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%I #33 Sep 09 2020 03:07:39
%S 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,
%T 42,40,56,30,72,32,51,48,54,48,91,38,60,56,90,42,96,44,84,78,72,48,68,
%U 57,93,72,98,54,120,72,120,80,90,60,168,62,96,104,119,84,144,68,126,96
%N 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.
%C If we group the exponents e in the Bower-Harris formula into the sets with d_k=0, 1, 2 and 3, we see that every n has a unique representation of the form n=prod q_i *prod (r_j)^2 *prod (s_k)^3, where each of q_i, r_j, s_k is a prime power of the form p^(k^4), p prime, k>=0. Using this representation, a(n)=prod (q_i+1)prod ((r_j)^2+r_j+1)prod ((s_k)^3+(s_k)^2+s_k+1) by simple expansion of the quotient on the right hand side of the Bower-Harris formula. - _Vladimir Shevelev_, May 08 2013
%H Reinhard Zumkeller, <a href="/A074847/b074847.txt">Table of n, a(n) for n = 1..10000</a>
%F 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_ and _Mitch Harris_, May 20 2005
%e 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.
%p 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:
%p seq(A074847(n),n=1..100) ; # _R. J. Mathar_, Oct 06 2010
%t f[p_, e_] := Module[{d = IntegerDigits[e, 4]}, m = Length[d]; Product[(p^((d[[j]] + 1)*4^(m - j)) - 1)/(p^(4^(m - j)) - 1), {j, 1, m}]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Sep 09 2020 *)
%o (Haskell) following Bower and Harris, cf. A049418:
%o a074847 1 = 1
%o a074847 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 (*) a000302_list $ map (+ 1) $ a030386_row e)
%o (map (subtract 1 . (p ^)) a000302_list)
%o -- _Reinhard Zumkeller_, Sep 18 2015
%Y Cf. A049417 (2-infinitary), A049418 (3-infinitary), A097863 (5-infinitary).
%Y Cf. A000302, A030386, A027748, A074848, A124010.
%K nonn,mult
%O 1,2
%A _Yasutoshi Kohmoto_, Sep 10 2002
%E More terms from _R. J. Mathar_, Oct 06 2010
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