|
%I #16 Mar 04 2021 01:43:47
%S 6,16,48,60,192,96,192,256,360,384,504,512,240,384,576,320,384,768,
%T 576,320,320,864,384,640,504,1152,960,1280,1280,576,576,768,960,768,
%U 1152,720,384,768,240,768,2048,2048,2304,384,1536,1920,3072,672,1152,1536,1280
%N a(n) is the number of divisors of prime(n)^6 - 1.
%C a(n) >= A309906(6) = 384 for n > 39.
%C 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.)
%C 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.)
%F a(n) = A000005(A000040(n)^6 - 1).
%e n prime(n) factorization of prime(n)^6 - 1 a(n)
%e -- -------- ----------------------------------- ----
%e 1 2 3^2 * 7 6
%e 2 3 2^3 * 7 * 13 16
%e 3 5 2^3 * 3^2 * 7 * 31 48
%e 4 7 2^4 * 3^2 * 19*43 60
%e 5 11 2^3 * 3^2 * 5 * 7 * 19*37 192
%e 6 13 2^3 * 3^2 * 7 * 61*157 96
%e 7 17 2^5 * 3^3 * 7 * 13*307 192
%e 8 19 2^3 * 3^3 * 5 * 7^3 * 127 256
%e 9 23 2^4 * 3^2 * 7 * 11*13^2*79 360
%e 10 29 2^3 * 3^2 * 5 * 7 * 13*67*271 384
%e 11 31 2^6 * 3^2 * 5 * 7^2 * 19*331 504
%e 12 37 2^3 * 3^3 * 7 * 19*31*43*67 512
%e 13 41 2^4 * 3^2 * 5 * 7 * 547*1723 240
%e 14 43 2^3 * 3^2 * 7 * 11*13*139*631 384
%e 15 47 2^5 * 3^2 * 7 * 23*37*61*103 576
%e 16 53 2^3 * 3^4 * 7 * 13*409*919 320
%e 17 59 2^3 * 3^2 * 5 * 7 * 29*163*3541 384
%e 18 61 2^3 * 3^2 * 5 * 7 * 13*31*97*523 768
%e 19 67 2^3 * 3^2 * 7^2 * 11*17*31*4423 576
%e 20 71 2^4 * 3^3 * 5 * 7 * 1657*5113 320
%e 21 73 2^4 * 3^3 * 7 * 37*751*1801 320
%e ...
%e 39 167 2^4 * 3^2 * 7 * 83*9241*28057 240
%t a[n_] := DivisorSigma[0, Prime[n]^6 - 1]; Array[a, 50] (* _Amiram Eldar_, Feb 26 2021 *)
%o (PARI) a(n) = numdiv(prime(n)^6-1); \\ _Michel Marcus_, Feb 26 2021
%Y Cf. A000005, A000040, A309906, A341655, A341656.
%K nonn
%O 1,1
%A _Jon E. Schoenfield_, Feb 25 2021
|
