, Ray, 2019). This is an improvement of an earlier result stating there are at least log(x)/log(log(x)) balancing non-Wieferich primes p == 1 (mod k) less than x (cf. Theorem 3.2 in Rout 2016). (End)

Examples

			PellQuotient(13) = 6214 = 13*478; PellQuotient(31) = 3470274850 = 31*111944350.
		

Crossrefs

Programs

  • Mathematica
    Select[Prime[Range[1000]], Mod[Fibonacci[# - JacobiSymbol[2, #], 2]/#, #] == 0 &]
  • PARI
    is(n)=isprime(n) && (Mod([2,1;1,0],n^2)^(n-kronecker(2,n)))[2,1]==0 \\ Charles R Greathouse IV, Mar 04 2014

Formula

The condition for p to be a member of this sequence is A000129(p-e)/p == F(p-e, 2)/p == 0 (mod p), where F(p-e, 2) is the p-e'th Fibonacci polynomial evaluated at the argument 2, and e = (2/p) is a Jacobi Symbol.
Let PellQuotient(p) = A000129(p-e)/p, q_2 = (2^(p-1) - 1)/p = A007663(p) be the corresponding Fermat quotient of base 2, H(floor(p/8)) be a harmonic number, and e = (2/p) be a Jacobi Symbol. Then a result of Williams (1991), as refined by Sun (1995), shows that 2*PellQuotient(p) == -4*q_2 - H(floor(p/8)) (mod p).

Extensions

Name amended by Felix Fröhlich, May 26 2019

A270096 Smallest m such that 2^m == 2^n (mod n).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 3, 2, 1, 2, 1, 2, 3, 4, 1, 6, 1, 4, 3, 2, 1, 4, 5, 2, 9, 4, 1, 2, 1, 5, 3, 2, 11, 6, 1, 2, 3, 4, 1, 6, 1, 4, 9, 2, 1, 4, 7, 10, 3, 4, 1, 18, 15, 5, 3, 2, 1, 4, 1, 2, 3, 6, 5, 6, 1, 4, 3, 10, 1, 6, 1, 2, 15, 4, 17, 6, 1, 4
Offset: 1

Views

Author

Thomas Ordowski, Mar 11 2016

Keywords

Comments

a(n) = 1 iff n is a prime or a pseudoprime (odd or even) to base 2.
We have a(n) <= n - phi(n) and a(n) <= phi(n), so a(n) <= n/2.
From Robert Israel, Mar 11 2016: (Start)
If n is in A167791, then a(n) = A068494(n).
If n is odd, a(n) = n mod A002326((n-1)/2).
a(n) >= A007814(n).
a(p^k) = p^(k-1) for all k >= 1 and all odd primes p not in A001220.
Conjecture: a(n) <= n/3 for all n > 8. (End)

Crossrefs

Cf. A276976 (a generalization on all integer bases).

Programs

  • Maple
    f:= proc(n) local d,b,t, m,c;
      d:= padic:-ordp(n,2);
      b:= n/2^d;
      t:= 2 &^ n mod n;
      m:= numtheory:-mlog(t,2,b,c);
      if m < d then m:= m + c*ceil((d-m)/c) fi;
      m
    end proc:
    f(1):= 0:
    map(f, [$1..1000]; # Robert Israel, Mar 11 2016
  • Mathematica
    Table[k = 0; While[PowerMod[2, n, n] != PowerMod[2, k, n], k++]; k, {n, 120}] (* Michael De Vlieger, Mar 15 2016 *)
  • PARI
    a(n) = {my(m = 0); while (Mod(2, n)^m != 2^n, m++); m; } \\ Altug Alkan, Sep 23 2016

Formula

a(n) < n/2 for n > 4.
a(2^k) = k for all k >= 0.
a(2*p) = 2 for all primes p.

Extensions

More terms from Michel Marcus, Mar 11 2016

A163209 Catalan pseudoprimes: odd composite integers n=2*m+1 satisfying A000108(m) == (-1)^m * 2 (mod n).

Original entry on oeis.org

5907, 1194649, 12327121
Offset: 1

Views

Author

Peter Luschny, Jul 24 2009

Keywords

Comments

Also, Wilson spoilers: composite n which divide A056040(n-1) - (-1)^floor(n/2). For the factorial function, a Wilson spoiler is a composite n that divides (n-1)! + (-1). Lagrange proved that no such n exists. For the swinging factorial (A056040), the situation is different.
Also, composite odd integers n=2*m+1 such that A000984(m) == (-1)^m (mod n).
Contains squares of A001220. In particular, a(2) = A001220(1)^2 = 1093^2 = 1194649 = A001567(274) and a(3) = A001220(2)^2 = 3511^2 = 12327121 = A001567(824).
See the Vardi reference for a binomial setup.
Aebi and Cairns 2008, page 9: a(4) either has more than 2 factors or is > 10^10. - Dana Jacobsen, May 27 2015
a(4) > 10^10. - Dana Jacobsen, Mar 03 2018

References

  • I. Vardi, Computational Recreations in Mathematica, 1991, p. 66.

Programs

Extensions

a(1) = 5907 = 3*11*179 was found by S. Skiena
Typo corrected Peter Luschny, Jul 25 2009
Edited by Max Alekseyev, Jun 22 2012

A172290 Prime divisors of 2^1092-1, listed with multiplicities.

