A347192 -id:A347192 - cp's OEIS frontend

A347191 Number of divisors of n^2-1.

2, 4, 4, 8, 4, 10, 6, 10, 6, 16, 4, 16, 8, 12, 8, 18, 4, 24, 8, 16, 8, 20, 6, 20, 12, 16, 8, 32, 4, 28, 8, 14, 16, 24, 8, 24, 8, 20, 8, 40, 4, 32, 12, 16, 12, 24, 6, 36, 12, 24, 8, 32, 8, 40, 16, 20, 8, 32, 4, 32, 12, 16, 24, 32, 8, 32, 8, 32, 8, 60, 4, 30, 12, 16, 24, 32, 8, 48, 10, 24

Offset: 2

Bernard Schott, Aug 22 2021

nonn

Inspired by problem A1885 in Diophante (see link).
As n^2-1 > 0 is never square, all terms are even.
a(n) = 2 iff n = 2.
a(n) = 4 iff n = 3 or iff n is average of twin prime pairs 'n-1' and 'n+1'; i.e. n is a member of ({3} Union A014574) or equivalently n is a term of A129297 \ {0,1,2}.
a(n) = 6 iff n is such that the two adjacent integers of n are a prime and a square of another prime: 8, 10, 24, 48, 168, 360, ... (A347194).

			a(5) = tau(5^2-1) = tau(24) = 8.
a(18) = tau(18^2-1) = tau(17*19) = 4, 18 is average of twin primes 17 and 19.

Amiram Eldar, Table of n, a(n) for n = 2..10000
Diophante, A1885, Cachés derrière leurs diviseurs (in French).
Adrian W. Dudek, On the number of divisors of n^2-1, Bulletin of the Australian Mathematical Society, Vol. 93, No. 2 (2016), pp. 194-198; arXiv preprint, arXiv:1507.08893 [math.NT], 2015.

Cf. A000005, A005563, A014574, A129296, A129297.

Cf. A347192 (records), A347193 (smallest k with a(k) = n), A347194 (a(n)=6).

with(numtheory):
seq(tau(n^2-1), n=2..81);

a[n_] := Length[Divisors[n^2 - 1]]; Table[a[n], {n, 2, 81}] (* Robert P. P. McKone, Aug 22 2021 *)
Table[DivisorSigma[0, n^2 - 1], {n, 2, 100}] (* Vaclav Kotesovec, Aug 23 2021 *)

a(n) = numdiv(n^2-1); \\ Michel Marcus, Aug 23 2021

a(n)=my(a=valuation(n-1,2),b=valuation(n+1,2)); numdiv((n-1)>>a)*numdiv((n+1)>>b)*(a+b+1) \\ Charles R Greathouse IV, Sep 17 2021

first(n)=my(v=vector(n-1),x=[1,factor(1)],y=[2,factor(2)]); forfactored(k=3,n+1,  my(e=max(valuation(x[1],2), valuation(k[1],2))); v[k[1]-2]=numdiv(k)*numdiv(x)*(e+2)/(2*e+2); x=y; y=k); v \\ Charles R Greathouse IV, Sep 17 2021

from math import prod
from sympy import factorint
def a(n):
    ft = factorint(n+1, multiple=True) + factorint(n-1, multiple=True)
    return prod((e + 1) for e in (ft.count(f) for f in set(ft)))
print([a(n) for n in range(2, 82)]) # Michael S. Branicky, Sep 17 2021

a(n) = A000005(A005563(n-1)).
a(n) = 2 * A129296(n).
Sum_{k=2..n} a(k) ~ (6/Pi^2) * n*log(n)^2 (Dudek, 2016). - Amiram Eldar, Apr 07 2023

A347193 a(n) is the smallest m such that A347191(m) = 2*n, where A347191(m) = tau(m^2 - 1).

Original entry on oeis.org

2, 3, 8, 5, 7, 15, 33, 11, 17, 23, 513, 19, 13841287200, 31, 73, 29, 650377879817809571042122834560, 49, 131073, 41, 97, 1537, 31381059608, 79, 50626, 10239, 127, 223, 459986536544739960976800, 71, 8193465725814765556554001028792218848, 109, 61953, 163839, 161

Offset: 1

Bernard Schott, Sep 17 2021

nonn

Generalization of the questions proposed in Diophante problem A1885 (see links).
When p is prime, as a(p) is the smallest m such that tau(m^2 - 1) = 2*p, hence m^2 - 1 is of the form q * r^(p-1) with q > r primes, so we must solve the Diophantine equation (m-1)*(m+1) = q * r^(p-1) to get the smallest m, when only p is known.
Two cases must be checked:
-> r = 2: if 2^(p-3) + 1 is prime, then m = 2^(p-2) + 1, and
          if 2^(p-3) - 1 is prime, then m = 2^(p-2) - 1.
If there is a solution m in the case r = 2, then a(p) is this smallest solution m (see example a(11)); if there is no solution m with r = 2, then try the 2nd case.
-> r odd prime: if r^(p-1) + 2 is prime, then m = r^(p-1) + 1, and
                if r^(p-1) - 2 is prime, then m = r^(p-1) - 1 (example a(13)).

