A352407 The number of terms before reaching zero when starting at n and iterating: f(n) = n, f(n+1) = n+1; f(n+k) = (f(n+k-2) * f(n+k-1)) (mod (n+k)), where k>=2.
2, 3, 14, 5, 5, 11, 4, 42, 54, 6, 17, 38, 6, 27, 12, 71, 20, 5, 6, 8, 12, 12, 42, 37, 36, 23, 22, 9, 5, 19, 10, 35, 31, 31, 60, 47, 33, 44, 46, 15, 8, 49, 14, 9, 12, 23, 35, 34, 28, 11, 86, 43, 20, 49, 18, 17, 12, 9, 22, 45, 26, 5, 31, 51, 72, 7, 6, 121, 120, 111, 86, 341, 56, 63, 12, 85, 12, 21
Offset: 0
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Examples
a(0) = 2 as starting at 0 and 1 gives 0*1 % 2 = 0, with two terms before reaching zero. This is the smallest possible value and the only term to equal 2. a(2) = 14 as starting at 2 and 3 gives 2*3 % 4 = 2, 3*2 % 5 = 1, 2*1 % 6 = 2, 1*2 % 7 = 2, 2*2 % 8 = 4, 2*4 % 9 = 8, 4*8 % 10 = 2, 8*2 % 11 = 5, 2*5 % 12 = 10, 5*10 % 13 = 11, 10*11 % 14 = 12, 11*12 % 15 = 12, 12*12 % 16 = 0, with fourteen terms before reaching zero. a(3) = 5 as starting at 3 and 4 gives 3*4 % 5 = 2, 4*2 % 6 = 2, 2*2 % 7 = 4, 2*4 % 8 = 0, with five terms before reaching zero.
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