Jon E. Schoenfield - cp's OEIS frontend

A374793 a(n) is the largest k such that tau(k)^n >= k.

2, 1260, 27935107200, 29564884570506808579056000

Offset: 1

Jon E. Schoenfield, Jul 20 2024

Let prime(j)# denote the product of the first j primes, A002110(j); then
  a(1)  = prime(1)# = 2,
  a(2)  = 6*prime(4)# = 1260,
  a(3)  = 2880*prime(8)# = 2.7935...*10^10,
  a(4)  = 907200*prime(16)# = 2.9564...*10^25,
  a(5) >= 259459200*prime(30)# = 8.2015...*10^54,
  a(6) >= 3238237626624000*prime(52)# = 3.4403...*10^111,
  a(7) >= 248818180782850398720000*prime(91)# = 5.4351...*10^218.

			27935107200 = 2^7 * 3^3 * 5^2 * 7^1 * 11^1 * 13^1 * 17^1 * 19^1,
so tau(27935107200) = (7+1)*(3+1)*(2+1)*(1+1)*(1+1)*(1+1)*(1+1)*(1+1) = 8*4*3*2*2*2*2*2 = 3072; 3072^3 = 28991029248 > 27935107200, and there is no larger number k such that tau(k)^3 >= k, so a(3) = 27935107200.

Cf. A000005, A035033, A056757, A056758.

A374592 Numbers k such that 3k^4 - 3k^2 + 1 is prime.

Original entry on oeis.org

2, 5, 7, 8, 9, 14, 15, 20, 23, 30, 36, 37, 43, 48, 49, 50, 54, 56, 57, 69, 71, 79, 85, 86, 91, 93, 97, 98, 106, 111, 112, 119, 124, 128, 131, 133, 134, 135, 140, 154, 159, 162, 167, 180, 181, 198, 204, 208, 212, 226, 232, 236, 246, 259, 278, 281, 285, 286, 288

Offset: 1

Jon E. Schoenfield, Jul 12 2024

nonn

Equivalently, numbers k such that there exists a prime of the form k^6 - m^3. Proof: Let d = k^2 - m. Then m = k^2 - d, so k^6 - m^3 = k^6 - (k^2 - d)^3 = k^6 - (k^6 - 3*k^4*d + 3*k^2*d^2 - d^3) = d*(3*k^4 - 3*k^2*d + d^2), which cannot be prime unless d = 1, i.e., k^6 - m^3 = 3*k^4 - 3*k^2 + 1.

A374403 Number of n-bit primes.

Original entry on oeis.org

0, 2, 2, 2, 5, 7, 13, 23, 43, 75, 137, 255, 464, 872, 1612, 3030, 5709, 10749, 20390, 38635, 73586, 140336, 268216, 513708, 985818, 1894120, 3645744, 7027290, 13561907, 26207278, 50697537, 98182656, 190335585, 369323305, 717267168, 1394192236, 2712103833, 5279763824

Offset: 1

Jon E. Schoenfield, Jul 07 2024

nonn

Number of primes whose binary expansion is n digits long.
a(n) is the number of primes in the half-open interval [2^(n-1), 2^n).
First differences of A185192.
See A007053 for additional information.

			a(1) = 0 because neither 0 nor 1 is a prime.
a(2) = 2 because the 2-bit primes are 10_2 = 2 and 11_2 = 3.
a(4) = 2 because the 4-bit primes are 1011_2 = 11 and 1101_2 = 13.

Essentially the same as A036378 and A162145.
Cf. A185192 (partial sums).
Cf. A000720, A007053.

a[n_]:=Sum[Boole[PrimeQ[i]],{i,2^(n-1),2^n-1}]; Array[a,38] (* Stefano Spezia, Jul 07 2024 *)

a(n) = A162145(n) for n >= 2. - Amiram Eldar, Jul 08 2024

A374264 a(n) is the smallest m such that A374167(m) = n, or -1 if no such m exists.

Original entry on oeis.org

2, 1, 15, 36, 3120, 551, 9608, 2521095, 17338312921

Offset: 2

Jon E. Schoenfield, Jul 02 2024

For n > 9, a(n) exceeds primepi(10^10) = 455052511 or is -1.

Conjecture: a(n) is never -1.

			a(5) = 36 because the first 5 that appears in A374167 is A374167(36).

Cf. A235266, A374167.

a(10) from Michael S. Branicky, Jul 05 2024

A374167 a(n) is the largest k such that the binary representation of prime(n), when interpreted as a base-b number, yields a prime for every b in 2..k.

Original entry on oeis.org

3, 2, 2, 3, 3, 3, 2, 2, 2, 2, 2, 2, 3, 2, 4, 2, 2, 2, 3, 2, 3, 3, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 5, 2, 2, 2, 5, 2, 5, 5, 3, 2, 2, 3, 3, 3, 4, 5, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 5, 2

Offset: 1

Jon E. Schoenfield, Jun 29 2024

a(n) = 2 iff prime(n) is not in A235266.

