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Search: id:A074847
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| A074847 |
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Sum of 4-infinitary divisors of n: if n=Product p(i)^r(i) and d=Product p(i)^s(i), each s(i) has a digit a<=b in its 4-ary expansion everywhere that the corresponding r(i) has a digit b, then d is a 4-infinitary-divisor of n. |
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+0
1
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| 1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 17, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 51, 48, 54, 48, 91, 38, 60, 56, 90
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OFFSET
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1,2
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COMMENT
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Multiplicative. If e = sum_{k >= 0} d_k 4^k (base 4 representation), then a(p^e) = prod_{k >= 0} (p^(4^k*{d_k+1}) - 1)/(p^(4^k) - 1). Christian G. Bower (bowerc(AT)usa.net) and Mitch Harris (Harris.Mitchell(AT)mgh.harvard.edu) May 20, 2005.
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EXAMPLE
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2^4*3 is a 4-infinitary-divisor of 2^5*3^2 because 2^4*3 = 2^10*3^1 and 2^5*3^2 = 2^11*3^2 in 4-ary expanded power. All corresponding digits satisfy the condition. 1<=1, 0<=1, 1<=2.
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CROSSREFS
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Sequence in context: A076887 A140782 A097011 this_sequence A097863 A097012 A143348
Adjacent sequences: A074844 A074845 A074846 this_sequence A074848 A074849 A074850
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KEYWORD
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nonn,mult
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AUTHOR
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Yasutoshi Kohmoto (zbi74583(AT)boat.zero.ad.jp), Sep 10 2002
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