A049418 3-i-sigma(n): sum of 3-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 ternary expansion everywhere that the corresponding r(i) has a digit b, then d is a 3-i-divisor of n.
1, 3, 4, 7, 6, 12, 8, 9, 13, 18, 12, 28, 14, 24, 24, 27, 18, 39, 20, 42, 32, 36, 24, 36, 31, 42, 28, 56, 30, 72, 32, 63, 48, 54, 48, 91, 38, 60, 56, 54, 42, 96, 44, 84, 78, 72, 48, 108, 57, 93, 72, 98, 54, 84, 72, 72, 80, 90, 60, 168, 62, 96, 104, 73, 84, 144, 68, 126, 96
Offset: 1
Examples
Let n = 28 = 2^2*7. Then a(n) = (2^2 + 2 + 1)*(7 + 1) = 56. - _Vladimir Shevelev_, May 07 2013
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- J. O. M. Pedersen, Tables of Aliquot Cycles, backup on web.archive.org of no more exiting page, as of May 2014
- J. O. M. Pedersen, Tables of Aliquot Cycles [Cached copy, pdf file only]
Crossrefs
Programs
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Haskell
following Bower and Harris: a049418 1 = 1 a049418 n = product $ zipWith f (a027748_row n) (a124010_row n) where f p e = product $ zipWith div (map (subtract 1 . (p ^)) $ zipWith (*) a000244_list $ map (+ 1) $ a030341_row e) (map (subtract 1 . (p ^)) a000244_list) -- Reinhard Zumkeller, Sep 18 2015 -
Maple
A049418 := proc(n) option remember; local ifa,a,p,e,d,k ; 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,3) ; for k from 0 to nops(d)-1 do a := a*(p^((1+op(k+1,d))*3^k)-1)/(p^(3^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(A049418(n),n=1..40) ; # R. J. Mathar, Oct 06 2010
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Mathematica
A049418[n_] := Module[{ifa = FactorInteger[n], a = 1, p, e, d, k}, If[ Length[ifa] == 1, p = ifa[[1, 1]]; e = ifa[[1, 2]]; d = Reverse[ IntegerDigits[e, 3] ]; For[k = 1, k <= Length[d], k++, a = a*(p^((1 + d[[k]])*3^(k - 1)) - 1)/(p^(3^(k - 1)) - 1)], Do[ a = a*A049418[ d[[1]]^d[[2]] ], {d, ifa}]]; Return[a] ]; A049418[1] = 1; Table[ A049418[n] , {n, 1, 69}] (* Jean-François Alcover, Jan 03 2012, after R. J. Mathar *)
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PARI
apply( {A049418(n)=vecprod([prod(k=1,#n=digits(f[2],3),(f[1]^(3^(#n-k)*(n[k]+1))-1)\(f[1]^3^(#n-k)-1))|f<-factor(n)~])}, [1..99]) \\ M. F. Hasler, Sep 21 2022
Formula
Multiplicative with a(p^e) = prod_{k >= 0} (p^(3^k*{d_k+1}) - 1)/(p^(3^k) - 1), where e = sum_{k >= 0} d_k 3^k (base 3 representation). - Christian G. Bower and Mitch Harris, May 20 2005. [Edited by M. F. Hasler, Sep 21 2022]
Denote P_3 = {p^3^k}, k = 0, 1, ..., p runs primes. Then every n has a unique representation of the form n = prod q_i prod (r_j)^2, where q_i, r_j are distinct elements of P_3. Using this representation, we have a(n) = prod (q_i+1)*prod ((r_j)^2+r_j+1). - Vladimir Shevelev, May 07 2013
Extensions
More terms from Naohiro Nomoto, Sep 10 2001