Conjecture: every integer > 1 eventually appears.

			a(1) = 3 because prime(1) = 2 = 10_2, and both 10_2 = 2 and 10_3 = 3 are primes, but 10_4 = 4 = 2*2.
a(36) = 5 because prime(36) = 151 = 10010111_2, and
  10010111_2 = 151 is a prime
  10010111_3 = 2281 is a prime,
  10010111_4 = 16661 is a prime, and
  10010111_5 = 78781 is a prime, but
  10010111_6 = 281275 = 5^2 * 11251.

Cf. A235266, A374264.

a(n) = my(v=binary(prime(n)), b=2); while (isprime(fromdigits(v, b)), b++); b-1; \\ Michel Marcus, Jul 02 2024

a(n) = max{ k : Sum_{j=0..D} b^j * d_j is a prime for b = 2..k } where the digits of the binary representation of n are d_D, d_{D-1}, ..., d_1, d_0.

A373756 Table read by antidiagonals: T(n,k) is the smallest m > 1 such that m^2 - 1 and m^2 + 1 have 2n and 2k divisors, respectively, or -1 if no such m exists.

Original entry on oeis.org

2, 4, -1, 10, 3, -1, 14, 8, 18, -1, 28560, 5, 168, 72, -1, 26, 9, 32, 360, 16068, -1, 25071688922457240, 15, 7, 68, 369465818568, 1620, -1, 56, 728, 332, 28398240, 182, 744768, 1407318, -1, 170, 11, 161245807967271241368, 98, 248872305817685706212070112080, 132, 4175536688568, 642, -1

Offset: 1

Jon E. Schoenfield, Jun 16 2024

m=1 is excluded because m^2 - 1 would be 0.
For all m > 1, both m^2 - 1 and m^2 + 1 are nonsquares, so each has an even number of divisors.
For k=1, m^2 + 1 is a prime, so T(n,1) == 0 (mod 2) for all n.
For n=1, m^2 - 1 = (m-1)*(m+1) is a prime, which occurs only at m=2; 2^2 + 1 = 5 is also a prime, so T(1,1) = 2 and T(1,k) = -1 for k > 1.
For n=2, m^2 - 1 = (m-1)*(m+1) has 4 divisors, so (except for T(2,2) = 3) T(2,k) is the average of a twin prime pair (A014574).
Is T(n,k) > 0 for all n > 1?

			T(5,1) is the smallest integer m > 1 such that m^2 - 1 and m^2 + 1 have 10 and 2 divisors, respectively; since m^2 - 1 cannot be the 9th power of a prime, this requires that p^4 * q + 1 = m^2 = r - 1, where p, q, and r are distinct primes. The smallest such m is 28560, which gives a solution with p = 13, q = 28559, r = 815673601.
T(5,5) is the smallest integer m > 1 such that m^2 - 1 and m^2 + 1 each have 10 divisors; since neither m^2 - 1 nor m^2 + 1 can be the 9th power of a prime, this is the smallest m such that p^4 * q + 1 = m^2 = r^4 * s - 1, where p, q, r, and s are distinct primes: 22335421^4 * 248872305817685706212070112079 + 1 = 248872305817685706212070112080^2 = 13^4 * 2168601400616633822685176617536070987718973054081571441 - 1.
The first eight antidiagonals of the table are shown below.
.
  n\k|                 1   2   3        4            5      6       7  8
  ---+------------------------------------------------------------------
   1 |                 2  -1  -1       -1           -1     -1      -1 -1
   2 |                 4   3  18       72        16068   1620 1407318
   3 |                10   8 168      360 369465818568 744768
   4 |                14   5  32       68          182
   5 |             28560   9   7 28398240
   6 |                26  15 332
   7 | 25071688922457240 728
   8 |                56

Cf. A000005, A002522, A005563, A014574, A069062, A193432, A347191.

Define f(m) = tau(m^2 - 1) and g(m) = tau(m^2 + 1), where tau is the number of divisors function (A000005). Then
    T(n,k) = min_{ m : f(m) = 2n and g(m) = 2k },
             or -1 if no such m exists.

A373213 Numbers k such that k^2 - 1 and k^2 + 1 have 6 divisors each.

Original entry on oeis.org

168, 1368, 97968, 10374840, 16104168, 44049768, 68674368, 100741368, 281803368, 486775968, 1177381968, 1262878368, 1336852968, 2321986968, 2404627368, 3476635368, 4374102768, 5102102040, 5142754368, 5182128168, 5385651768, 6035269968, 9218496168, 10657878168

Offset: 1

Jon E. Schoenfield, Jun 21 2024

nonn

Each term is a number of the form k = sqrt(p^2 * q + 1) such that q = p^2 - 2 and k^2 + 1 = r^2 * s, where p, q, r, and s are distinct primes.

			168 is a term: both 168^2 - 1 = 28223 = 13^2 * 167 and 168^2 + 1 = 28225 = 5^2 * 1129 have 6 divisors.

Cf. A000005, A002522, A005563, A069062, A108278, A193432, A347191, A373209.

{ k : tau(k^2 - 1) = tau(k^2 + 1) = 6}, where tau() is the number of divisors function, A000005.

