OFFSET
0,2
COMMENTS
It appears that 3 and 5 divide a(n) for n>4. Note that a(18)<a(17). - T. D. Noe, Jul 19 2005
Conjecture: a(n)==0 (mod 3) for n > 2. Then n-2^k is not == 0 (mod 3) and a prime is more probable. - Robert G. Wilson v, Jul 21 2005
Conjecture: a(n+15)==0 (mod 30) for n > 4. - Robert G. Wilson v, Jul 21 2005
a(n) > 10^10 for n >= 21. - Donovan Johnson, Jan 21 2009
EXAMPLE
a(4) = 21, 21-2 =19, 21-4 = 17, 21-8 = 13, 21-16 = 5, 21 is the smallest number that gives four such primes.
MATHEMATICA
t=Table[cnt=0; r=1; While[r<n, If[PrimeQ[n-r], cnt++ ]; r=2r]; cnt, {n, 250000}]; Table[First[Flatten[Position[t, n]]], {n, 13}] (Noe)
f[n_] := Count[ PrimeQ[n - 2^Range[0, Floor[ Log[2, n]]]], True]; t = Table[0, {30}]; Do[ a = f[n]; If[ t[[a+1]] == 0, t[[a+1]] = n], {n, 4*10^8}]; t (* Robert G. Wilson v *)
CROSSREFS
KEYWORD
hard,more,nonn
AUTHOR
Amarnath Murthy, Jul 17 2005
EXTENSIONS
Edited, corrected and extended by T. D. Noe and Robert G. Wilson v, Jul 19 2005
a(20) from Donovan Johnson, Jan 21 2009
STATUS
approved