A309906 -id:A309906 - cp's OEIS frontend

A349937 Odd numbers k > 1 such that A309906(k-1) < A309906(k) > A309906(k+1).

315, 525, 693, 765, 825, 945, 1125, 1155, 1323, 1395, 1575, 1683, 1725, 1755, 1785, 1815, 1845, 1995, 2205, 2275, 2277, 2415, 2457, 2475, 2535, 2565, 2691, 2695, 2793, 2805, 2835, 2907, 3003, 3045, 3285, 3315, 3375, 3465, 3591, 3645, 3675, 3705, 3735, 3825, 3885

Offset: 1

Jianing Song, Dec 05 2021

nonn

Conjecturally, odd numbers k > 1 such that liminf_{n->oo} d(p(n)^(k-1)-1) < liminf_{n->oo} d(p(n)^k-1) > liminf_{n->oo} d(p(n)^(k+1)-1), where p(n) = prime(n), d = A000005.

If k is odd, then A079612(k) = 2, so A309906(k) is usually smaller than either A309906(k-1) or A309906(k+1) (or both). This sequence lists the exceptions.

			A309906(314) = 128 < A309906(315) = 8192 > A309906(316) = 2560, so 315 is a term.

Cf. A309906, A349938, A349941.

isA349937(k) = (k%2&&k>1) && A309906(k)>A309906(k-1) && A309906(k)>A309906(k+1) \\ See A309906 for its program

A349938 Odd numbers k > 1 such that A309906(k-1) < A309906(k) > A309906(k+1) < A309906(k+2) > A309906(k+3).

Original entry on oeis.org

2275, 11275, 16443, 34263, 42775, 42955, 47955, 49075, 49383, 53163, 55683, 58075, 61623, 69795, 70315, 70735, 71643, 76323, 77875, 83235, 88443, 90963, 100375, 102555, 103383, 107523, 108295, 110955, 112723, 113155, 113575, 120783, 124315, 127015, 128945, 136323

Offset: 1

Jianing Song, Dec 05 2021

nonn

Conjecturally, odd numbers k > 1 such that liminf_{n->oo} d(p(n)^(k-1)-1) < liminf_{n->oo} d(p(n)^k-1) > liminf_{n->oo} d(p(n)^(k+1)-1) < liminf_{n->oo} d(p(n)^(k+2)-1) > liminf_{n->oo} d(p(n)^(k+3)-1), where p(n) = prime(n), d = A000005.
Odd numbers k such that both k and k+2 are in A349937.
What's the smallest term congruent to 5 modulo 6? That is to say, what's the smallest k such that both k and k+2 are in A349941?

Jianing Song, Table of n, a(n) for n = 1..414 (all terms <= 10^6)

Cf. A309906, A349937.

isA349938(k) = if(k%2&&k>1, my(v=vector(5, n, A309906(k-2+n))); v[2]>v[1] && v[2]>v[3] && v[4]>v[3] && v[4]>v[5], 0) \\ See A309906 for its program

A349941 Terms of A349937 that are not divisible by 3: numbers k > 1 not divisible by 2 or 3 such that A309906(k-1) < A309906(k) > A309906(k+1).

Original entry on oeis.org

2275, 2695, 6125, 6545, 7735, 11165, 11275, 16445, 18473, 21175, 22253, 24115, 26455, 27115, 28985, 30485, 31255, 32585, 34265, 34675, 34925, 35035, 35275, 36725, 37037, 37625, 38525, 38885, 39715, 40565, 42775, 42955, 43225, 44275, 45175, 45353, 47047, 47957, 49075, 49385

Offset: 1

Jianing Song, Dec 05 2021

nonn

Conjecturally, numbers k > 1 not divisible by 2 or 3 such that liminf_{n->oo} d(p(n)^(k-1)-1) < liminf_{n->oo} d(p(n)^k-1) > liminf_{n->oo} d(p(n)^(k+1)-1), where p(n) = prime(n), d = A000005.

			A309906(2274) = 6144 < A309906(2275) = 8192 > A309906(2276) = 1280, 2275 is not divisible by 2 or 3, so 2275 is a term.
A309906(18472) = 6144 < A309906(18473) = 8192 > A309906(18474) = 6144, 18473 is not divisible by 2 or 3, so 18473 is a term.

