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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
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)
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)))); };
KEYWORD
nonn
AUTHOR
Antti Karttunen and David A. Corneth, May 26 2019, with better name from Charlie Neder, Jun 02 2019
STATUS
approved