B A = ( 1 1 ) 2 + ( 2 1 + 4 1 ) 2 + ( 3 1 + 6 1 + 9 1 ) 2 + ( 4 1 + 8 1 + 1 2 1 + 1 6 1 ) 2 + ⋯ ( 2 1 ) 2 + ( 3 1 + 6 1 ) 2 + ( 4 1 + 8 1 + 1 2 1 ) 2 + ( 5 1 + 1 0 1 + 1 5 1 + 2 0 1 ) 2 + ⋯
The equation above holds true for positive coprime integers A and B . Find the arithmetic mean of A and B .
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S A = ( 2 1 ) 2 + ( 3 1 + 6 1 ) 2 + ( 4 1 + 8 1 + 1 2 1 ) 2 + ⋯ = k = 1 ∑ ∞ H k 2 ( k + 1 ) − 2 = s h ( 2 , 2 ) = 3 6 0 1 1 π 4 where s h ( m , n ) is Euler sum. See note 1.
S B = ( 1 1 ) 2 + ( 2 1 + 4 1 ) 2 + ( 3 1 + 6 1 + 9 1 ) 2 + ⋯ = k = 1 ∑ ∞ H k 2 k − 2 = s h ( 2 , 2 ) + 2 s h ( 1 , 3 ) + ζ ( 4 ) = 3 6 0 1 1 π 4 + 2 × 3 6 0 π 4 + 9 0 π 4 = 3 6 0 1 7 π 4 See note 2. See note.
Therefore, S B S A = 1 7 1 1 , then 2 A + B = 2 1 1 + 1 7 = 1 4 .
See reference: Euler sum
Note 1: s h ( 2 , 2 ) = 2 3 ζ ( 4 ) + 2 1 [ ζ ( 2 ) ] 2 = 3 6 0 1 1 π 4 . . . ( 2 8 )
Note 2: k = 1 ∑ ∞ H k 2 k − n = s h ( 2 , n ) + 2 s h ( 1 , n + 1 ) + ζ ( n + 2 ) . . . ( 7 )
Note 3: s h ( 1 , n ) = 2 1 n ζ ( n + 1 ) − 2 1 k = 1 ∑ n − 2 ζ ( n − k ) ζ ( k + 1 ) . . . ( 2 1 ) ⟹ s h ( 1 , 3 ) = 2 3 ζ ( 4 ) − 2 1 ζ ( 2 ) ζ ( 2 ) = 3 6 0 π 4