OFFSET
0,3
COMMENTS
Fn is defined by: Fn(0) = Fn(1) = ... = Fn(n-2) = 0, Fn(n-1) = 1, and Fn(k+1) = Fn(k) + Fn(k-1) + ... + Fn(k-n+1).
In general, Fn(k) is odd iff k == -1 or -2 (mod n+1), therefore a(n) = k*(n+1) - (1 or 2) for all n. Since Fn(n-1) = F(n) = 1, we must have a(n) >= 2n. Since Fn(k) = 2^(k-n) for n <= k < 2n, Fn(2n) = 2^n-1, so a(n) = 2n exactly for the Mersenne prime exponents A000043, while a(n) = 2n+1 when n is not in A000043 but n+1 is in A050414. - M. F. Hasler, Apr 18 2018
Further terms of the sequence: a(38) > 62000, a(39) > 72000, a(40) = 285, a(41) > 178000, a(42) = 558, a(44) = 19529, a(46) = 33369, a(47) = 239, a(48) = 6368, a(53) = 2860, a(54) = 2418, a(58) = 176, a(59) = 18418, a(60) = 1463, a(61) = 122, a(62) = 8755, a(63) = 5118, a(64) = 25089, a(65) = 988, a(66) = 333, a(67) = 406, a(70) = 1632, a(74) = 374, a(76) = 13704, a(77) = 4991, a(86) = 347, a(89) = 178, a(92) = 1114, a(93) = 187, a(98) = 395, a(100) > 80000; a(n) > 10^4 for all other n up to 100. - Jacques Tramu and M. F. Hasler, Apr 18 2018
FORMULA
a(n) == -1 or -2 (mod n+1). a(n) >= 2n, with equality iff n is in A000043. a(n) <= 2n+1 for n+1 in A050414. - M. F. Hasler, Apr 18 2018
EXAMPLE
a(2) = 4 because F2 (Fibonacci) = 0, 1, 1, 2, 3, 5, 8, ... and F2(4) = 3 is prime.
a(3) = 6 because F3 (tribonacci) = 0, 0, 1, 1, 2, 4, 7, 13, ... and F3(6) = 7 is prime.
a(4) = 9 because F4 (tetranacci) = 0, 0, 0, 1, 1, 2, 4, 8, 15, 29, 56, ... and F4(9) = 29 is prime.
From M. F. Hasler, Apr 18 2018: (Start)
We see that Fn(k) = 2^(k-n) for n <= k < 2n and thus Fn(2n) = 2^n-1, so a(n) = 2n exactly for the Mersenne prime exponents A000043.
a(n) = 2n + 1 when 2^(n+1) - 3 is prime (n+1 in A050414) but 2^n-1 is not, i.e., n = 4, 8, 9, 11, 21, 23, 28, 93, 115, 121, 149, 173, 212, 220, 232, 265, 335, 451, 544, 688, 693, 849, 1735, ...
For other primes we have: a(29) = 687*30 - 2, a(37) = 20*38 - 2, a(41) > 10^4, a(43) > 10^4, a(47) = 5*48 - 1, a(53) = 53*54 - 2, a(59) = 307*60 - 2, a(67) = 6*67 - 1. (End)
PROG
(PARI) A302990(n, L=oo, a=vector(n+1, i, if(i<n, 2^i, 1)))={n>1 && for(i=-2+2*n+=1, L, ispseudoprime(a[i%n+1]=2*a[(i-1)%n+1]-a[i%n+1]) && return(i))} \\ Testing primality only for i%n>n-3 is not faster, even for large n. - M. F. Hasler, Apr 17 2018; improved Apr 18 2018
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Jacques Tramu, Apr 17 2018
EXTENSIONS
a(29) from Jacques Tramu, Apr 19 2018
a(33) from Daniel Suteu, Apr 20 2018
a(36) from Jacques Tramu, Apr 25 2018
STATUS
approved