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 1-7 of 7 results.

A341658 Primes p such that p^2 - 1 has 32 divisors.

Original entry on oeis.org

29, 43, 53, 59, 61, 67, 83, 107, 157, 173, 193, 227, 277, 283, 317, 347, 563, 653, 733, 787, 877, 907, 997, 1213, 1237, 1283, 1307, 1523, 1867, 2083, 2693, 2797, 2803, 3253, 3413, 3517, 3643, 3677, 3733, 3803, 4253, 4363, 4547, 4723, 5387, 5443, 5483, 5717
Offset: 1

Views

Author

Jon E. Schoenfield, Feb 26 2021

Keywords

Comments

Conjecture: sequence is infinite.
All terms are primes p such that p^2 - 1 is of the form 24*q*r = 2^3 * 3 * q * r (where q and r are distinct primes), with only three exceptions: 53, 107, and 193 (see Example section).
For primes p > 3, p^2 - 1 = (p-1)*(p+1) will have p-1 and p+1 as consecutive even numbers (so one of them is divisible by 4, so their product is divisible by 8), and one of p-1 and p+1 will be divisible by 3. For each term other than 53, 107, and 193, the factors p-1 and p+1 are, in some order, numbers of the forms 2*q and 12*r or 4*q and 6*r.

Examples

			      p =               factorization
   n  a(n)  p^2 - 1      of (p^2 - 1)
  --  ----  -------  -------------------
   1   29      840   2^3 * 3   *  5 *  7
   2   43     1848   2^3 * 3   *  7 * 11
   3   53     2808   2^3 * 3^3 * 13
   4   59     3480   2^3 * 3   *  5 * 29
   5   61     3720   2^3 * 3   *  5 * 31
   6   67     4488   2^3 * 3   * 11 * 17
   7   83     6888   2^3 * 3   *  7 * 41
   8  107    11448   2^3 * 3^3 * 53
   9  157    24648   2^3 * 3   * 13 * 79
  10  173    29928   2^3 * 3   * 29 * 43
  11  193    37248   2^7 * 3   * 97

Crossrefs

Cf. A000005, A000040, A309906, A341655.

Programs

  • Mathematica
    Select[Range[6000], PrimeQ[#] && DivisorSigma[0, #^2 - 1] == 32 &] (* Amiram Eldar, Feb 26 2021 *)
  • PARI
    isok(p) = isprime(p) && (numdiv(p^2-1) == 32); \\ Michel Marcus, Feb 26 2021

A341670 a(n) is conjecturally the largest prime p such that, for every prime q > p, q^n - 1 has more divisors than does p^n - 1, or -1 if no such prime p exists.

Original entry on oeis.org

5, 73, 5, 101, 3, 167, 5, 71, 3, 43, 5, 167, 5, 73, 3, 19, 2, 17, 2, 19, 3, 23, 2, 71, 2, 7, 2, 29, -1, 13, 2, 11, 2, 7, 2, 13, 2, 3, 2, 11, 2, 7, -1, 23, 2, 17, 5, 17, 5, 7, -1, 7, -1, 11, -1, 5, 2, 7, 2, 11, 2, 2, 3, 3, 2, 5, 2, 2, 2, 3, 3, 11, -1, 2, -1, 7
Offset: 1

Views

Author

Jon E. Schoenfield, Feb 26 2021

Keywords

Comments

Conjecture: a(n) is also the largest prime p for which there exists no prime q > p such that p^n - 1 and q^n - 1 have the same number of divisors.

Examples

			It is conjectured that there are infinitely many primes q such that q-1 has exactly A309906(1)=4 divisors (see also A005385), and p=5 is the largest prime p such that p-1 has fewer than 4 divisors, so a(1)=5.
Similarly, there appear to be infinitely many primes q such that q^2 - 1 has exactly A309906(2)=32 divisors (e.g., primes q such that, of the two factors q-1 and q+1 of q^2 - 1, one is twice a prime > 3 and the other is 12 times a prime > 3), and p=73 is the largest prime p such that p^2 - 1 has fewer than 32 divisors (see A341655), so a(2)=73.
a(4)=101 is the largest prime p such that p^4 - 1 has fewer than 160 divisors, and there are conjecturally infinitely many primes q such that q^4 - 1 has exactly A309906(4)=160 divisors.
a(29)=-1 because there are conjecturally infinitely many primes q such that q^29 - 1 has exactly A309906(29)=8 divisors, and there exists no prime p such that p^29 - 1 has fewer than 8 divisors.

Crossrefs

Cf. A000005, A000040, A005385, A309906, A341655, A341656, A341657, A341658, A341659.

A341657 a(n) is the number of divisors of prime(n)^6 - 1.

