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A074847 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. 7
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; text; internal format)
OFFSET

1,2

COMMENTS

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

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

FORMULA

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

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.

MAPLE

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) ; # R. J. Mathar, Oct 06 2010

MATHEMATICA

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 *)

PROG

(Haskell)  following Bower and Harris, cf. A049418:

a074847 1 = 1

a074847 n = product $ zipWith f (a027748_row n) (a124010_row n) where

   f p e = product $ zipWith div

           (map (subtract 1 . (p ^)) $

                zipWith (*) a000302_list $ map (+ 1) $ a030386_row e)

           (map (subtract 1 . (p ^)) a000302_list)

-- Reinhard Zumkeller, Sep 18 2015

CROSSREFS

Cf. A049417 (2-infinitary), A049418 (3-infinitary), A097863 (5-infinitary).

Cf. A000302, A030386, A027748, A074848, A124010.

Sequence in context: A140782 A284587 A097011 * A325317 A325316 A227131

Adjacent sequences:  A074844 A074845 A074846 * A074848 A074849 A074850

KEYWORD

nonn,mult

AUTHOR

Yasutoshi Kohmoto, Sep 10 2002

EXTENSIONS

More terms from R. J. Mathar, Oct 06 2010

STATUS

approved

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Last modified November 30 03:48 EST 2021. Contains 349417 sequences. (Running on oeis4.)