Coprime triangles' row-sums

[Leroy Quet]

I just submitted the following 2 sequences:

>%S A114718 1,2,6,6,10,13,17,43,59,69
>%N A114718 a(n) = sum of terms in nth row of triangle A112599.
>%e A114718 The 7th row of triangle A112599 is [6,1,3,1,2,1,3].
>So a(7) = 6+1+3+1+2+1+3 = 17.
>%Y A114718 A112599,A114719
>%O A114718 1
>%K A114718 ,more,nonn,
>%A A114718 Leroy Quet (qq-quet at mindspring.com), Dec 27 2005



>%S A114719 0,1,4,10,14,15,21,34,40
>%N A114719 a(n) = sum of terms in nth row of triangle A112592.
>%e A114719 The 7th row of triangle A112592 is [6,3,3,3,0,3,3].
>So a(7) = 6+3+3+3+0+3+3 = 21.
>%Y A114719 A112592,A114718
>%O A114719 1
>%K A114719 ,more,nonn,
>%A A114719 Leroy Quet (qq-quet at mindspring.com), Dec 27 2005


I refer to triangle A112599 and A112592, which are here (first few rows, 
formatted in triangle form):

A112599:
         1
        1 1
       2 2 2
      3 0 3 0
     4 2 0 2 2
    5 0 4 0 4 0
   6 1 3 1 2 1 3
  7 5 4 5 7 3 7 5
 8 7 7 7 5 6 5 7 7
9 7 8 7 7 7 4 7 8 5
>%N A000001 Triangle where a(1,1) = 1, a(n,m) = number of terms of row 
>(n-1) which are coprime to m.
>%C A000001 GCD(m,0) is considered here to be m, so 0 is coprime to no 
>positive integer but 1.
>%e A000001 Row 6 of the triangle is [5,0,4,0,4,0]. Among these terms there 
>are 6 terms coprime to 1, 1 term coprime to 2, 3 terms coprime to 3, 1 
>term coprime to 4, 2 terms coprime to 5, 1 term coprime to 6, and 3 terms 
>coprime to 7. So row 7 is [6,1,3,1,2,1,3].
>%O A000001 1
>%K A000001 ,more,nonn,tabl,
>%A A000001 Leroy Quet (qq-quet at mindspring.com), Dec 21 2005

A112592:
        0,
       1,0,
      2,1,1,
     3,2,3,2,
    4,2,2,2,4,
   5,0,5,0,5,0,
  6,3,3,3,0,3,3,
 7,5,0,5,6,0,6,5,
8,4,4,4,3,4,5,4,4
>%N A000001 Triangle where a(1,1) = 0, a(n,m) = number of terms of row 
>(n-1) which are coprime to m.
>%C A000001 GCD(m,0) is considered here to be m, so 0 is coprime to no 
>positive integer but 1.
>%e A000001 Row 6 of the triangle is [5,0,5,0,5,0]. Among these terms there 
>are 6 terms coprime to 1, 3 terms coprime to 2, 3 terms coprime to 3, 3 
>terms coprime to 4, 0 terms coprime to 5, 3 terms coprime to 6, and 3 
>terms coprime to 7. So row 7 is [6,3,3,3,0,3,3].
>%Y A000001 A112599
>%O A000001 1
>%K A000001 ,more,nonn,tabl,
>%A A000001 Leroy Quet (qq-quet at mindspring.com), Dec 24 2005


Is there a simple way to calculate the row-sums?
All I get is, if b(n,k) is number of k's in nth row of each respective 
triangle
(so the sum of the nth row = sum{k>=0} b(n,k)*k, of course), then (I 
think)
the sum of the nth row also = sum{k>=0} b(n-1,k)*phi(n,k),

where phi(n,k) = number of positive integers <= n which are coprime to k.
(phi(n,0) = 1 for n = any positive integer, because 0 is only coprime to 
1.)

Is there any more that can be said about this?

thanks,
Leroy Quet