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A296075 Sum of deficiencies of divisors of n. 6
1, 2, 3, 3, 5, 4, 7, 4, 8, 8, 11, 1, 13, 12, 13, 5, 17, 6, 19, 7, 19, 20, 23, -10, 24, 24, 22, 13, 29, 4, 31, 6, 31, 32, 33, -16, 37, 36, 37, -2, 41, 12, 43, 25, 30, 44, 47, -37, 48, 34, 49, 31, 53, 8, 53, 6, 55, 56, 59, -49, 61, 60, 46, 7, 63, 28, 67, 43, 67, 36, 71, -78, 73, 72, 58, 49, 75, 36, 79, -27, 63, 80, 83, -47, 83 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
a(n)=0 for n in A066218. Are 1 and 12 the only solutions to a(n)=1? - Robert Israel, Dec 04 2017
LINKS
FORMULA
a(n) = Sum_{d|n} A033879(d).
a(n) = A296074(n) + A033879(n).
If m and n are coprime, a(m*n) = 2*a(m)*A000203(n)+2*a(n)*A000203(m)-a(m)*a(n)-2*A000203(m)*A000203(n). - Robert Israel, Dec 04 2017
a(n) = 2*A000203(n) - A007429(n). - Ridouane Oudra, Jul 29 2019
Sum_{k=1..n} a(k) ~ (Pi^2/6 - Pi^4/72) * n^2. - Amiram Eldar, Dec 04 2023
EXAMPLE
For n = 6, whose divisors are 1, 2, 3, 6, their deficiencies are 1, 1, 2, 0, thus a(6) = 1 + 1 + 2 + 0 = 4.
For n = 24, whose divisors are 1, 2, 3, 4, 6, 8, 12, 24, their deficiencies are 1, 1, 2, 1, 0, 1, -4, -12, thus a(24) = 1 + 1 + 2 + 1 + 0 + 1 + -4 + -12 = -10.
MAPLE
f:= n -> add(2*t-numtheory:-sigma(t), t=numtheory:-divisors(n)):
map(f, [$1..100]); # Robert Israel, Dec 04 2017
MATHEMATICA
f1[p_, e_] := (p^(e+1)-1)/(p-1); f2[p_, e_] := (p*(p^(e+1)-1) - (p-1)*(e+1))/(p-1)^2; a[1] = 1; a[n_] := Module[{f = FactorInteger[n]}, 2 * Times @@ f1 @@@ f - Times @@ f2 @@@ f]; Array[a, 100] (* Amiram Eldar, Dec 04 2023 *)
PROG
(PARI)
A033879(n) = ((2*n)-sigma(n));
A296075(n) = sumdiv(n, d, A033879(d));
CROSSREFS
Sequence in context: A275314 A161857 A135533 * A318678 A119674 A064920
KEYWORD
sign,easy
AUTHOR
Antti Karttunen, Dec 04 2017
STATUS
approved

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Last modified March 29 06:34 EDT 2024. Contains 371265 sequences. (Running on oeis4.)