Original entry on oeis.org

3, 3, 5, 7, 7, 13, 13, 29, 43, 53, 79, 113, 127, 157, 313, 337, 547, 911, 1093, 1093, 1249, 1429, 1613, 2731, 3121, 4733, 5419, 8191, 14449, 21841, 121369, 224771, 503413, 1210483, 1948129, 22366891, 108749551, 112901153, 23140471537, 25829691707, 105310750819, 467811806281, 4093204977277417, 8861085190774909, 556338525912325157, 86977595801949844993, 275700717951546566946854497, 292653113147157205779127526827, 3194753987813988499397428643895659569
Offset: 1

Views

Author

Artur Jasinski, Jan 30 2010

Keywords

Comments

Up to now only two primes p such that p^2 divide 2^(p-1)-1 are known (these two are Wieferich primes, see A001220).
The sequence is finite with A001222(2^1092-1) = 49 terms; A001221(2^1092-1) = 45. - Reinhard Zumkeller, May 14 2010
Terms appearing more than once (in fact twice) are 3, 7, 13, and 1093.

Crossrefs

Extensions

Missing terms a(34) and a(35) inserted by Reinhard Zumkeller, May 14 2010
Definition clarified and terms corrected by Joerg Arndt, Apr 25 2011

A243905 Multiplicative order of 2 modulo prime(n)^2 for n >= 2.

Original entry on oeis.org

6, 20, 21, 110, 156, 136, 342, 253, 812, 155, 1332, 820, 602, 1081, 2756, 3422, 3660, 4422, 2485, 657, 3081, 6806, 979, 4656, 10100, 5253, 11342, 3924, 3164, 889, 17030, 9316, 19182, 22052, 2265, 8164, 26406, 13861, 29756, 31862, 32580, 18145, 18528, 38612
Offset: 2

Views

Author

Felix Fröhlich, Jun 14 2014

Keywords

Comments

p=prime(n) is in A001220 if and only if a(n) is equal to A014664(n). So far this is known to hold only for p=1093 and p=3511.
This happens for n=183 and 490, that is for p=prime(183)=1093 and p=prime(490)=3511, with values 364 and 1755 (see b-files). - Michel Marcus, Jun 29 2014
If 2^q-1 is p=prime(n), i.e., for n in A016027, then a(n)=pq and lpf(2^a(n)-1)=p. - Thomas Ordowski, Feb 04 2019

Crossrefs

Programs

  • Maple
    seq(numtheory:-order(2, ithprime(i)^2), i=2..1000); # Robert Israel, Jul 08 2014
  • Mathematica
    Table[MultiplicativeOrder[2, Prime[n]^2], {n, 2, 100}] (* Jean-François Alcover, Jul 08 2014 *)
  • PARI
    forprime(p=3, 10^2, print1(znorder(Mod(2, p^2)), ", "))

Formula

a(n) = prime(n)*A014664(n) for all odd primes that are not Wieferich. - Thomas Ordowski, Feb 04 2019

A244550 a(n) = first odd Wieferich prime to base a(n-1) for n > 1, with a(1) = 2.

Original entry on oeis.org

2, 1093, 5, 20771, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71, 3, 11, 71
Offset: 1

Views

Author

Felix Fröhlich, Jun 29 2014

Keywords

Comments

a(2) = 1093 since 1093 is the smallest odd Wieferich prime to base 2.
a(3) = 5 since 5 is the smallest odd Wieferich prime to base 1093.
Subsequence starting at a(5) is periodic with period 3, repeating the terms {3, 11, 71}.
Do values for a(1) exist such that the resulting sequence does not eventually become periodic?
The following table lists the values for a(1) and the resulting cycles those values produce. An entry of the form x-y in first column means all terms from x up to and including y reach the corresponding cycle. An entry of the form {t_1, t_2, t_3, ..., t_n} in second column means the listed terms form a repeating cycle. Entries in second column without curly braces mean the listed terms are reached in order and the term following the last listed term is unknown. A question mark means no further terms have been found in the resulting trajectory of a(1).
a(1) | resulting terms
----------------------------------
2-13, 15-20, | {3, 11, 71}
22-28, 30-40, |
42-46, 48-59, |
62-71, 73-82, |
84-87, 89-118, |
120-132, 134-136,|
138, 140-155, |
157-185, 188, |
190-195, 197-199 |
|
14, 41, 60, 137, | 29
196 |
|
21, 29, 47, 61, | ?
72, 139, 186-187 |
|
83 | {4871, 83}
|
88 | 2535619637, 139
|
119 | 1741
|
133 | 5277179
|
156 | 347
|
189 | 1847
|
Notes
------
The terms of the cycle reached from 83 correspond to A124121(4) and A124122(4), so those terms form a double Wieferich prime pair.

Crossrefs

Programs

  • Magma
    [2, 1093, 5, 20771] cat &cat [[3, 11, 71]^^30]; // Wesley Ivan Hurt, Jun 30 2016
  • Maple
    2,1093,5,20771,seq(op([3, 11, 71]), n=5..50); # Wesley Ivan Hurt, Jun 30 2016
  • Mathematica
    Join[{2, 1093, 5, 20771},LinearRecurrence[{0, 0, 1},{3, 11, 71},66]] (* Ray Chandler, Aug 25 2015 *)
  • PARI
    i=0; a=2; print1(a, ", "); while(i<100, forprime(p=2, 10^6, if(Mod(a, p^2)^(p-1)==1 && p%2!=0, print1(p, ", "); i++; a=p; break({n=1}))))
    

Formula

From Wesley Ivan Hurt, Jun 30 2016: (Start)
G.f.: x*(2+1093*x+5*x^2+20769*x^3-1090*x^4+6*x^5-20700*x^6) / (1-x^3).
a(n) = a(n-3) for n>7.
a(n) = (85 - 52*cos(2*n*Pi/3) + 68*sqrt(3)*sin(2*n*Pi/3))/3 for n>4. (End)
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