(i) If n is an odd positive integer, then the number of Sylow 2-subgroups of the dihedral group of order 2n is n. Therefore all terms of this sequence are even.
(ii) Let p be a prime and let q be a power of p. The number of Sylow p-subgroups of PGL(2,q) is q+1.
(iii) Let q be a power of a prime and let r be a prime factor of q-1. Then the group of all affine transformations from the finite field F_q to itself has exactly q Sylow r-subgroups. This implies that q is a nonmember of the sequence, as long as q-1 has prime factors. Thus it works for all q except 2. (iv) The Third Sylow Theorem: If G is a finite group and p is a prime factor of |G|, then the number of Sylow p-subgroups of G is congruent to 1 mod p. 2 is in the sequence, since 2 is not congruent to 1 modulo any primes. It is never again this easy. Therefore we need an easier equivalent form of the membership criterion:
Easier Test. n > 2 is in this sequence iff the symmetric group S_n lacks a transitive subgroup with exactly n Sylow p-subgroups.
Proof. If S_n has a transitive subgroup with exactly n Sylow p-subgroups, then n trivially is a nonmember of this sequence.
For the other direction, suppose that G has exactly n Sylow p-subgroups. Then we will show that S_n has a transitive subgroup with exactly n Sylow p-subgroups. Let P be a Sylow p-subgroup of G and let N be its normalizer in G. Let G act on its Sylow p-subgroups by conjugation. This gives a homomorphism phi: G --> S_n. We want to show that phi(G) is also a group with exactly n Sylow p-subgroups. It is transitive by the Second Sylow Theorem.
First, phi(P) is a nontrivial subgroup of G: By properties of Sylow subgroups, P normalizes none of the other Sylow p-subgroups of G. Therefore phi(P) is a p-subgroup of G. Also, [phi(G):phi(P)] divides [G:P], which is a nonmultiple of p. Therefore phi(P) is a Sylow p-subgroup of phi(G). Since N normalizes P, phi(N) normalizes phi(P). Finally, nothing outside phi(N) can normalize phi(P), by the definition of phi.
Likewise phi(N) is a point-stabilizer in phi(G). Since phi(G) is a transitive subgroup of S_n, the index of phi(N) in phi(G) is n. Since phi(N) is the normalizer of phi(P) in phi(G), this means the number of Sylow p-subgroups of phi(G) is n and we are done.
We continue the notation used above with G, N and P in what follows. (For these comments, isolated examples are given that show different variations on the reasoning used to prove membership in this sequence.)
22 is in the sequence: If G is a finite group with exactly 22 Sylow p-subgroups, then p= 3 or 7. (Since there is no need to divide into cases based on the value of p, in what follows p= 3 or 7.) 22 has no proper divisors congruent to 1 mod p, so N is a maximal subgroup of G. Then the Easier Test can be strengthened to say "primitive" where it says "transitive". There are only 4 nonisomorphic primitive subgroups of S_22: M_22, Aut(M_22), A_22 and S_22. All of these have more than 22 Sylow p-subgroups.
34 is in the sequence: If G is a finite group with exactly 34 Sylow p-subgroups, then p= 3 or 11. 34 has no proper divisors congruent to 1 mod p and (proceeding as before) the only primitive subgroups of S_34 are S_34 and A_34. Both have more than 34 Sylow p-subgroups.
88 is a member: the only primitive permutation groups of degree 88 are S_88 and A_88. Neither of these has exactly 88 Sylow 29-subgroups and 88 has no proper divisors congruent to 1 mod 29. The divisors of 88 congruent to 1 mod 3 are 1, 4, 22 and 88. Then a minimal extension K of a Sylow 3-subgroup normalizer N must have [K:N]=4 (22 was already ruled out, as was 88 by the fact that S_88 and A_88 have more than 88 Sylow 3-subgroups). K will be maximal in G and have [G:K]=22. All of M_22, Aut(M_22), A_22 or S_22 have more than 88 Sylow 3-subgroups and G itself will have at least as many.