A309906 -id:A309906 - cp's OEIS frontend

A342062 a(n) is the number of divisors of prime(n)^8 - 1.

8, 24, 48, 84, 96, 192, 192, 288, 224, 192, 576, 576, 672, 2304, 1024, 768, 768, 192, 768, 336, 672, 1024, 3072, 1344, 864, 576, 448, 1152, 1536, 512, 2112, 768, 1792, 768, 1152, 1344, 2304, 960, 896, 1536, 1728, 1152, 2560, 1280, 1728, 504, 1536, 2304, 1536

Offset: 1

Jon E. Schoenfield, Feb 27 2021

nonn

a(n) >= 384 for n > 20.

			   n  prime(n)      factorization of prime(n)^8 - 1      a(n)
  --  --------  ---------------------------------------  ----
   1      2           3   * 5   * 17                        8
   2      3     2^5       * 5   * 41                       24
   3      5     2^5 * 3         * 13 * 313                 48
   4      7     2^6 * 3   * 5^2 * 1201                     84
   5     11     2^5 * 3   * 5   * 61 * 7321                96
   6     13     2^5 * 3   * 5   * 7 * 17 * 14281          192
   7     17     2^7 * 3^2 * 5   * 29 * 41761              192
   8     19     2^5 * 3^2 * 5   * 17 * 181 * 3833         288
   9     23     2^6 * 3   * 5   * 11 * 53 * 139921        224
  10     29     2^5 * 3   * 5   * 7 * 421 * 353641        192
  11     31     2^8 * 3   * 5   * 13 * 37 * 409 * 1129    576
  12     37     2^5 * 3^2 * 5   * 19 * 89 * 137 * 10529   576
  13     41     2^6 * 3   * 5   * 7 * 29^2 * 137 * 10313  672
  ...
  20     71     2^6 * 3^2 * 5   * 7 * 2521 * 12705841     336

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000005, A000040, A309906, A342063, A342064.

f:= n -> numtheory:-tau(ithprime(n)^8-1):
map(f, [$1..100]); # Robert Israel, Feb 28 2021

a[n_] := DivisorSigma[0, Prime[n]^8 - 1]; Array[a, 50] (* Amiram Eldar, Feb 27 2021 *)

a(n) = numdiv(prime(n)^8-1); \\ Michel Marcus, Feb 27 2021

a(n) = A000005(A000040(n)^8 - 1).

A350780 Numbers that are the number of divisors of p^2 - 1 for some prime p.

Original entry on oeis.org

2, 4, 8, 10, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 48, 54, 56, 60, 64, 70, 72, 80, 84, 88, 90, 96, 100, 104, 108, 112, 120, 126, 128, 132, 136, 140, 144, 150, 152, 156, 160, 162, 168, 176, 180, 182, 184, 192, 196, 198, 200, 204, 208, 210, 216, 220, 224, 228

Offset: 1

Jon E. Schoenfield, May 02 2022

For all primes p > 73, tau(p^2 - 1) >= A309906(2) = 32.

