A188999 Bi-unitary sigma: sum of the bi-unitary divisors of n.
1, 3, 4, 5, 6, 12, 8, 15, 10, 18, 12, 20, 14, 24, 24, 27, 18, 30, 20, 30, 32, 36, 24, 60, 26, 42, 40, 40, 30, 72, 32, 63, 48, 54, 48, 50, 38, 60, 56, 90, 42, 96, 44, 60, 60, 72, 48, 108, 50, 78, 72, 70, 54, 120, 72, 120, 80, 90, 60, 120, 62, 96, 80, 119, 84, 144, 68, 90, 96, 144, 72, 150, 74, 114, 104, 100
Offset: 1
Examples
The divisors of n=16 are d=1, 2, 4, 8 and 16. The greatest common unitary divisor of (1,16) is 1, of (2,8) is 1, of (4,4) is 4, of (8,2) is 1, of (16,1) is 1 (see A165430). So 1, 2, 8 and 16 are bi-unitary divisors of 16, which sum to a(16) = 1 + 2 + 8 + 16 = 27.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- Krishnaswami Alladi, On arithmetic functions and divisors of higher order, J. Austral. Math. Soc. 23 (series A) (1977), 9-27.
- József Sándor and Borislav Crstici, Perfect numbers: Old and new issues; perspectives, in Handbook of number theory, II, p. 45.
- László Tóth, On the bi-unitary analogues of Euler's arithmetical function and the gcd-sum function J. Int. Seq. 12 (2009), Article 09.5.2.
- Charles R. Wall, Bi-unitary perfect numbers, Proc. Am. Math. Soc. 33 (1) (1972), 39-42.
- Eric Weisstein's World of Mathematics, Biunitary Divisor.
- Tomohiro Yamada, 2 and 9 are the only biunitary superperfect numbers, arXiv:1705.00189 [math.NT], 2017.
Programs
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Haskell
a188999 n = product $ zipWith f (a027748_row n) (a124010_row n) where f p e = (p ^ (e + 1) - 1) `div` (p - 1) - (1 - m) * p ^ e' where (e', m) = divMod e 2 -- Reinhard Zumkeller, Mar 04 2013 -
Maple
A188999 := proc(n) local a,e,p,f; a :=1 ; for f in ifactors(n)[2] do e := op(2,f) ; p := op(1,f) ; if type(e,'odd') then a := a*(p^(e+1)-1)/(p-1) ; else a := a*((p^(e+1)-1)/(p-1)-p^(e/2)) ; end if; end do: a ; end proc: seq( A188999(n),n=1..80) ;
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Mathematica
f[n_] := Select[Divisors[n], Function[d, CoprimeQ[d, n/d]]]; Table[DivisorSum[n, # &, Last@ Intersection[f@ #, f[n/#]] == 1 &], {n, 76}] (* Michael De Vlieger, May 07 2017 *) a[n_] := If[n==1, 1, Product[{p, e} = pe; If[OddQ[e], (p^(e+1)-1)/(p-1), ((p^(e+1)-1)/(p-1)-p^(e/2))], {pe, FactorInteger[n]}]]; Array[a, 80] (* Jean-François Alcover, Sep 22 2018 *) -
PARI
udivs(n) = {my(d = divisors(n)); select(x->(gcd(x, n/x)==1), d); } gcud(n, m) = vecmax(setintersect(udivs(n), udivs(m))); biudivs(n) = select(x->(gcud(x, n/x)==1), divisors(n)); a(n) = vecsum(biudivs(n)); \\ Michel Marcus, May 07 2017 -
PARI
a(n) = {f = factor(n); for (i=1, #f~, p = f[i,1]; e = f[i,2]; f[i,1] = if (e % 2, (p^(e+1)-1)/(p-1), (p^(e+1)-1)/(p-1) -p^(e/2)); f[i,2] = 1;); factorback(f);} \\ Michel Marcus, Nov 09 2017 -
Python
from math import prod from sympy import factorint def A188999(n): return prod((p**(e+1)-1)//(p-1)-(0 if e&1 else p**(e>>1)) for p,e in factorint(n).items()) # Chai Wah Wu, Dec 28 2024
Formula
Multiplicative with a(p^e) = (p^(e+1)-1)/(p-1) if e is odd, a(p^e) = (p^(e+1)-1)/(p-1) -p^(e/2) if e is even.
Dirichlet g.f.: zeta(s-1) * zeta(s) * zeta(2*s-1) * Product_{p prime} (1 - 2/p^(2*s-1) + 1/p^(3*s-2) + 1/p^(3*s-1) - 1/p^(4*s-2)). - Amiram Eldar, Aug 28 2023
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