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Now with four numbers computed and found, I searched the OEIS and found the corresponding sequence. Since the Online Encyclopedia already has this sequence, I'm hanging up my computational hat and heading off to work. :)

http://oeis.org/A086560

Start of first run of n successive numbers in which i-th number has exactly i distinct prime divisors for i = 1..n

2, 5, 64, 1867, 491851, 17681491, 35565206671

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 64, p. 23, Ellipses, Paris 2008.

You missed the excitement of 1998, with $1999=1999, 2000=2^4 \cdot 5^3, 2001=3 \cdot 23 \cdot 29,$ and $2002=2 \cdot 7 \cdot 11 \cdot 13$

A quick set of thoughts, too long to fit in a comment:

this requires finding a "prime gap" of length $k-1$, since $s(n+1)=1$ means that $n+1$ is prime, but the next $k-1$ digits are composite since s(n+x)>1 for $2 \le x \le k$. This also means that $s(n+2)=2$ only because $s(n+2)$ is even, thus $2$ is one of the factors and implies that $(n+2)/2$ is prime (or that $(n+2)/2^j$ is prime for some $j \in \mathbb{Z}$), since $n+2$ only has two factors and one of them is $2$ (or $2^j$).

For $s(n+k)$ to have $k$ distinct prime factors means that it has to be at minimum a product of the first $k$ prime numbers, while it definitely has to be a multiple of the product of $k$ prime numbers. So the two key restrictions are that s(n+k) is $k$-composite (has $k$ prime factors) and that both (n+1) and (n+2)/2 are prime numbers.

Hmm, I thought something about the fact that either s(n+2) or s(n+4) would be divisible by $4$ while the other would be divisible by $2$ but not by $4$ would play some role in this.

Here are some quick results from running "bash", "factor", and "sed" and "awk" at the command line:

If you want an ascending run of 1,2, and 3 prime factors, the smallest example starts at $n=63$, with $64=2^6, 65=5 \cdot 13, 66=2 \cdot 3 \cdot 11$

If you want an ascending run of 1,2,3, and 4 prime factors, we already missed the exciting years of $n=1866$ and $n=1998$

1867: 1867
1868: 2 2 467
1869: 3 7 89
1870: 2 5 11 17

1999: 1999
2000: 2 2 2 2 5 5 5
2001: 3 23 29
2002: 2 7 11 13

And the next few years with ascending runs of 1,2,3, and 4 prime factors will start after the years 3216, 4056, and 4176 with 3217, 4057, and 4177 as prime years. Unfortunately, these computational results are not giving me the germ of any shortcut or understanding. There are also some descending sequences in terms of the number of prime factors, and their placement also does not help.

You missed the excitement of 1998, with $1999=1999, 2000=2^4 \cdot 5^3, 2001=3 \cdot 23 \cdot 29,$ and $2002=2 \cdot 7 \cdot 11 \cdot 13$

A quick set of thoughts, too long to fit in a comment:

this requires finding a "prime gap" of length $k-1$, since $s(n+1)=1$ means that $n+1$ is prime that $n+1$ is either prime or a power of a prime, but the next $k-1$ digits are composite since s(n+x)>1 for $2 \le x \le k$. This also means that $s(n+2)=2$ only because $s(n+2)$ is even, thus $2$ is one of the factors and implies that $(n+2)/2$ is prime (or that $(n+2)/2^j$ is prime for some $j \in \mathbb{Z}$), since $n+2$ only has two factors and one of them is $2$ (or $2^j$).

For $s(n+k)$ to have $k$ distinct prime factors means that it has to be at minimum a product of the first $k$ prime numbers, while it definitely has to be a multiple of the product of $k$ prime numbers. So the two key restrictions are that s(n+k) is $k$-composite (has $k$ prime factors) and that both (n+1) and (n+2)/2 are prime numbers.

Hmm, I thought something about the fact that either s(n+2) or s(n+4) would be divisible by $4$ while the other would be divisible by $2$ but not by $4$ would play some role in this.

