A233655 - OEIS

%I #35 Dec 23 2013 06:13:04

%S 1,2,4,4,9,9,11,8,17,12,26,17,26,26,26,16,33,26,48,26,45,45,63,33,48,

%T 45,63,48,63,63,57,32,65,50,92,40,97,97,115,50,97,54,120,97,120,120,

%U 140,65,92,97,115,97,120,120,140,92,115,120,140,115,140,140,120,64

%N Sum of parts power divisors of canonical representation of n (A233569).

%C If the canonical representation of n is A233569(n)=(1)^k_1[*](10)^k_2[*]...[*](10...0)^k_t, where [*] means   concatenation, then we say that a number (1)^r_1[*](10)^r_2[*]...[*](10...0)^r_t is a parts power divisor of canonical representation of n, iff all r_i<=k_i.

%C Note that, by agreement, (10...0)^0 means the absence of the corresponding part.

%F a((10...0[m zeros])^k) = 2^m/(2^(m+1)-1)^2 * (2^((m+1)*(k+1)) - 1) - (k+1)*2^m/(2^(m+1)-1). For example, a(101010)[here m=1,k=3] = 2/9*(2^8-1) - 4*2/3 = 54.

%F Thus a(42)=54. What is a general formula for a(n)?

%e Since A233569(5)=6, then the canonical representation of 5 is (1)^1[*](10)^1 which has parts power divisors 0, (1)^1, (10)^1, (1)^1[*](10)^1. Converting to decimal, they are 0,1,2,6 with sum 9. So a(5)=9. Note that 6 is a parts power divisor of 5, but not a c-divisors of 5 (see comment in A124771).

%e Analogously, 12 = (1)^1[*](10)^0[*](100)^1 is a parts power divisor of 52 = (1)^1[*](10)^1[*](100)^1, but not a c-divisor of 52.

%Y Cf. A233394, A233569, A124771.

%K nonn

%O 1,2

%A _Vladimir Shevelev_, Dec 14 2013