Original entry on oeis.org

6, 16, 48, 60, 192, 96, 192, 256, 360, 384, 504, 512, 240, 384, 576, 320, 384, 768, 576, 320, 320, 864, 384, 640, 504, 1152, 960, 1280, 1280, 576, 576, 768, 960, 768, 1152, 720, 384, 768, 240, 768, 2048, 2048, 2304, 384, 1536, 1920, 3072, 672, 1152, 1536, 1280
Offset: 1

Views

Author

Jon E. Schoenfield, Feb 25 2021

Keywords

Comments

a(n) >= A309906(6) = 384 for n > 39.
p^6 - 1 = A*B*C*D where A=(p-1), B=(p+1), C=(p^2 - p + 1), and D=(p^2 + p + 1), and A, B, C, and D are pairwise coprime except that 2 may divide both A and B and that 3 may divide both A and D or both B and C. For prime p > 7, A and B are consecutive even numbers (so one of them is divisible by 4), so 8|AB; 3 divides both A and D or both B and C, so 9|ABCD; and 7 divides exactly one of A, B, C, and D. Thus, 8*9*7 = 2^3 * 3^2 * 7^1 = 504|ABCD = p^6 - 1. Generally, for sufficiently large primes p, the factors of ABCD, counted with multiplicity, include at least three 2's, two 3's, one 7, and at least four distinct larger primes, so tau(ABCD) = A000005(ABCD) >= (3+1)*(2+1)*(1+1)*(1+1)^4 = 384. (For sufficiently large primes p such that one of A, B, C, or D has no prime factors other than 2, 3, or 7, ABCD will still have at least four distinct prime factors > 7 unless the other three of A, B, C, and D have only one such larger prime factor each; in every such case where p > 167 (e.g., at p = 193, 383, 1373, and 6047), even though ABCD has only 3 distinct prime factors > 7, the multiplicities of 2, 3, and 7 in ABCD are collectively large enough that ABCD nevertheless has at least 384 divisors.)
The largest prime p at which tau(p^6 - 1) < 384 is p = prime(39) = 167: the prime factorizations of A, B, C, and D are A = 166 = 2 * 83, B = 168 = 2^3 * 3 * 7, C = 27723 = 3 * 9241, and D = 28057, so p^6 - 1 = ABCD = 2^4 * 3^2 * 7 * 83 * 9241 * 28057, and thus tau(p^6 - 1) = (4+1)*(2+1)*(1+1)*(1+1)*(1+1)*(1+1) = 5*3*2*2*2*2 = 240. (Note that the prime factorization of 167^6 - 1 contains four 2's, two 3's, one 7, and only 3 distinct primes > 7; B = 168 is 7-smooth.)

Examples

			   n  prime(n)    factorization of prime(n)^6 - 1      a(n)
  --  --------  -----------------------------------    ----
   1      2           3^2     * 7                         6
   2      3     2^3           * 7   * 13                 16
   3      5     2^3 * 3^2     * 7   * 31                 48
   4      7     2^4 * 3^2           * 19*43              60
   5     11     2^3 * 3^2 * 5 * 7   * 19*37             192
   6     13     2^3 * 3^2     * 7   * 61*157             96
   7     17     2^5 * 3^3     * 7   * 13*307            192
   8     19     2^3 * 3^3 * 5 * 7^3 * 127               256
   9     23     2^4 * 3^2     * 7   * 11*13^2*79        360
  10     29     2^3 * 3^2 * 5 * 7   * 13*67*271         384
  11     31     2^6 * 3^2 * 5 * 7^2 * 19*331            504
  12     37     2^3 * 3^3     * 7   * 19*31*43*67       512
  13     41     2^4 * 3^2 * 5 * 7   * 547*1723          240
  14     43     2^3 * 3^2     * 7   * 11*13*139*631     384
  15     47     2^5 * 3^2     * 7   * 23*37*61*103      576
  16     53     2^3 * 3^4     * 7   * 13*409*919        320
  17     59     2^3 * 3^2 * 5 * 7   * 29*163*3541       384
  18     61     2^3 * 3^2 * 5 * 7   * 13*31*97*523      768
  19     67     2^3 * 3^2     * 7^2 * 11*17*31*4423     576
  20     71     2^4 * 3^3 * 5 * 7   * 1657*5113         320
  21     73     2^4 * 3^3     * 7   * 37*751*1801       320
  ...
  39    167     2^4 * 3^2     * 7   * 83*9241*28057     240

Crossrefs

Cf. A000005, A000040, A309906, A341655, A341656.

Programs

  • Mathematica
    a[n_] := DivisorSigma[0, Prime[n]^6 - 1]; Array[a, 50] (* Amiram Eldar, Feb 26 2021 *)
  • PARI
    a(n) = numdiv(prime(n)^6-1); \\ Michel Marcus, Feb 26 2021

Formula

a(n) = A000005(A000040(n)^6 - 1).

A341660 Primes p such that p^2 - 1 has fewer than 32 divisors.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 31, 37, 47, 73
Offset: 1

Views

Author

Jon E. Schoenfield, Feb 26 2021

Keywords

Comments

For all primes p > 73, p^2 - 1 has at least A309906(2)=32 divisors.

