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A035092
Smallest k - dependent on n - such that (n^2)*k+1 is prime where k is the subscript of the progressions.
a(40) = 1 because in 1600k + 1 at k = 1, 1601 is the smallest prime; a(61)=46 because in 46*46*k+1 sequence the first prime appears at k=46, it is 171167.