A097863 Sum of 5-infinitary divisors of n.
1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 31, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 33, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 124, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 99, 84, 144, 68, 126, 96, 144, 72, 195, 74, 114, 124
Offset: 1
Examples
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).
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Programs
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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
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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:
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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 *)
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
Comments