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A324213 Number of k with 0 <= k <= sigma(n) such that n-k and 2n-sigma(n) are relatively prime. 17
2, 4, 3, 8, 4, 2, 4, 16, 12, 9, 6, 14, 6, 12, 8, 32, 10, 26, 8, 21, 14, 18, 12, 20, 30, 16, 18, 2, 14, 24, 10, 64, 16, 24, 22, 88, 14, 30, 26, 36, 18, 32, 14, 42, 26, 28, 24, 54, 56, 80, 20, 32, 26, 40, 36, 60, 38, 42, 30, 56, 18, 42, 48, 128, 42, 48, 22, 50, 28, 72, 26, 122, 26, 54, 58, 46, 48, 56, 26, 86, 120, 60, 42, 96, 54 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Number of ways to form the sum sigma(n) = x+y so that n-x and n-y are coprime, with x and y in range 0..sigma(n).
From Antti Karttunen, May 28 - Jun 08 2019: (Start)
Empirically, it seems that a(n) >= A034444(n) and also that a(n) >= A034444(A000203(n)) unless n is in A000396.
Specifically, if it could be proved that a(n) >= A034444(n)/2 for n >= 2, which in turn would imply that a(n) >= A001221(n) for all n, then we would know that no odd perfect numbers could exist. Note that a(n) must be 2 on all perfect numbers, whether even or odd. See also A325819.
(End)
LINKS
FORMULA
a(n) = Sum_{i=0..sigma(n)} [1 == gcd(n-i,n-(sigma(n)-i))], where [ ] is the Iverson bracket and sigma(n) is A000203(n).
a(A000396(n)) = 2.
a(n) = A325815(n) + A034444(n).
a(n) = 1+A000203(n) - A325816(n).
a(A228058(n)) = A325819(n).
EXAMPLE
For n=1, sigma(1) = 1, both gcd(1-0, 1-(1-0)) = gcd(1,0) = 1 and gcd(1-1, 1-(1-1)) = gcd(0,1) = 1, thus a(1) = 2.
--
For n=3, sigma(3) = 4, we have 5 cases to consider:
gcd(3-0, 3-(4-0)) = 1 = gcd(3-4, 3-(4-4)),
gcd(3-1, 3-(4-1)) = 2 = gcd(3-3, 3-(4-3)),
gcd(3-2, 3-(4-2)) = 1,
of which three cases give 1 as a result, thus a(3) = 3.
--
For n=6, sigma(6) = 12, we have 13 cases to consider:
gcd(6-0, 6-(12-0)) = 6 = gcd(6-12, 6-(12-12)),
gcd(6-1, 6-(12-1)) = 5 = gcd(6-11, 6-(12-11)),
gcd(6-2, 6-(12-2)) = 4 = gcd(6-10, 6-(12-10)),
gcd(6-3, 6-(12-3)) = 3 = gcd(6-9, 6-(12-9)),
gcd(6-4, 6-(12-4)) = 2 = gcd(6-8, 6-(12-8))
gcd(6-5, 6-(12-5)) = 1 = gcd(6-7, 6-(12-7)),
gcd(6-6, 6-(12-6)) = 0,
of which only two give 1 as a result, thus a(6) = 2.
--
For n=10, sigma(10) = 18, we have 19 cases to consider:
gcd(10-0, 10-(18-0)) = 2 = gcd(10-18, 10-(18-18)),
gcd(10-1, 10-(18-1)) = 1 = gcd(10-17, 10-(18-17)),
gcd(10-2, 10-(18-2)) = 2 = gcd(10-16, 10-(18-16)),
gcd(10-3, 10-(18-3)) = 1 = gcd(10-15, 10-(18-15)),
gcd(10-4, 10-(18-4)) = 2 = gcd(10-14, 10-(18-14)),
gcd(10-5, 10-(18-5)) = 1 = gcd(10-13, 10-(18-13)),
gcd(10-6, 10-(18-6)) = 2 = gcd(10-12, 10-(18-12)),
gcd(10-7, 10-(18-7)) = 1 = gcd(10-11, 10-(18-11)),
gcd(10-8, 10-(18-8)) = 2 = gcd(10-10, 10-(18-10)),
gcd(10-9, 10-(18-9)) = 1,
of which 9 cases give 1 as a result, thus a(10) = 9.
--
For n=15, sigma(15) = 24, we have 25 cases to consider:
gcd(15-0, 15-(24-0)) = 3 = gcd(15-24, 15-(24-24)),
gcd(15-1, 15-(24-1)) = 2 = gcd(15-23, 15-(24-23)),
gcd(15-2, 15-(24-2)) = 1 = gcd(15-22, 15-(24-22)),
gcd(15-3, 15-(24-3)) = 6 = gcd(15-21, 15-(24-21)),
gcd(15-4, 15-(24-4)) = 1 = gcd(15-20, 15-(24-20)),
gcd(15-5, 15-(24-5)) = 2 = gcd(15-19, 15-(24-19)),
gcd(15-6, 15-(24-6)) = 3 = gcd(15-18, 15-(24-18)),
gcd(15-7, 15-(24-7)) = 2 = gcd(15-17, 15-(24-17)),
gcd(15-8, 15-(24-8)) = 1 = gcd(15-16, 15-(24-16)),
gcd(15-9, 15-(24-9)) = 6 = gcd(15-15, 15-(24-15)),
gcd(15-10, 15-(24-10)) = 1 = gcd(15-14, 15-(24-14)),
gcd(15-11, 15-(24-11)) = 2 = gcd(15-13, 15-(24-13)),
gcd(15-12, 15-(24-12)) = 3,
of which 2*4 = 8 cases give 1 as a result, thus a(15) = 8.
MATHEMATICA
Array[Sum[Boole[1 == GCD[#1 - i, #1 - (#2 - i)]], {i, 0, #2}] & @@ {#, DivisorSigma[1, #]} &, 85] (* Michael De Vlieger, Jun 09 2019 *)
PROG
(PARI) A324213(n) = { my(s=sigma(n)); sum(i=0, s, (1==gcd(n-i, n-(s-i)))); };
CROSSREFS
Sequence in context: A108503 A331595 A347288 * A052131 A329486 A051145
KEYWORD
nonn
AUTHOR
Antti Karttunen and David A. Corneth, May 26 2019, with better name from Charlie Neder, Jun 02 2019
STATUS
approved

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Last modified March 28 16:12 EDT 2024. Contains 371254 sequences. (Running on oeis4.)