			184 is a term: p = 111149057 is a prime, and p^2 - 1 = (p-1)*(p+1) = 2^22 * 3 * 53 * 18524843, which has 23*2*2*2 = 184 divisors.
190 is not a term: 190 = 2 * 5 * 19, so a number with 190 divisors must be of the form q^189, q^94 * r, q^37 * r^4, q^18 * r^9, or q^18 * r^4 * s, and for every prime p > 3, p^2 - 1 is a multiple of 24 = 2^3 * 3, so all the forms with 190 divisors are easily ruled out except for q^18 * r^4 * s. If p^2 - 1 = q^18 * r^4 * s, then it's one of the products 2^18 * 3^4 * s, 2^18 * r^4 * 3, 3^18 * 2^4 * s, or q^18 * 2^4 * 3. Each of these can be shown to be impossible by examining all possible ways of factoring the product into two even factors (p-1 and p+1) that differ by exactly two.
From _Jianing Song_, Feb 11 2025: (Start)
Let Omega = A001222, k be an even number and p be a prime.
 - Omega(k) <= 2. If odd p != 3, 7 (not necessarily prime) satisfies tau(p^2 - 1) = k = 2q for prime q, then p^2 - 1 = 2^(q-1)*P for some prime P, so (p-1,p+1) = (2^(q-2),2P) or (2P,2^(q-2)), which means that P = 2^(q-3) +- 1. Note that "-" is impossible since q-3 is even, so we have q = 2^r + 3, P = 2^2^r + 1, and p = 2^(2^r+1) + 1 for some r. In particular, p must be divisible by 3, so p cannot be prime.
 - Omega(k) = 3, or k = 16, 24, 36, or 54. Then tau(p^2 - 1) = k has finitely many solutions p == 1, 5 (mod 6) (not necessarily prime). See my first pdf link in the Links section for a proof. In fact, it seems that if we require p to be prime, then k <= 518, and the complete list of (k,p), Omega(k) = 3 is (k,p) = (8,5), (18,17), (20,23), (28,31), (30,73), (42,97), (70,2593), (182,1492993), and (518,4803028329503971873=32*3^36+1).
 - If k/2 has only prime factors congruent to 1 modulo 4, then tau(p^2 - 1) = k has no solutions for odd p. See my second pdf link in the Links section for a proof.
 - If k/2 has only prime factors congruent to 1 modulo 2*r for some odd r >= 3, then tau(p^2 - 1) = k for odd p implies that p is of the form p = 2^(2*r-1)*M^(2*r) + 1 for some M.
 - In general, if d(x) = k, then the largest exponent in the canonical factorization of x must be at least gpf(k)-1, where gpf = A006530 is the largest prime factor function. So if d(p^2 - 1) = k, then one of p-1 and p+1 must be divisible by M^(gpf(k)-1) for some odd prime M or by 2^(gpf(k)-2).
Conjecture: if Omega(k) >= 4, k != 16, 24, 36, or 54, and k/2 has a prime factor not congruent to 1 modulo 4, then tau(p^2 - 1) = k has infinitely many solutions. (End)

Jianing Song, Table of n, a(n) for n = 1..192
Jianing Song, Notes on A350780 and A358881 (1)
Jianing Song, Notes on A350780 and A358881 (2)
Jianing Song, PARI program for A350780 and A358881

Cf. A000005, A309906, A341660, A358881.

PARI
```
\\ See Links. Jianing Song, Feb 16 2025
```

A342063 Primes p such that p^8 - 1 has fewer than 384 divisors.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 61, 71

Offset: 1

Jon E. Schoenfield, Feb 27 2021

For all primes p > 71, p^8 - 1 has at least A309906(8)=384 divisors.

			      p =
   n  a(n)        factorization of p^8 - 1         a(n)
  --  ----  -------------------------------------  ----
   1    2         3   * 5   * 17                      8
   2    3   2^5       * 5   * 41                     24
   3    5   2^5 * 3         * 13 * 313               48
   4    7   2^6 * 3   * 5^2 * 1201                   84
   5   11   2^5 * 3   * 5   * 61 * 7321              96
   6   13   2^5 * 3   * 5   * 7 * 17 * 14281        192
   7   17   2^7 * 3^2 * 5   * 29 * 41761            192
   8   19   2^5 * 3^2 * 5   * 17 * 181 * 3833       288
   9   23   2^6 * 3   * 5   * 11 * 53 * 139921      224
  10   29   2^5 * 3   * 5   * 7 * 421 * 353641      192
  11   61   2^5 * 3   * 5   * 31 * 1861 * 6922921   192
  12   71   2^6 * 3^2 * 5   * 7 * 2521 * 12705841   336

Cf. A000005, A000040, A309906, A342062, A342064.

A342064 Primes p such that p^8 - 1 has 384 divisors.

Original entry on oeis.org

821, 997, 2819, 6619, 17827, 20947, 24917, 42709, 43411, 46141, 49261, 51691, 80077, 108803, 158981, 159539, 161341, 171659, 202667, 228611, 268573, 304477, 315803, 350971, 420781, 447683, 463459, 816709, 848227, 887989, 953773, 991811, 1056829, 1131379

Offset: 1

Jon E. Schoenfield, Feb 27 2021

nonn

Conjecture: sequence is infinite.