Cf. A309906, A349937.

isA349941(k) = (k%2&&k%3&&k>1) && A309906(k)>A309906(k-1) && A309906(k)>A309906(k+1) \\ See A309906 for its program

A341655 a(n) is the number of divisors of prime(n)^2 - 1.

Original entry on oeis.org

2, 4, 8, 10, 16, 16, 18, 24, 20, 32, 28, 24, 40, 32, 24, 32, 32, 32, 32, 60, 30, 48, 32, 60, 42, 48, 40, 32, 64, 48, 54, 64, 40, 64, 48, 60, 32, 40, 40, 32, 48, 96, 64, 32, 72, 90, 64, 56, 32, 64, 60, 96, 72, 96, 40, 40, 64, 96, 32, 80, 32, 48, 96, 80, 40, 32

Offset: 1

Jon E. Schoenfield, Feb 25 2021

nonn

a(n) >= A309906(2) = 32 for n > 21.

			        p =                factorization
   n  prime(n)  p^2 - 1      of p^2 - 1      a(n)
  --  --------  -------  ------------------  ----
   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     29       840    2^3 * 3 * 5 * 7      32
  11     31       960    2^6 * 3 * 5          28
  12     37      1368    2^3 * 3^2 * 19       24
  13     41      1680    2^4 * 3 * 5 * 7      40
  14     43      1848    2^3 * 3 * 7 * 11     32
  15     47      2208    2^5 * 3 * 23         24
  16     53      2808    2^3 * 3^3 * 13       32
  17     59      3480    2^3 * 3 * 5 * 29     32
  18     61      3720    2^3 * 3 * 5 * 31     32
  19     67      4488    2^3 * 3 * 11 * 17    32
  20     71      5040    2^4 * 3^2 * 5 * 7    60
  21     73      5328    2^4 * 3^2 * 37       30
  22     79      6240    2^5 * 3 * 5 * 13     48
  23     83      6888    2^3 * 3 * 7 * 41     32
  24     89      7920    2^4 * 3^2 * 5 * 11   60

Cf. A000005, A000040, A084920, A309906.

Table[DivisorSigma[0,Prime[n]^2-1],{n,66}] (* Stefano Spezia, Feb 25 2021 *)

a(n) = numdiv(prime(n)^2-1); \\ Michel Marcus, Feb 25 2021

a(n) = A000005(A000040(n)^2 - 1) = A000005(A084920(n)).

A341656 a(n) is the number of divisors of prime(n)^4 - 1.

Original entry on oeis.org

4, 10, 20, 36, 40, 80, 84, 60, 96, 80, 128, 120, 144, 240, 224, 160, 80, 80, 160, 144, 288, 112, 320, 288, 192, 120, 192, 240, 320, 224, 240, 160, 192, 160, 240, 288, 480, 200, 192, 320, 240, 240, 576, 288, 360, 216, 320, 256, 160, 320, 576, 560, 336, 720, 264

Offset: 1

Jon E. Schoenfield, Feb 25 2021

nonn

a(n) >= A309906(4) = 160 for n > 26.

			        p =                       factorization
   n  prime(n)  p^4 - 1             of p^4 - 1            a(n)
  --  -------- ---------  ------------------------------  ----
   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     43      3418800  2^4 * 3 * 5^2 * 7 * 11 * 37      240
  15     47      4879680  2^6 * 3 * 5 * 13 * 17 * 23       224
  16     53      7890480  2^4 * 3^3 * 5 * 13 * 281         160
  17     59     12117360  2^4 * 3 * 5 * 29 * 1741           80
  18     61     13845840  2^4 * 3 * 5 * 31 * 1861           80
  19     67     20151120  2^4 * 3 * 5 * 11 * 17 * 449      160
  20     71     25411680  2^5 * 3^2 * 5 * 7 * 2521         144
  21     73     28398240  2^5 * 3^2 * 5 * 13 * 37 * 41     288
  22     79     38950080  2^6 * 3 * 5 * 13 * 3121          112
  23     83     47458320  2^4 * 3 * 5 * 7 * 13 * 41 * 53   320
  24     89     62742240  2^5 * 3^2 * 5 * 11 * 17 * 233    288
  25     97     88529280  2^7 * 3 * 5 * 7^2 * 941          192
  26    101    104060400  2^4 * 3 * 5^2 * 17 * 5101        120

