|
%I #36 Sep 21 2022 21:25:57
%S 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,
%T 42,28,56,30,72,32,63,48,54,48,91,38,60,56,54,42,96,44,84,78,72,48,
%U 108,57,93,72,98,54,84,72,72,80,90,60,168,62,96,104,73,84,144,68,126,96
%N 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.
%H Reinhard Zumkeller, <a href="/A049418/b049418.txt">Table of n, a(n) for n = 1..10000</a>
%H J. O. M. Pedersen, <a href="http://web.archive.org/web/20140502102524/http://amicable.homepage.dk/tables.htm">Tables of Aliquot Cycles</a>, backup on web.archive.org of no more exiting page, as of May 2014
%H J. O. M. Pedersen, <a href="/A063990/a063990.pdf">Tables of Aliquot Cycles</a> [Cached copy, pdf file only]
%F 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]
%F 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
%e Let n = 28 = 2^2*7. Then a(n) = (2^2 + 2 + 1)*(7 + 1) = 56. - _Vladimir Shevelev_, May 07 2013
%p 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:
%p seq(A049418(n),n=1..40) ; # _R. J. Mathar_, Oct 06 2010
%t 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_ *)
%o (Haskell) following Bower and Harris:
%o a049418 1 = 1
%o a049418 n = product $ zipWith f (a027748_row n) (a124010_row n) where
%o f p e = product $ zipWith div
%o (map (subtract 1 . (p ^)) $
%o zipWith (*) a000244_list $ map (+ 1) $ a030341_row e)
%o (map (subtract 1 . (p ^)) a000244_list)
%o -- _Reinhard Zumkeller_, Sep 18 2015
%o (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
%Y Cf. A049417 (2-infinitary), A074847 (4-infinitary), A097863 (5-infinitary).
%Y Cf. A000244, A030341, A027748, A124010.
%K nonn,nice,easy,mult
%O 1,2
%A _Yasutoshi Kohmoto_
%E More terms from _Naohiro Nomoto_, Sep 10 2001
|
