|
|
A320146
|
|
a(n) = 2*prime(n) modulo (prime(n-1) + prime(n+1)).
|
|
1
|
|
|
6, 0, 14, 2, 26, 2, 38, 46, 4, 62, 2, 2, 86, 94, 0, 4, 122, 2, 2, 146, 2, 166, 178, 4, 2, 206, 2, 218, 226, 10, 262, 4, 278, 8, 302, 0, 2, 334, 0, 4, 362, 8, 386, 2, 398, 0, 8, 2, 458, 466, 4, 482, 4, 0, 0, 4, 542, 2, 2, 566, 586, 10, 2, 626, 634, 8, 674, 8, 698, 706, 718, 2, 0, 2, 766, 778, 4, 802, 818, 8, 842
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
2,1
|
|
COMMENTS
|
This sequence has to do with the relative position of primes with respect to their adjacent primes:
(i) if prime(n) is closer to its predecessor than to its successor, then a(n) = 2*prime(n);
(ii) if prime(n) is closer to its successor than to its predecessor, then a(n) = 2*prime(n) - prime(n-1) - prime(n+1); and
(iii) if prime(n) is equidistant from its predecessor and its successor, then a(n) = 0.
Is lim_{n -> infinity} (Sum_{i=1..n} a(i))/(Sum_{i=1..n} prime(i)) finite? If so, what is its value?
|
|
LINKS
|
|
|
MAPLE
|
seq(modp(2*ithprime(n), (ithprime(n-1)+ithprime(n+1))), n=2..90); # Muniru A Asiru, Oct 07 2018
|
|
MATHEMATICA
|
Table[Mod[2*Prime[n], Prime[n-1] + Prime[n+1]], {n, 2, 120}]
Mod[2#[[2]], #[[1]]+#[[3]]]&/@Partition[Prime[Range[90]], 3, 1] (* Harvey P. Dale, Jan 03 2019 *)
|
|
PROG
|
(PARI) a(n) = 2*prime(n) % (prime(n-1) + prime(n+1)); \\ Michel Marcus, Oct 18 2018
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|