Cf. A000005, A000040, A309906.

a[n_] := DivisorSigma[0, Prime[n]^4 - 1]; Array[a, 50] (* Amiram Eldar, Feb 25 2021 *)

a(n) = numdiv(prime(n)^4-1); \\ Michel Marcus, Feb 25 2021

from sympy import prime, divisor_count
def A341656(n): return divisor_count(prime(n)**4-1) # Chai Wah Wu, Feb 25 2021

a(n) = A000005(A000040(n)^4 - 1).

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

Jon E. Schoenfield, Feb 26 2021

nonn

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.

			      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

Cf. A000005, A000040, A309906, A341655.

Select[Range[6000], PrimeQ[#] && DivisorSigma[0, #^2 - 1] == 32 &] (* Amiram Eldar, Feb 26 2021 *)

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

Jon E. Schoenfield, Feb 26 2021

sign

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.

			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.

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

Jon E. Schoenfield, Feb 25 2021

nonn

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.)

			   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

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

a[n_] := DivisorSigma[0, Prime[n]^6 - 1]; Array[a, 50] (* Amiram Eldar, Feb 26 2021 *)

a(n) = numdiv(prime(n)^6-1); \\ Michel Marcus, Feb 26 2021

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

A341659 Primes p such that p^3 - 1 has 8 divisors.

Original entry on oeis.org

59, 167, 383, 839, 1487, 4259, 5087, 6047, 6599, 6719, 8543, 8963, 9743, 12227, 12647, 13163, 14087, 14867, 18947, 20123, 22643, 23099, 23159, 24083, 24239, 24659, 25583, 27107, 27299, 30203, 30803, 32507, 34319, 37463, 37799, 38603, 41879, 42839, 44519, 44687

Offset: 1

Jon E. Schoenfield, Feb 26 2021

nonn

Intersection of A005385 (Safe primes p: (p-1)/2 is also prime) and A053182 (Primes p such that p^2 + p + 1 is prime).
For each term p, p^3 - 1  = (p-1)*(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^2 + p + 1 = r.
Conjecture: sequence is infinite.

			     p =                    factorization
  n  a(n)    p^3 - 1         of (p^3 - 1)
  -  ----  ------------  -------------------
  1    59        205378  2 *   29 *     3541
  2   167       4657462  2 *   83 *    28057
  3   383      56181886  2 *  191 *   147073
  4   839     590589718  2 *  419 *   704761
  5  1487    3288008302  2 *  743 *  2212657
  6  4259   77254345978  2 * 2129 * 18143341
  7  5087  131639193502  2 * 2543 * 25882657
  8  6047  221115865822  2 * 3023 * 36572257
  9  6599  287365339798  2 * 3299 * 43553401

David A. Corneth, Table of n, a(n) for n = 1..6183

Cf. A000005, A000040, A005385, A053182, A309906.

Select[Range[50000], PrimeQ[#] && DivisorSigma[0, #^3 - 1] == 8 &] (* Amiram Eldar, Feb 26 2021 *)
Select[Prime[Range[5000]],DivisorSigma[0,#^3-1]==8&] (* Harvey P. Dale, Apr 17 2025 *)

isok(p) = isprime(p) && (numdiv(p^3-1) == 8); \\ Michel Marcus, Feb 26 2021

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

Jon E. Schoenfield, Feb 26 2021

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

			      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

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

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

cp's OEIS Frontend

A349937 Odd numbers k > 1 such that A309906(k-1) < A309906(k) > A309906(k+1).

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A349938 Odd numbers k > 1 such that A309906(k-1) < A309906(k) > A309906(k+1) < A309906(k+2) > A309906(k+3).

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A349941 Terms of A349937 that are not divisible by 3: numbers k > 1 not divisible by 2 or 3 such that A309906(k-1) < A309906(k) > A309906(k+1).

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A341655 a(n) is the number of divisors of prime(n)^2 - 1.

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A341656 a(n) is the number of divisors of prime(n)^4 - 1.

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

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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.

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

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A341659 Primes p such that p^3 - 1 has 8 divisors.