OFFSET
1,1
LINKS
V. Shevelev, Compact integers and factorials, Acta Arithmetica 126 (2007), no. 3, 195-236.
FORMULA
For sufficiently large n, 2^n - 1 <= a(n) <= 2^ceiling(40*n/19). Let k >= n. Put g = g(n,k) = min{odd j >= 2^(k-n): 2^k - j is prime} and h(n) = min{k: k - n = floor(log_2(g))}. Then a(n) = 2^h(n) - g(n,h(n)).
EXAMPLE
By the formula, for n=6, consider k >= 6. If k=6, then g(6,6) = 3, but 6 does not equal to 6 - floor(log_2(3)); if k=7, then g=15, but 6 does not equal to 7 - floor(log_2(15)); if k=8, then g=5 and we see that 6 = 8 - floor(log_2(5)). Therefore a(6) = 2^8 - 5 = 251.
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, May 07 2010
STATUS
approved