For every term p, p^8 - 1 is of the form 2^5 * 3 * 5 * q * r * s * t, where q, r, s, and t are distinct primes > 5 (see Example section).

			   p =
n  a(n)               factorization of p^8 - 1
- ----- -----------------------------------------------------
1  821  2^5 * 3 * 5 *  41 *  137 *   337021 *    227165634841
2  997  2^5 * 3 * 5 *  83 *  499 *    99401 *    494026946041
3 2819  2^5 * 3 * 5 *  47 * 1409 *  3973381 *  31575505195561
4 6619  2^5 * 3 * 5 * 331 * 1103 * 21905581 * 959708914083961

Cf. A000005, A000040, A309906, A342062, A342063.

Select[Prime[Range[90000]],DivisorSigma[0,#^8-1]==384&] (* Harvey P. Dale, Jul 08 2025 *)

A341661 Primes p such that p^4 - 1 has fewer than 160 divisors.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 59, 61, 71, 79, 101

Offset: 1

Jon E. Schoenfield, Feb 26 2021

For all primes p > 101, p^4 - 1 has at least A309906(4)=160 divisors.

			      p =
   n  a(n)   p^4 - 1   factorization of p^4 - 1  tau(p^4 - 1)
  --  ----  ---------  ------------------------- ------------
   1    2          15  3 * 5                           4
   2    3          80  2^4 * 5                        10
   3    5         624  2^4 * 3 * 13                   20
   4    7        2400  2^5 * 3 * 5^2                  36
   5   11       14640  2^4 * 3 * 5 * 61               40
   6   13       28560  2^4 * 3 * 5 * 7 * 17           80
   7   17       83520  2^6 * 3^2 * 5 * 29             84
   8   19      130320  2^4 * 3^2 * 5 * 181            60
   9   23      279840  2^5 * 3 * 5 * 11 * 53          96
  10   29      707280  2^4 * 3 * 5 * 7 * 421          80
  11   31      923520  2^7 * 3 * 5 * 13 * 37         128
  12   37     1874160  2^4 * 3^2 * 5 * 19 * 137      120
  13   41     2825760  2^5 * 3 * 5 * 7 * 29^2        144
  14   59    12117360  2^4 * 3 * 5 * 29 * 1741        80
  15   61    13845840  2^4 * 3 * 5 * 31 * 1861        80
  16   71    25411680  2^5 * 3^2 * 5 * 7 * 2521      144
  17   79    38950080  2^6 * 3 * 5 * 13 * 3121       112
  18  101   104060400  2^4 * 3 * 5^2 * 17 * 5101     120

Cf. A000005, A000040, A309906, A341656.

Select[Range[101], PrimeQ[#] && DivisorSigma[0, #^4 - 1] < 160 &] (* Amiram Eldar, Feb 26 2021 *)

A341662 Primes p such that p^4 - 1 has 160 divisors.

Original entry on oeis.org

53, 67, 131, 139, 227, 277, 283, 347, 383, 641, 653, 661, 821, 877, 997, 1069, 1181, 1213, 1811, 2083, 2389, 2459, 2819, 3803, 4021, 4253, 4723, 6619, 6829, 7213, 7933, 8069, 9013, 9187, 10589, 11261, 16139, 17827, 18133, 18587, 19309, 19541, 20477, 20947

Offset: 1

Jon E. Schoenfield, Feb 26 2021

nonn

Conjecture: sequence is infinite.

