OFFSET
1,2
COMMENTS
If n=Product p_i^r_i and d=Product p_i^s_i, each s_i has a digit a<=b in its 5-ary expansion everywhere that the corresponding r_i has a digit b, then d is a 5-infinitary-divisor of n.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
FORMULA
Denote by P_5={p^5^k} the two-parameter set when k=0,1,... and p runs prime values. Then every n has a unique representation of the form n=prod q_i prod (r_j)^2 prod (s_k)^3 prod (t_m)^4, where q_i, r_j, s_k, t_m are distinct elements of P_5. Using this representation, we have a(n)=prod (q_i+1)prod ((r_j)^2+r_j+1)prod ((s_k)^3+(s_k)^2+s_k+1) prod ((t_m)^4+(t_m)^3+(t_m)^2+t_m+1). - Vladimir Shevelev, May 08 2013
EXAMPLE
a(32) = a(2^10) = 2^10 + 2^0 = 32 + 1 = 33, in 5-ary expansion. This is the first term which is different from sigma(n).
MAPLE
A097863 := 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, 5) ; for k from 0 to nops(d)-1 do a := a*(p^((1+op(k+1, d))*5^k)-1)/(p^(5^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:
MATHEMATICA
f[p_, e_] := Module[{d = IntegerDigits[e, 5]}, m = Length[d]; Product[(p^((d[[j]] + 1)*5^(m - j)) - 1)/(p^(5^(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:
a097863 1 = 1
a097863 n = product $ zipWith f (a027748_row n) (a124010_row n) where
f p e = product $ zipWith div
(map (subtract 1 . (p ^)) $
zipWith (*) a000351_list $ map (+ 1) $ a031235_row e)
(map (subtract 1 . (p ^)) a000351_list)
-- Reinhard Zumkeller, Sep 18 2015
CROSSREFS
KEYWORD
nonn,mult
AUTHOR
STATUS
approved