cp's OEIS Frontend

Numbers k such that A322906(k) = 4.

Also numbers k such that A214027(k) is odd.

Interestingly, the Lucas numbers separate the primes into three disjoint sets: (A053028) primes that do not divide any Lucas number, (A053027) primes that divide Lucas numbers of even index and (A053032) primes that divide Lucas numbers of odd index.

Primes of the form 2x^2+2xy+13y^2. Discriminant = -100. - T. D. Noe, May 02 2008

Primes of the form a^2 + b^2 such that a^2 == b^2 (mod 5). - Thomas Ordowski, May 18 2015

If a prime p divides a term a(k) of this sequence, then k must be less than the period of the sequence mod p. Hence these primes are found by computing A001644(k) mod p for increasing k and stopping when either A001644(k) mod p = 0 or the end of the period is reached. Interestingly, for all of these primes except 211, the period of the sequence A001644(k) mod p is (p-1)/d, where d is a small integer. The only other exceptional primes less than 1000000 are 23977 and 47093.

The tribonacci numbers are indexed so that trib(0) = trib(1) = 0, trib(2) = 1, for n>2: trib(n) = trib(n-1) + trib(n-2) + trib(n-3). See A112618 for another version.

Brenner proves that every prime divides some tribonacci number T(n). For the similar 3-step Lucas sequence A001644, there are primes (A106299) that do not divide any term.

The -1 terms are for the primes in A053028. Note that 2 divides the zeroth Lucas number. In the plots, the uppermost line consists of the odd primes in A000057. Note that when a(n) > 0, then a(n) = A001602(n)/2.

If one takes L_k, for k >= 1, that is A000204, then a(1) = 1 and a(2) = 3 followed by the given numbers. This fits then with A106291(n) = A253808(n)*a(n), n >= 1 (where in A253808 a negative entry at position n indicates, as in the present sequence, that the Lucas numbers are not divisible by n. For odd primes not dividing any Lucas numbers see A053028. No power 2^m, m >= 3 divides any Lucas number, see, e.g., Vajda, p. 81). - Wolfdieter Lang, Jan 20 2015

If a prime p divides a term a(k) of this sequence, then k must be less than the period of the sequence mod p. Hence these primes are found by computing A073817(k) mod p for increasing k and stopping when either A073817(k) mod p = 0 or the end of the period is reached. Interestingly, for all of these primes, the period of the sequence A073817(k) mod p appears to be (p-1)/d, where d is a small integer.

If a prime p divides a term a(k) of this sequence, then k must be less than the period of the sequence mod p. Hence these primes are found by computing A074048(k) mod p for increasing k and stopping when either A074048(k) mod p = 0 or the end of the period is reached. Interestingly, for all of these primes, the period of the sequence A074048(k) mod p appears to be (p-1)/d, where d is a small integer.

These a(n) values may be called Lucas "entry points".

a(1)=0, corresponding to prime(1)= 2, must be viewed as a special case here, because 0, in this case only, is the correct smallest index.

For all other zeros in a(n) (e.g., for a(3), a(6), a(7) corresponding to prime(3)=5, prime(6)=13, prime(7)=17), there are no Lucas numbers divisible by these primes. The full set of primes not divisible into any Lucas number are given by A053028.

For a(n) > 0, a(n) is always equal to either (prime(n)-1)/k or (prime(n)+1)/k, for some k >= 1. This is similar to the Fibonacci entry points given by A001602.

For each value of a(n) > 0 there is an infinite, periodic subsequence within the Lucas numbers divisible by prime(n), given by Lucas(a(n) + 2*i*a(n)), for all i >= 0. Again this is analogous to the A001602 for Fibonacci. The periodicity of the subsequence is 2*a(n), twice the entry point vs. equal to the entry point for Fibonacci.

Conjecture: Infinite Lucas subsequences divisible by the powers of odd primes, p(n), for which a(n) > 0, are given by:

Lucas(a(n) + a(n)*(p(n)-1)*(Sum_{j=1..m-1} p(n)^(j-1)) + 2*i*a(n)*p(n)^(m-1)) is divisible by p(n)^m, where p(n) > 2, i >= 0, m > 1.

Note: the formula above also works for m=1 if the "Sum" is assumed to be zero when the upper summation index limit is less than initial summation index. See second Mathematica example below which works in this way for all m >= 1, to demonstrate the rule for p = p(n) with corresponding a = a(n).

The divisibility of Lucas numbers by powers of 2 is limited to 2^1 and 2^2, as follows: Lucas(3*i) is divisible by 2 and Lucas(3+6i) is divisible by 4, for all i >= 0.