For every term p, p^4 - 1 is of the form 2^4 * 3 * 5 * q * r * s, where q, r, and s are distinct primes > 5, with three exceptions: p = 53, 383, and 641 (see Example section).

			      p =
   n  a(n)     p^4 - 1    factorization of p^4 - 1
  --  ----  ------------  -------------------------------
   1    53       7890480  2^4 * 3^3 * 5 * 13 * 281
   2    67      20151120  2^4 * 3 * 5 * 11 * 17 * 449
   3   131     294499920  2^4 * 3 * 5 * 11 * 13 * 8581
   4   139     373301040  2^4 * 3 * 5 * 7 * 23 * 9661
   5   227    2655237840  2^4 * 3 * 5 * 19 * 113 * 5153
   6   277    5887339440  2^4 * 3 * 5 * 23 * 139 * 7673
   7   283    6414247920  2^4 * 3 * 5 * 47 * 71 * 8009
   8   347   14498327280  2^4 * 3 * 5 * 29 * 173 * 12041
   9   383   21517662720  2^9 * 3 * 5 * 191 * 14669
  10   641  168823196160  2^9 * 3 * 5 * 107 * 205441
  11   653  181824635280  2^4 * 3 * 5 * 109 * 163 * 42641

Cf. A000005, A000040, A309906, A341656, A341661.

Select[Range[21000], PrimeQ[#] && DivisorSigma[0, #^4 - 1] == 160 &] (* Amiram Eldar, Feb 26 2021 *)

isok(p) = isprime(p) && (numdiv(p^4-1) == 160); \\ Michel Marcus, Feb 26 2021

A341664 a(n) is the number of divisors of prime(n)^5 - 1.

Original entry on oeis.org

2, 6, 12, 8, 12, 12, 10, 24, 8, 12, 48, 72, 24, 16, 32, 48, 64, 72, 32, 96, 24, 16, 8, 64, 96, 96, 32, 16, 96, 80, 24, 48, 64, 32, 48, 32, 96, 160, 32, 24, 16, 108, 96, 28, 72, 48, 96, 128, 8, 48, 32, 16, 120, 60, 36, 8, 36, 96, 24, 192, 64, 24, 96, 48, 64, 48

Offset: 1

Jon E. Schoenfield, Feb 26 2021

nonn

a(n) >= A309906(5) = 8 for n > 2.

			       p =                factorization
  n  prime(n)  p^5 - 1      of p^5 - 1      a(n)
  -  --------  -------  ------------------  ----
  1      2          31  31                    2
  2      3         242  2 * 11^2              6
  3      5        3124  2^2 * 11 * 71        12
  4      7       16806  2 * 3 * 2801          8
  5     11      161050  2 * 5^2 * 3221       12
  6     13      371292  2^2 * 3 * 30941      12
  7     17     1419856  2^4 * 88741          10
  8     19     2476098  2 * 3^2 * 151 * 911  24
  9     23     6436342  2 * 11 * 292561       8
  ...

Cf. A000005, A000040, A309906, A341665.

a[n_] := DivisorSigma[0, Prime[n]^5 - 1]; Array[a, 50] (* Amiram Eldar, Feb 26 2021 *)

a(n) = numdiv(prime(n)^5-1); \\ Michel Marcus, Feb 26 2021

a(n) = A000005(A000040(n)^5 - 1).

A341665 Primes p such that p^5 - 1 has 8 divisors.

Original entry on oeis.org

7, 23, 83, 227, 263, 359, 479, 503, 563, 1187, 2999, 3803, 4703, 4787, 4919, 5939, 6599, 8819, 10667, 14159, 16139, 16187, 18119, 21227, 22943, 25847, 26003, 26903, 27827, 29123, 29339, 29663, 36263, 43403, 44519, 44963, 46199, 47123, 48947, 49103, 49499

Offset: 1

Jon E. Schoenfield, Feb 26 2021

nonn

For each term p, p^5 - 1 = (p-1)*(p^4 + p^3 + p^2 + p + 1) is a number of the form 2*q*r (where q and r are distinct primes): p-1 = 2*q and p^4 + p^3 + p^2 + p + 1 = r.

Conjecture: sequence is infinite.

			     p =                       factorization
  n  a(n)      p^5 - 1          of (p^5 - 1)
  -  ----  --------------  ---------------------
  1     7           16806  2 *   3 *        2801
  2    23         6436342  2 *  11 *      292561
  3    83      3939040642  2 *  41 *    48037081
  4   227    602738989906  2 * 113 *  2666986681
  5   263   1258284197542  2 * 131 *  4802611441
  6   359   5963102065798  2 * 179 * 16656709681
  7   479  25216079618398  2 * 239 * 52753304641
  8   503  32198817702742  2 * 251 * 64141071121
  ...

