login
a(1) = 1, a(2) = 1, a(n) = a(n-p) + a((n+1)-p), n > 2, where p is the largest prime less than n.
0

%I #8 Jan 11 2020 17:26:26

%S 1,1,2,2,3,2,3,2,3,4,5,2,3,2,3,4,5,2,3,2,3,4,5,2,3,4,5,5,5,2,3,2,3,4,

%T 5,5,5,2,3,4,5,2,3,2,3,4,5,2,3,4,5,5,5,2,3,4,5,5,5,2,3,2,3,4,5,5,5,2,

%U 3,4,5,2,3,2,3,4,5,5,5,2,3,4,5,2,3,4,5,5,5,2,3,4,5,5,5,5,5,2,3,4,5,2,3,2,3

%N a(1) = 1, a(2) = 1, a(n) = a(n-p) + a((n+1)-p), n > 2, where p is the largest prime less than n.

%e a(16) = a(16-13) + a(17-13) = a(3) + a(4) = 2 + 2 = 4.

%p with(numtheory): a := proc(n) option remember: if n <= 2 and n >=1 then RETURN(1) fi: a(n-prevprime(n))+ a(n+1-prevprime(n)): end: for n from 1 to 250 do printf(`%d,`,a(n)) od:

%K nonn

%O 1,3

%A Brian Wallace (wallacebrianedward(AT)yahoo.co.uk), Mar 04 2001

%E More terms from _James A. Sellers_, Mar 15 2001