Examples

			      p =            factorization
   n  a(n)  p^2 - 1    of p^2 - 1    tau(p^2 - 1)
  --  ----  -------  --------------  ------------
   1    2        3   3                     2
   2    3        8   2^3                   4
   3    5       24   2^3 * 3               8
   4    7       48   2^4 * 3              10
   5   11      120   2^3 * 3 * 5          16
   6   13      168   2^3 * 3 * 7          16
   7   17      288   2^5 * 3^2            18
   8   19      360   2^3 * 3^2 * 5        24
   9   23      528   2^4 * 3 * 11         20
  10   31      960   2^6 * 3 * 5          28
  11   37     1368   2^3 * 3^2 * 19       24
  12   47     2208   2^5 * 3 * 23         24
  13   73     5328   2^4 * 3^2 * 37       30

Crossrefs

Cf. A000005, A000040, A309906, A341655, A341658.

Programs

  • Mathematica
    Select[Range[100], PrimeQ[#] && DivisorSigma[0, #^2 - 1] < 32 &] (* Amiram Eldar, Feb 26 2021 *)

A358881 a(n) is the smallest prime p such that p^2 - 1 has 2*n divisors, or -1 if no such prime exists.

Original entry on oeis.org

2, 3, -1, 5, 7, -1, -1, 11, 17, 23, -1, 19, -1, 31, 73, 29, -1, 383, -1, 41, 97, -1, -1, 79, -1, -1, 127, 223, -1, 71, -1, 109, -1, -1, 2593, 197, -1, -1, -1, 281, -1, 1439, -1, 34303, 199, -1, -1, 181, -1, 647, -1, 6143, -1, 7057, -1, 929, -1, -1, -1, 521, -1
Offset: 1

Views

Author

Jon E. Schoenfield, Dec 04 2022

Keywords

Comments

See A350780 for a discussion about the prime solution to d(p^2 - 1) = 2*n for n in certain cases. - Jianing Song, Feb 15 2025

Examples

			For p = 11, p^2 - 1 = 121 - 1 = 120 = 2^3 * 3 * 5 has 16 divisors. 11 is the smallest prime p such that p^2 - 1 has 16 = 2*8 divisors, so a(8) = 11.
There does not exist any prime p such that p^2 - 1 has 6 = 2*3 divisors, so a(3) = -1.

Links

Crossrefs

Cf. A000005, A000040, A341655, A341658, A341660, A350780.

Programs

A335325 Primes p such that d(p^2-1) sets a record, where d(n) is the number of divisors of n.

Original entry on oeis.org

2, 3, 5, 7, 11, 17, 19, 29, 41, 71, 109, 181, 379, 449, 701, 881, 1429, 1871, 2729, 3079, 4159, 5851, 11969, 22679, 23561, 23869, 40699, 65449, 90271, 104651, 188189, 226799, 244529, 252449, 388961, 403649, 815671, 825551, 1276001, 2380951, 2408561
Offset: 1

Views

Author

Austin Nguyen Tran, May 31 2020

Keywords

Examples

			7^2-1 = 48 has 10 factors, which is the largest for any prime <= 7 (5^2-1 has 8 factors, 3^2-1 has 4 factors, and 2^2-1 has 2 factors).

Links

Crossrefs

Cf. A000005, A341655.

Programs

  • Mathematica
    seq[len_] := Module[{s = {}, p = 2, dm = 0, c = 0, d}, While[c < len, If[(d = DivisorSigma[0, p^2 - 1]) > dm, dm = d; c++; AppendTo[s, p]]; p = NextPrime[p]]; s]; seq[30] (* Amiram Eldar, Jul 07 2022 *)
  • PARI
    my(r=0,d);forprime(p=2,3*10^6,if((d=numdiv(p^2-1))>r,r=d;print1(p,", "))); \\ Joerg Arndt, Jun 01 2020

A358879 Primes p such that p^2 + 1 has more divisors than p^2 - 1.

Original entry on oeis.org

2917, 5443, 7187, 9133, 10357, 12227, 12967, 13043, 14243, 17047, 20507, 20743, 21767, 25657, 27893, 27997, 28163, 30307, 32323, 32443, 33493, 33623, 34157, 34367, 34897, 35537, 37783, 37957, 39827, 41387, 41893, 42793, 43633, 44357, 49109, 49993, 56597, 56857
Offset: 1

Views

Author

Jon E. Schoenfield, Dec 04 2022

Keywords

Comments

Fewer than 1.2% of the first million primes have this property.
For all primes p > 3, p^2 - 1 is divisible by 24 (since it is factorable as (p-1)*(p+1)), but p^2 + 1, although it is even, is divisible by neither 4 nor 3.

Examples

			2917 is a term:
2917^2 - 1 = 8508888 = 2^3 * 3^6 * 1459 has 56 divisors, but
2917^2 + 1 = 8508890 = 2 * 5 * 13 * 29 * 37 * 61 has 64.
399173 is a term:
399173^2 - 1 = 159339083928 = 2^3 * 3 * 66529 * 99793 has 32 divisors, but
399173^2 + 1 = 159339083930 = 2 * 5 * 13 * 17 * 29 * 53 * 61 * 769 has 256.

Crossrefs

Cf. A000005, A000040, A341655, A341658, A341660.
Showing 1-7 of 7 results.

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