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

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

Author

Bernard Schott, Sep 17 2021

Keywords

• nonn

Comments

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)).

Examples

			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.
		

Links

• 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.

Crossrefs

Cf. A000005, A347191, A347192, A347194.

Extensions

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