A337297 a(n) = sigma(n)*(tau(n) - 1).
0, 3, 4, 14, 6, 36, 8, 45, 26, 54, 12, 140, 14, 72, 72, 124, 18, 195, 20, 210, 96, 108, 24, 420, 62, 126, 120, 280, 30, 504, 32, 315, 144, 162, 144, 728, 38, 180, 168, 630, 42, 672, 44, 420, 390, 216, 48, 1116, 114, 465, 216, 490, 54, 840, 216, 840, 240, 270, 60, 1848, 62, 288
Offset: 1
Examples
a(3) = 4; The divisors of 3 are {1,3}. If we form all ordered pairs (d1,d2) such that d1 < d2, we have: (1,3). The sum of the coordinates gives 1+3 = 4.
a(4) = 14; The divisors of 4 are {1,2,4}. If we form all ordered pairs (d1,d2) such that d1<d2, we have: (1,2), (1,4), (2,4). The sum of all the coordinates gives 1+2+1+4+2+4 = 14.
Programs
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Mathematica
Table[Sum[Sum[(i + k)*(1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k - 1}], {k, n}], {n, 100}] -
PARI
a(n) = my(d = divisors(n)); sum(i=1, #d, sum(j=1, i-1, d[i]+d[j])); \\ Michel Marcus, Aug 22 2020
Formula
a(n) = Sum_{d1|n, d2|n, d1
a(p^k) = k*(p^(k+1)-1)/(p-1) for p prime and k >= 1. - Wesley Ivan Hurt, Aug 23 2025
Extensions
New name using formula from Ridouane Oudra, Jul 31 2025
A337298 Sum of the coordinates of all relatively prime pairs of divisors of n, (d1,d2), such that d1 <= d2.
2, 5, 6, 10, 8, 21, 10, 19, 16, 29, 14, 46, 16, 37, 36, 36, 20, 61, 22, 64, 46, 53, 26, 91, 34, 61, 44, 82, 32, 141, 34, 69, 66, 77, 64, 136, 40, 85, 76, 127, 44, 181, 46, 118, 106, 101, 50, 176, 60, 133, 96, 136, 56, 173, 92, 163, 106, 125, 62, 316, 64, 133, 136, 134, 106, 261, 70
Offset: 1
Examples
a(4) = 10; There are 3 divisors of 4: {1,2,4}. If we list the relatively prime pairs (d1,d2), where d1 <= d2, we get (1,1), (1,2), (1,4). The sum of the coordinates from all pairs is 1+1+1+2+1+4 = 10.
a(5) = 8; There are 2 divisors of 5: {1,5}. The relatively prime pairs (d1,d2), where d1 <= d2 are: (1,1) and (1,5). The sum of the coordinates is then 1+1+1+5 = 8.
Programs
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Mathematica
Table[Sum[Sum[(i + k) KroneckerDelta[GCD[i, k], 1] (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k}], {k, n}], {n, 100}] -
PARI
a(n) = my(d = divisors(n)); sum(i=1, #d, sum(j=1, i, if (gcd(d[i],d[j])==1, d[i]+d[j]))); \\ Michel Marcus, Aug 22 2020
Formula
a(n) = Sum_{i|n, k|n, i<=k, gcd(i,k)=1} (i+k).
Comments