A318760 -id:A318760 - cp's OEIS frontend

A318761 Composite k that divides 2^(k-2) + 3^(k-2) + 6^(k-2) - 1.

4, 6, 8, 12, 24, 25, 125, 174, 228, 276, 325, 348, 451, 1032, 1105, 1128, 1729, 2388, 2465, 2701, 2821, 3721, 5272, 5365, 6601, 8911, 10585, 12025, 12673, 15841, 18721, 22681, 23585, 23725, 29341, 31621, 32376, 35016, 35425, 41041, 41125, 46632, 46657, 47125

Offset: 1

Jianing Song, Sep 02 2018

nonn

Note that for primes p >= 5, p always divides 2^(p-2) + 3^(p-2) + 6^(p-2) - 1 (see A318760).

It's interesting to study the squares of primes in this sequence. For primes p >= 5, x^(p^2-2) == x^(p-2) (mod p^2) for any integer x, so p^2 is a term if and only if p^2 divides 2^(p-2) + 3^(p-2) + 6^(p-2) - 1. It's easy to see that for any prime p, p^2 is a term of this sequence if and only if p is in A238201 and p != 3 (p = 2, 5, 61, 1680023, 7308036881, there are no others up to 7*10^10). - Jianing Song, Dec 25 2018

			(2^10 + 3^10 + 6^10 - 1)/12 = 5403854 which is an integer, so 12 is a term.
(2^22 + 3^22 + 6^22 - 1)/24 = 5484238967813377 which is also an integer, so 24 is a term.

Amiram Eldar, Table of n, a(n) for n = 1..4127 (terms below 10^10)

Cf. A238201, A318760.

A052155 is a proper subsequence.

Select[Range[48000], CompositeQ[#] && Mod[Sum[PowerMod[k, #-2, #],{k, {2, 3, 6}}], #] == 1 &] (* Amiram Eldar, Jul 17 2024 *)

b(n) = lift(Mod(2, n)^(n-2) + Mod(3, n)^(n-2) + Mod(6, n)^(n-2));
for(n=2, 30000, if(isprime(n)==0&&b(n)==1, print1(n, ", ")))

A330170 a(n) = 2^n + 3^n + 6^n - 1.

Original entry on oeis.org

10, 48, 250, 1392, 8050, 47448, 282250, 1686432, 10097890, 60526248, 362976250, 2177317872, 13062296530, 78368963448, 470199366250, 2821153019712, 16926788715970, 101560344351048, 609360902796250, 3656161927895952, 21936961102828210

Offset: 1

Bernard Schott, Dec 04 2019

This sequence is the subject of the 4th problem, proposed by Poland, of the 46th International Mathematical Olympiad in 2005 at Mérida (Mexico) [see the link IMO].
Answer to the question: 1 is the only positive integer that is relatively prime to every term of the sequence.
Proof: p=2 divides a(1) = 10, p=3 divides a(2) = 48, and if p prime >= 5, then p divides a(p-2). So, for every prime p, there exists n >= 1 such that p divides a(n).

			a(9) = 2^9 + 3^9 + 6^9 - 1 = 10097890 = 11 * 917990.

Colin Barker, Table of n, a(n) for n = 1..1000
IMO, IMO 2005 - Problem 4
Index entries for linear recurrences with constant coefficients, signature (12,-47,72,-36).

Cf. A000079 (2^n), A000244 (3^n), A000400 (6^n), A318760 (a(p-2)/p).

Maple
```
A330170 := seq(2^n+3^n+6^n-1, n=1..50);
```

Table[2^n + 3^n + 6^n - 1, {n, 1, 21}] (* Amiram Eldar, Dec 04 2019 *)

Vec(2*x*(5 - 36*x + 72*x^2 - 36*x^3) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 6*x)) + O(x^40)) \\ Colin Barker, Dec 04 2019

a(n) = A000079(n) + A000244(n) + A000400(n) - 1.
From Colin Barker, Dec 04 2019: (Start)
G.f.: 2*x*(5 - 36*x + 72*x^2 - 36*x^3) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 6*x)).
a(n) = 12*a(n-1) - 47*a(n-2) + 72*a(n-3) - 36*a(n-4) for n>5.
(End)

cp's OEIS Frontend

A318761 Composite k that divides 2^(k-2) + 3^(k-2) + 6^(k-2) - 1.

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