			tau(2^2 - 1) = 2 = 2*1, so a(1) = 2.
tau(3^2 - 1) = 4 = 2*2, so a(2) = 3.
tau(4^2 - 1) = 4 = 2*2, tau(5^2 - 1) = 8 = 2*4 so a(4) = 5.
For a(11): if r = 2, 2^8 + 1 = 257 is prime, while 2^8 - 1 is not prime, hence a(11) = 2^9 + 1 = 513.
For a(13):
  if r = 2, 2^10 +- 1 are not prime, so not possible;
  if r = 3, 3^12 +- 2 are not prime, so not possible;
  if r = 5, 5^12 +- 2 are not prime, so not possible;
  if r = 7, 7^12 - 2 = 13841287199 is prime, while 7^12 + 2 is not prime, then a(13) = 13841287199+1 = 13841287200.

Diophante, A1885, Cachés derrière leurs diviseurs (in French).
Adrian Dudek, On the Number of Divisors of n^2-1, arXiv:1507.08893 [math.NT], 2015.

Cf. A000005, A347191, A347192, A347194.

a(31)-a(35) from Jinyuan Wang, Sep 23 2021

A347194 Numbers such that the two adjacent integers are a prime and the square of another prime.

Original entry on oeis.org

8, 10, 24, 48, 168, 360, 840, 1368, 1848, 2208, 3720, 5040, 7920, 10608, 11448, 16128, 17160, 19320, 29928, 36480, 44520, 49728, 54288, 57120, 66048, 85848, 97968, 113568, 128880, 177240, 196248, 201600, 218088, 241080, 273528, 292680, 323760, 344568, 368448, 426408, 458328, 516960, 528528, 537288, 552048, 564000, 573048, 579120

Offset: 1

Bernard Schott, Sep 23 2021

nonn

-> Equivalently, numbers k such that tau(k^2-1) = A347191(k) = 6 (see example; used for Maple code).
Proof: tau(k^2-1) = 6 <==> k^2-1 = p^5 or k^2-1 = p*q^2 with p <> q primes; but k^2-p^5 = 1 is impossible, as a consequence of the Catalan-Mihăilescu theorem; now, (k-1)*(k+1) = p*q^2 ==> (k-1 = p and k+1 = q^2) or (k-1 = q^2 and k+1 = p), because k-1 = q and k+1 = p*q is not possible, otherwise 2 = q*(p-1), which would contradict p <> q.
-> There are two possible configurations with p, q primes: (q^2 < a(n) < p) or (p < a(n) < q^2).
The unique configuration q^2 < a(n) < p is for q = 3, a(2) = 10 and  p = 11.
All the other configurations, for n = 1 or n >= 3, are of the form p < a(n) < q^2 with p = A049002(n) and q = A062326(n).
-> Note that there is only one integer such that the two adjacent integers are a prime and the square of that prime: it is 3, which lies between 2 and 2^2; in this case, tau(3^2-1) = 4.

			8 is a term since 8 lies between 7 (prime) and 9 = 3^2 (square of prime); also tau(8^2-1) = tau(63) = 6.
10 is a term since 10 lies between 9 = 3^2 (square of prime) and 11 (prime); also tau(10^2-1) = tau(99) = 6.
24 is a term since 24 lies between 23 (prime) and 25 = 5^2 (square of prime); also tau(24^2-1) = tau(575) = 6.

Diophante, A1885, Cachés derrière leurs diviseurs (in French).
Adrian Dudek, On the Number of Divisors of n^2-1, arXiv:1507.08893 [math.NT], 2015.
Wikipedia, Catalan's conjecture.

Cf. A049002, A062326, A146981.
Cf. A347191, A347192, A347193.
Subsequence of A163492 (between prime and a perfect square).

with(numtheory):
filter := q-> tau(q^2-1) = 6 : select(filter, [$2..580000]);

q[n_] := Module[{e1 = FactorInteger[n - 1][[;; , 2]], e2 = FactorInteger[n + 1][[;; , 2]]}, (e1 == {1} && e2 == {2}) || (e1 == {2} && e2 == {1})]; Select[Range[4, 600000], q] (* Amiram Eldar, Sep 23 2021 *)

isok(m) = my(pa, pb); (isprimepower(m-1, &pa)*isprimepower(m+1, &pb) == 2) && (pa != pb); \\ Michel Marcus, Sep 23 2021

upto(n) = { my(res = List()); forprime(i = 3, sqrtint(n-1), if(isprime(i^2 - 2), listput(res, i^2-1); ); if(isprime(i^2 + 2), listput(res, i^2 + 1); ) ); res } \\ David A. Corneth, Sep 23 2021

For n >= 3: a(n) = A049002(n) + 1 = a(n) = A146981(n) - 1 = (A049002(n) + A146981(n))/2 = A062326(n)^2 - 1.

cp's OEIS Frontend

A347191 Number of divisors of n^2-1.

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A347193 a(n) is the smallest m such that A347191(m) = 2*n, where A347191(m) = tau(m^2 - 1).

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A347194 Numbers such that the two adjacent integers are a prime and the square of another prime.

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Keywords

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Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula