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A233655
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Sum of parts power divisors of canonical representation of n (A233569).
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3
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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, 45, 63, 48, 63, 63, 57, 32, 65, 50, 92, 40, 97, 97, 115, 50, 97, 54, 120, 97, 120, 120, 140, 65, 92, 97, 115, 97, 120, 120, 140, 92, 115, 120, 140, 115, 140, 140, 120, 64
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OFFSET
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1,2
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COMMENTS
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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.
Note that, by agreement, (10...0)^0 means the absence of the corresponding part.
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LINKS
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FORMULA
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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.
Thus a(42)=54. What is a general formula for a(n)?
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EXAMPLE
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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).
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.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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