In the list of consecutive positive base-ten integers: 1, 2, 3, ...
the 1111111111st occurrence of 1 is the first 1 in 1111111111,
the 2222222222nd occurrence of 2 is the second 2 in 2222222222,
the 3333333333rd occurrence of 3 is the third 3 in 3333333333,
the 4444444444th occurrence of 4 is the fourth 4 in 4444444444,
the 5555555555th occurrence of 5 is the fifth 5 in 5555555555,
the 6666666666th occurrence of 6 is the sixth 6 in 6666666666,
the 7777777777th occurrence of 7 is the seventh 7 in 7777777777,
the 8888888888th occurrence of 8 is the eighth 8 in 8888888888, and
the 9999999999th occurrence of 9 is the ninth 9 in 9999999999.
In other bases, the n-th occurrence of 1 as the first 1 in n is given in A023037 as a base-ten equivalent. Be careful about the offset (base two is the third term). Multiply the number by d for the n-th occurrence of d as the d-th d in n, up to d = base - 1. These n-th occurrence in n assertions are special solutions to the more general problem but they are not the only ones (all base-ten solutions are illustrated here).