A297402 a(n) = gcd_{k=1..n} (prime(k+1)^n-1)/2.

1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 32, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 64, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 32, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 128, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 32, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 64, 1, 4, 1, 8

Offset: 1

Author

Frank M Jackson, Dec 29 2017

Keywords

• nonn
• easy
• mult

Comments

If p is an odd prime and p^n is the length of the odd leg of a primitive Pythagorean triangle it constrains the other leg and hypotenuse to be (p^(2n)-1)/2 and (p^(2n)+1)/2. The resulting triangle has a semiperimeter of p^n(p^n+1)/2, an area of (p^n-1)p^n(p^n+1)/4 and an inradius of (p^n-1)/2. a(n) equals the GCD of the inradius terms (p^n-1)/2 for at least the first n odd primes.

Conjecture: a(n) equals the GCD of the inradius terms (p^n-1)/2 for all odd primes, i.e. a(n) = GCD_{k=1..oo} (prime(k+1)^n-1)/2.

From David A. Corneth, Dec 29 2017: (Start)

All terms are powers of 2. Proof: suppose p | a(n) for some odd prime p. Then p | (p^n - 1) / 2 and so p | (p^n - 1) which isn't the case.

If n is odd then a(n) = 1. Proof: 2 | (p^k - 1) for all k and odd primes p. 3^n - 1 = 3 * 9^k - 1 = 3 - 1 = 2 (mod 4), so 3^n - 1 is of the form 2*m for some odd m, hence the GCD of all (p^n - 1) / 2 is 1 for odd n. (End)

This is the even bisection of A059159. - Rémy Sigrist, Dec 30 2017

a(n) is the size of the group Z_2*/(Z_2*)^n, where Z_2 is the ring of 2-adic integers. We have that Z_2*/(Z_2*)^n is the inverse limit of (Z/2^iZ)*/((Z/2^iZ)*)^n as i tends to infinity. If n is odd, then the group is trivial. If n = 2^e * n' is even, where n' is odd, then the group is the product of a cyclic group of order 2^e and a cyclic group of order 2. See A370050. - Jianing Song, May 12 2024

Examples

			a(4)=8 because for n=4 and for the first 4 odd primes {3, 5, 7, 11}, the term (p^n-1)/2 gives {40, 312, 1200, 7320} with a GCD of 8.
		

Links

• Frank M Jackson, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001620, A006519, A059159, A074723, A127922, A171977, A202318.

Programs

Mathematica

a[n_] := GCD @@ Array[(Prime[# +1]^n -1)/2 &, n]; Array[a, 90] (* slightly modified by Robert G. Wilson v, Jan 01 2018 *)
a[n_] := If[EvenQ[n], 2^(FactorInteger[n][[1]][[2]] + 1), 1]; Array[a, 90] (* Frank M Jackson, Jul 28 2018 *)

PARI

a(n) = gcd(vector(n, i, (prime(i+1)^n-1)/2)) \\ Iain Fox, Dec 29 2017

PARI

a(n)=if(n%2,1,2)<Charles R Greathouse IV, Jan 06 2018

Formula

It appears that for k > 0, a(2^k) = 2^(k+1).

a(n) = A006519(2n) for even n and a(n) = 1 for odd n. - David A. Corneth, Dec 29 2017

a(n) = A074723(n)/2. - Iain Fox, Dec 30 2017

Multiplicative with a(2^e) = 2^(e+1), a(p^e) = 1 for odd prime p. - Andrew Howroyd, Jul 25 2018

It appears that for m > 0, a(2m-1) = 1 (proved in comments) and a(2m) = 2^(k+1) where k is the exponent of the even prime in the prime factorization of 2m. - Frank M Jackson, Jul 28 2018

From Amiram Eldar, Nov 24 2023: (Start)

Dirichlet g.f.: zeta(s) * (1 + 1/2^s + 1/(2^(s-1) - 1)).

Sum_{k=1..n} a(k) ~ (n/log(2)) * (log(n) + gamma + log(2) - 1), where gamma is Euler's constant (A001620). (End)