A325968 a(n) is the sum k of such a subset of divisors of n with the largest sum and for which n-k and n-(sigma(n)-k) are relatively prime.

1, 3, 4, 7, 6, 7, 8, 15, 13, 17, 12, 27, 14, 23, 20, 31, 18, 38, 20, 41, 32, 35, 24, 59, 31, 39, 40, 29, 30, 71, 32, 63, 44, 53, 48, 91, 38, 59, 56, 89, 42, 95, 44, 83, 74, 69, 48, 123, 57, 93, 68, 95, 54, 119, 72, 119, 80, 89, 60, 167, 62, 95, 104, 127, 84, 143, 68, 125, 92, 143, 72, 194, 74, 113, 124, 137, 96, 167, 80, 185, 121, 125

Offset

1,2

Links

Antti Karttunen, Table of n, a(n) for n = 1..25000

Index entries for sequences related to sigma(n)

Formula

a(n) = A000203(n) - A325967(n).

a(n) = n + A325969(n).

For all n:

a(A000040(n)) = A000040(n)+1.

a(A000396(n)) = A000396(n)+1.

a(n) <= A325818(n).

Example

For n=15, its divisors are [1, 3, 5, 15]. If we take the full set [1, 3, 5, 15] and its complement [], their sums are 24 and 0, but gcd(15-0, 24-15) = gcd(15, 9) = 3 > 1. If we take subsets [1] and [3, 5, 15], then their sums are 1 and 23, but gcd(15-1, 23-15) = gcd(14,8) = 2 > 1. If we take subsets [3] and [1, 5, 15], their sums are 3 and 21, but gcd(15-3, 21-15) = gcd(12, 6) = 6 > 1. Only when we take the subset with the four smallest sum, [1, 3] and its complement [5, 15], we get such sums 4 and 20 for which gcd(15-4, 20-15) = gcd(11, 5) = 1. Thus a(15) = 20, the size of the subset with larger sum.

Prog

(PARI)
A325968(n) = { my(divs=divisors(n), s=sigma(n), r, ms=0); for(b=0, (2^(length(divs)))-1, r=sumbybits(divs, b); if(1==gcd(n-(s-r), n-r), ms=max(r, ms))); (ms); };
sumbybits(v, b) = { my(s=0, i=1); while(b>0, s += (b%2)*v[i]; i++; b >>= 1); (s); };

Crossrefs

Cf. A000040, A000203, A000396, A009194, A325807, A325818, A325967, A325969.

Sequence in context: A331694 A290270 A168275 * A325818 A120224 A210471

Adjacent sequences: A325965 A325966 A325967 * A325969 A325970 A325971

Keyword

nonn

Author

Antti Karttunen, May 29 2019

Status

approved