A193432 -id:A193432 - cp's OEIS frontend

A361904 Odd numbers k such that for all even divisors d of k^2+1, d^2+1 is a prime number.

1, 3, 5, 45, 65, 175, 277, 345, 435, 573, 673, 695, 715, 875, 955, 985, 1095, 1255, 1405, 1495, 1515, 1845, 1915, 2035, 2135, 2315, 2375, 2525, 2687, 2805, 2837, 2965, 3035, 3665, 3715, 3725, 4185, 4225, 4265, 4345, 4495, 4635, 4865, 5987, 6195, 6205, 6315

Offset: 1

Michel Lagneau, Mar 28 2023

nonn

Conjecture: the sequence is infinite.
We observe two subsequences each having specific properties:
(i) a subsequence of numbers divisible by 5 such that a(n)^2+1 contains 4 divisors {1, 2, p, 2*p} including two even divisors 2 and 2*p = a(n)^2+1 where p is prime.
(ii) a subsequence of numbers > 3 not divisible by 5 such that a(n)^2+1 contains eight divisors {1, 2, 5, 10, q, 2*q, 5*q, 10*q} including four even divisors 2, 10, 2*q = (a(n)^2+1)/5 and 10*q = a(n)^2+1, where q is prime.

			5 is in the sequence because the divisors of 5^2+1 are {1, 2, 13, 26} and 2^2+1 = 5 is prime, 26^2+1 = 677 is prime.
277 is in the sequence because the divisors of 277^2+1 are {1, 2, 5, 10, 7673, 15346, 38365, 76730} and 2^2+1 = 5 is prime, 10^2+1 = 101 is prime, 15346^2+1 = 235499717 is prime, 76730^2+1 = 5887492901 is prime.

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

Cf. A000203, A002522, A193432.

q:= k-> andmap(d-> d::odd or isprime(d^2+1), numtheory[divisors](k^2+1)):
select(q, [2*i-1$i=1..4000])[];  # Alois P. Heinz, Mar 30 2023

Lst={};A=Table[Length[Select[Divisors[n^2+1],EvenQ]],{n,1,10000}];B=Array[DivisorSum[#^2+1,1&,And[EvenQ@#,PrimeQ[#^2+1]]&]&,10000];Do[If[A[[m]]==B[[m]]||B[m]!=0||B[m]!=0,AppendTo[Lst,m]],{m,1,10000,2}];Lst

isok(k) = if (k%2, my(d=select(x->!(x%2), divisors(k^2+1))); for (i=1, #d, if (!isprime(d[i]^2+1), return(0))); return(1)); \\ Michel Marcus, Mar 28 2023

from sympy import divisors, isprime
def ok(n): return n&1 and all(d&1 or isprime(d**2+1) for d in divisors(n**2+1))
print([k for k in range(7000) if ok(k)]) # Michael S. Branicky, Apr 17 2023

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.

A373903 Numbers k such that k^2 - 1 has fewer divisors than k^2 + 1.

Original entry on oeis.org

18, 72, 132, 138, 182, 192, 228, 242, 268, 278, 282, 327, 348, 360, 378, 382, 408, 418, 432, 438, 618, 632, 642, 660, 682, 684, 693, 718, 772, 788, 798, 822, 843, 858, 882, 918, 948, 957, 1032, 1048, 1068, 1092, 1113, 1118, 1143, 1152, 1227, 1228, 1230, 1282, 1292

Offset: 1

Amiram Eldar, Jun 22 2024

Numbers k such that A347191(k) < A193432(k).

			18 is a term since 18^2 - 1 = 323 has 4 divisors (1, 17, 19 and 323) while 18^2 + 1 = 325 has 6 divisors (1, 5, 13, 25, 65 and 325).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000005, A002522, A005563, A069062, A188629, A193432, A347191.

Select[Range[2, 1300], DivisorSigma[0, #^2 - 1] < DivisorSigma[0, #^2 + 1] &]

is(k) = k > 1 && numdiv(k^2 - 1) < numdiv(k^2 + 1);

A321519 Let d(n,i), i = 1..k be the k divisors of n^2 + 1 (the number 1 is not counted). a(n) is the number of ordered pairs d(n,i) < d(n,j) such that gcd(d(n,i), d(n,j)) = 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 1, 1, 6, 0, 1, 0, 6, 2, 1, 0, 6, 1, 6, 0, 1, 0, 6, 1, 1, 1, 6, 2, 6, 1, 1, 0, 6, 2, 1, 0, 2, 1, 11, 1, 1, 1, 25, 1, 1, 1, 1, 1, 6, 0, 6, 0, 16, 1, 1, 1, 1, 1, 6, 1, 1, 0, 6, 3, 1, 2, 1, 6, 25, 0, 6, 1, 6, 1, 1, 1, 6, 2, 25, 0, 1, 1

Offset: 1

Michel Lagneau, Nov 12 2018

nonn

Terms only depends on prime signature of n^2+1. - David A. Corneth, Nov 14 2018
We observe an interesting statistic for n <= 10^5: the four values of a(n) = 0, 1, 6, 25 represent more than 82% (see the table below).
a(A005574(n)) = 0, a(A085722(n)) = 1, a(A272078(n)) = 6, a(A316351(n)) = 25.
In the general case, a(k) = m if k^2+1 = p*q^m, m = 1, 2, 3, ... with p, q primes.
+--------------+-----------------------+------------+
|              | number of occurrences |            |
|     a(n)     |     for n <= 10^5     | percentage |
+--------------+-----------------------+------------+
|       0      |          6656         |    6.656%  |
|       1      |         23255         |   23.255%  |
|       6      |         31947         |   31.947%  |
|      25      |         20461         |   20.461%  |
| other values |         17681         |   17.681%  |
+--------------+-----------------------+------------+

			a(13) = 6 because the divisors {d(i)} of 13^2 + 1 = 170 (without the number 1)  are  {2, 5, 10, 17, 34, 85, 170}, and gcd(d(i), d(j)) = 1 for the 6 following pairs of elements of {d(i)}: (2, 5), (2, 17), (2, 85), (5, 17), (5, 34) and (10, 17).

Cf. A005574, A002522, A085722, A089233, A193432, A272078, A316351.

with(numtheory):nn:=10^3:
for n from 1 to nn do:
  it:=0:d:=divisors(n^2+1):n0:=nops(d):
   for k from 2 to n0-1 do:
    for l from k+1 to n0 do:
     if gcd(d[k],d[l])= 1
      then
      it:=it+1
      else
     fi:
   od:
  od:
  printf(`%d, `,it):
od:

f[n_] := (DivisorSigma[0, n^2] - 1)/2 - DivisorSigma[0, n] + 1; Map[f, Range[0,100]^2+1] (* Amiram Eldar, Nov 14 2018 after Robert G. Wilson v at A089233 *)

a(n) = {my(d=divisors(n^2+1)); sum(k=2, #d, sum(j=2, k-1, gcd(d[k], d[j]) == 1));} \\ Michel Marcus, Nov 12 2018

a(n) = A089233(n^2+1). - Michel Marcus, Nov 13 2018

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.

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.

cp's OEIS Frontend

A361904 Odd numbers k such that for all even divisors d of k^2+1, d^2+1 is a prime number.

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

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A373903 Numbers k such that k^2 - 1 has fewer divisors than k^2 + 1.

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A321519 Let d(n,i), i = 1..k be the k divisors of n^2 + 1 (the number 1 is not counted). a(n) is the number of ordered pairs d(n,i) < d(n,j) such that gcd(d(n,i), d(n,j)) = 1.

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

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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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