Here are some quick results from running "bash", "factor", and "sed" and "awk" at the command line:

If you want an ascending run of 1,2, and 3 prime factors, the smallest example starts at $n=63$, with $64=2^6, 65=5 \cdot 13, 66=2 \cdot 3 \cdot 11$

If you want an ascending run of 1,2,3, and 4 prime factors, we already missed the exciting years of $n=1866$ and $n=1998$

1867: 1867
1868: 2 2 467
1869: 3 7 89
1870: 2 5 11 17

1999: 1999
2000: 2 2 2 2 5 5 5
2001: 3 23 29
2002: 2 7 11 13

And the next few years with ascending runs of 1,2,3, and 4 prime factors will start after the years 3216, 4056, and 4176 with 3217, 4057, and 4177 as prime years. Unfortunately, these computational results are not giving me the germ of any shortcut or understanding. There are also some descending sequences in terms of the number of prime factors, and their placement also does not help.

If you want an ascending run of 1,2,3,4, and 5 prime factors, we have to wait almost half-a-million years to get to the exciting years of $n=491850$ and $n=521880$ for $k=5$

491851: 491851
491852: 2 2 122963
491853: 3 19 8629
491854: 2 11 79 283
491855: 5 7 13 23 47

521881: 521881
521882: 2 260941
521883: 3 3 3 3 17 379
521884: 2 2 11 29 409
521885: 5 7 13 31 37

A quick set of thoughts, too long to fit in a comment:

this requires finding a "prime gap" of length $k-1$, since $s(n+1)=1$ means that $n+1$ is prime, but the next $k-1$ digits are composite since s(n+x)>1 for $2 \le x \le k$. This also means that $s(n+2)=2$ only because $s(n+2)$ is even, thus $2$ is one of the factors and implies that $(n+2)/2$ is prime (or that $(n+2)/2^j$ is prime for some $j \in \mathbb{Z}$), since $n+2$ only has two factors and one of them is $2$ (or $2^j$).

For $s(n+k)$ to have $k$ distinct prime factors means that it has to be at minimum a product of the first $k$ prime numbers, while it definitely has to be a multiple of the product of $k$ prime numbers. So the two key restrictions are that s(n+k) is $k$-composite (has $k$ prime factors) and that both (n+1) and (n+2)/2 are prime numbers.

Hmm, I thought something about the fact that either s(n+2) or s(n+4) would be divisible by $4$ while the other would be divisible by $2$ but not by $4$ would play some role in this.

You missed the excitement of 1998, with $1999=1999, 2000=2^4 \cdot 5^3, 2001=3 \cdot 23 \cdot 29,$ and $2002=2 \cdot 7 \cdot 11 \cdot 13$

A quick set of thoughts, too long to fit in a comment:

this requires finding a "prime gap" of length $k-1$, since $s(n+1)=1$ means that $n+1$ is prime, but the next $k-1$ digits are composite since s(n+x)>1 for $2 \le x \le k$. This also means that $s(n+2)=2$ only because $s(n+2)$ is even, thus $2$ is one of the factors and implies that $(n+2)/2$ is prime (or that $(n+2)/2^j$ is prime for some $j \in \mathbb{Z}$), since $n+2$ only has two factors and one of them is $2$ (or $2^j$).

For $s(n+k)$ to have $k$ distinct prime factors means that it has to be at minimum a product of the first $k$ prime numbers, while it definitely has to be a multiple of the product of $k$ prime numbers. So the two key restrictions are that s(n+k) is $k$-composite (has $k$ prime factors) and that both (n+1) and (n+2)/2 are prime numbers.

Hmm, I thought something about the fact that either s(n+2) or s(n+4) would be divisible by $4$ while the other would be divisible by $2$ but not by $4$ would play some role in this.

Here are some quick results from running "bash", "factor", and "sed" and "awk" at the command line:

If you want an ascending run of 1,2, and 3 prime factors, the smallest example starts at $n=63$, with $64=2^6, 65=5 \cdot 13, 66=2 \cdot 3 \cdot 11$

If you want an ascending run of 1,2,3, and 4 prime factors, we already missed the exciting years of $n=1866$ and $n=1998$

1867: 1867
1868: 2 2 467
1869: 3 7 89
1870: 2 5 11 17

1999: 1999
2000: 2 2 2 2 5 5 5
2001: 3 23 29
2002: 2 7 11 13

And the next few years with ascending runs of 1,2,3, and 4 prime factors will start after the years 3216, 4056, and 4176 with 3217, 4057, and 4177 as prime years. Unfortunately, these computational results are not giving me the germ of any shortcut or understanding. There are also some descending sequences in terms of the number of prime factors, and their placement also does not help.