|
|
A145236
|
|
a(n) is the least positive integer such that if p_n is the n-th prime then (ceiling(sqrt(a(n)*p_n)))^2 - a(n)*p_n is a perfect square.
|
|
7
|
|
|
2, 1, 1, 3, 5, 5, 9, 9, 13, 17, 19, 23, 25, 27, 31, 35, 41, 41, 47, 51, 51, 57, 61, 65, 73, 75, 77, 81, 83, 85, 99, 101, 107, 109, 117, 119, 125, 129, 133, 139, 145, 145, 155, 157, 161, 163, 173, 183, 187, 189, 193, 199, 201, 209, 215, 221, 225, 227, 233, 237, 239, 247
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Conjectures: 1) for n >= 2, the sequence does not decrease; 2) for n > 1, a(n) is odd; 3) a(n) can be equal to a(n+1) only for twins: p_(n+1) - p_n = 2 (although there also exist twins for which a(n) < a(n+1)).
All these conjectures are proved using the formula a(n) = p_n - 2*floor(sqrt(2p_n)) + 2, n > 1. See also A145701 and A145714. - Vladimir Shevelev, Oct 18 2008
|
|
LINKS
|
|
|
MAPLE
|
A145236 := proc(n) local p, k, a ; p := ithprime(n) ; for k from 1 do ceil(sqrt(ceil(k*p))) ; a := %^2-k*p ; if issqr(a) then return k ; end if; end do: end proc:
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|