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

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

Author

Jon E. Schoenfield, Feb 25 2021

Keywords

• nonn

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).