A373209 Numbers k such that k^2 - 1 and k^2 + 1 have 8 divisors each.

Original entry on oeis.org

68, 112, 128, 162, 200, 212, 252, 294, 318, 336, 338, 372, 448, 450, 498, 502, 542, 578, 592, 598, 612, 648, 672, 678, 708, 752, 762, 808, 812, 852, 878, 888, 938, 952, 992, 996, 1012, 1038, 1098, 1102, 1116, 1122, 1188, 1202, 1212, 1248, 1258, 1328, 1362, 1380

Offset: 1

Jon E. Schoenfield, Jun 21 2024

nonn

Among the first 10000 terms (from a(1) = 68 through a(10000) = 697578), k^2 - 1 and k^2 + 1 are each the product of three distinct primes, except for
     125 terms for which k^2 + 1 =  5^3 times a prime
       6 terms for which k^2 + 1 = 13^3 times a prime
       1 terms for which k^2 + 1 = 17^3 times a prime
       1 terms for which k^2 + 1 = 29^3 times a prime, and
       4 terms for which k^2 - 1 =  p^3 * (p^3 +/- 2) (with p = 19, 29, 37, 83, respectively).
The first term for which both k^2 - 1 and k^2 + 1 are of the form p^3 * q is k = 41457661182: k^2 - 1 = 3461^3 * 41457661183, while k^2 + 1 = 5^3 * 13749901365452077097.

			68 is a term: both 68^2 - 1 = 4623 = 3 * 23 * 67 and 68^2 + 1 = 4625 = 5^3 * 37 have 8 divisors.

Cf. A000005, A002522, A005563, A069062, A108278, A193432, A347191.

{ k : tau(k^2 - 1) = tau(k^2 + 1) = 8}, where tau() is the number of divisors function, A000005.

A373560 a(n) is the smallest multiple of prime(n)^2 that starts a run of 5 consecutive integers with 6 divisors, or -1 if no such multiple exists.

Original entry on oeis.org

-1, -1, -1, 10093613546512321, -1, -1, 7700031346933907521, -1, 5344962129269790721, -1, 20453982425165652721, -1, 8163195338222675521, -1, 2467958104789157112721, -1, -1, -1, -1, 14666767069023896053921, 212170739123852995921, 287954235303137500060321, -1, 84769922583214545304321

Offset: 1

Jon E. Schoenfield, Jun 09 2024

sign

Terms were obtained using the b-file at A141621.
a(n) = -1 if prime(n) is not in A001132.
Conjecture: the converse is also true.

			a(1) = a(2) = a(3) = -1 because the first of five consecutive integers having six divisors is never a multiple of 2^2, 3^2, or 5^2.
a(4) = 10093613546512321 because it is the smallest term in A141621 that is a multiple of prime(4)^2 = 49.
a(9) = 5344962129269790721 because it is the smallest term in A141621 that is a multiple of prime(9)^2 = 23^2.

Cf. A001132, A030515, A042999, A054753, A141621.

A373348 Squarefree numbers k such that there exists a k-digit multiple of k that has k divisors.

Original entry on oeis.org

1, 11, 206, 500015

Offset: 1

Jon E. Schoenfield, Jun 01 2024

Squarefree numbers k such that A182749(k) and A182750(k) are nonzero.
a(2) = 11 is the only prime term.
Conjecture: a(3) = 206 = 2*103 and a(4) = 500015 = 5*100003 are the only semiprime terms.

			The table below lists the known terms k and, for each k, the corresponding multiple:
.
  n   a(n) = k   k-digit multiple of k having k divisors
  -   --------   -----------------------------------------
  1          1   1
  2         11   11^10 = 25937424601
  3        206   2 * 103^102 = 4.0778...*10^205
  4     500015   5^4 * 100003^100002 = 1.2553...*10^500014

Cf. A182749, A182750.

cp's OEIS Frontend

User: Jon E. Schoenfield

A374793 a(n) is the largest k such that tau(k)^n >= k.

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A374592 Numbers k such that 3*k^4 - 3*k^2 + 1 is prime.

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A374403 Number of n-bit primes.

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A374264 a(n) is the smallest m such that A374167(m) = n, or -1 if no such m exists.

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A374167 a(n) is the largest k such that the binary representation of prime(n), when interpreted as a base-b number, yields a prime for every b in 2..k.

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A373756 Table read by antidiagonals: T(n,k) is the smallest m > 1 such that m^2 - 1 and m^2 + 1 have 2n and 2k divisors, respectively, or -1 if no such m exists.

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A373213 Numbers k such that k^2 - 1 and k^2 + 1 have 6 divisors each.

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A373209 Numbers k such that k^2 - 1 and k^2 + 1 have 8 divisors each.

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A373560 a(n) is the smallest multiple of prime(n)^2 that starts a run of 5 consecutive integers with 6 divisors, or -1 if no such multiple exists.

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A373348 Squarefree numbers k such that there exists a k-digit multiple of k that has k divisors.

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A374592 Numbers k such that 3k^4 - 3k^2 + 1 is prime.