Cf. A000005, A000040, A309906, A341664.

Select[Range[50000], PrimeQ[#] && DivisorSigma[0, #^5 - 1] == 8 &] (* Amiram Eldar, Feb 26 2021 *)

isok(p) = isprime(p) && (numdiv(p^5-1) == 8); \\ Michel Marcus, Feb 26 2021

A341666 Primes p such that p^6 - 1 has 384 divisors.

Original entry on oeis.org

29, 43, 59, 83, 157, 193, 317, 1093, 1373, 1523, 2803, 3557, 3677, 3733, 12227, 13093, 20507, 25933, 28163, 29243, 32443, 33493, 38603, 53917, 100523, 109883, 122117, 134363, 140197, 190573, 236723, 242773, 249397, 256757, 258403, 274237, 299723, 333283

Offset: 1

Jon E. Schoenfield, Feb 26 2021

nonn

Conjecture: sequence is infinite.

For every term p, p^6 - 1 is of the form 2^3 * 3^2 * 7 * q * r * s * t, where q, r, s, and t are distinct primes > 7, with four exceptions: p = 29, 59, 193, and 1373 (see Example section).

			  p =
n a(n)                factorization of p^6 - 1
- ---- ------------------------------------------------------
1   29 2^3 * 3^2 * 5 * 7   *  13 *     67 *     271
2   43 2^3 * 3^2     * 7   *  11 *     13 *     139 *     631
3   59 2^3 * 3^2 * 5 * 7   *  29 *    163 *    3541
4   83 2^3 * 3^2     * 7   *  19 *     41 *     367 *    2269
5  157 2^3 * 3^2     * 7   *  13 *     79 *    3499 *    8269
6  193 2^7 * 3^2     * 7   *  97 *   1783 *   37057
7  317 2^3 * 3^2     * 7   *  53 *     79 *   14401 *   33391
8 1093 2^3 * 3^2     * 7   *  13 *    547 *  398581 * 1193557
9 1373 2^3 * 3^2     * 7^3 * 229 * 627919 * 1886503

Cf. A000005, A000040, A309906, A341657, A341667.

Select[Range[350000], PrimeQ[#] && DivisorSigma[0, #^6 - 1] == 384 &] (* Amiram Eldar, Feb 27 2021 *)

isok(p) = isprime(p) && (numdiv(p^6-1) == 384); \\ Michel Marcus, Feb 27 2021

A341667 Primes p such that p^6 - 1 has fewer than 384 divisors.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 41, 53, 71, 73, 167

Offset: 1

Jon E. Schoenfield, Feb 26 2021

For all primes p > 167, p^6 - 1 has at least 384 divisors.

			      p =
   n  a(n)      factorization of p^6 - 1       tau(p^6 - 1)
  --  ----  ---------------------------------  ------------
   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    41  2^4 * 3^2 * 5 * 7 * 547 * 1723          240
  11    53  2^3 * 3^4 * 7 * 13 * 409 * 919          320
  12    71  2^4 * 3^3 * 5 * 7 * 1657 * 5113         320
  13    73  2^4 * 3^3 * 7 * 37 * 751 * 1801         320
  14   167  2^4 * 3^2 * 7 * 83 * 9241 * 28057       240

Cf. A000005, A000040, A309906, A341657, A341666.

Select[Range[200], PrimeQ[#] && DivisorSigma[0, #^6 - 1] < 384 &] (* Amiram Eldar, Feb 27 2021 *)

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A342062 a(n) is the number of divisors of prime(n)^8 - 1.

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A350780 Numbers that are the number of divisors of p^2 - 1 for some prime p.

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A342063 Primes p such that p^8 - 1 has fewer than 384 divisors.

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A342064 Primes p such that p^8 - 1 has 384 divisors.

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A341661 Primes p such that p^4 - 1 has fewer than 160 divisors.

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A341662 Primes p such that p^4 - 1 has 160 divisors.

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A341664 a(n) is the number of divisors of prime(n)^5 - 1.

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A341665 Primes p such that p^5 - 1 has 8 divisors.

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