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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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2
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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, 42, 96, 44, 84, 78, 72, 48, 68, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 119, 84, 144, 68, 126, 96
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| 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
| 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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MAPLE
| Contribution from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 06 2010: (Start)
A074847 := proc(n) option remember; ifa := ifactors(n)[2] ; a := 1 ; if nops(ifa) = 1 then p := op(1, op(1, ifa)) ; e := op(2, op(1, ifa)) ; d := convert(e, base, 4) ; for k from 0 to nops(d)-1 do a := a*(p^((1+op(k+1, d))*4^k)-1)/(p^(4^k)-1) ; end do: else for d in ifa do a := a*procname( op(1, d)^op(2, d)) ; end do: return a; end if; end proc:
seq(A074847(n), n=1..100) ; (End)
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CROSSREFS
| Cf. A049417 (2-infinitary), A049418 (3-infinitary). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 06 2010]
Sequence in context: A076887 A140782 A097011 * A097863 A097012 A143348
Adjacent sequences: A074844 A074845 A074846 * A074848 A074849 A074850
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KEYWORD
| nonn,mult
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AUTHOR
| Yasutoshi Kohmoto (zbi74583(AT)boat.zero.ad.jp), Sep 10 2002
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EXTENSIONS
| More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 06 2010
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