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The number of divisions of $\mathbb{R}^3$ by $k \ge 0$ planes in general position starts 1,2,4,8, then 15, etc. For $\mathbb{R}^6$ it is 1,2,4,8,16,32,64 then 127. In general for $\mathbb{R}^N$ it is the sum of the binomial coefficients from $\binom{k}{0}$ up to $\binom{k}{N}$ and hence it agrees with $2^k$ for terms 0,1,2, up to N before starting to fall off.

other answers Of course for prime p, $2^{p-1}=1 \mod p$ but there are only 2 known cases $p=1093$ and $3511$ where $2^{p-1}=1 \mod p^2$. SO primes and primes with $2^{p-1} \ne 1 mod p^2$ agree for the first 182 primes.

For "listed in the OEIS" there are a couple which go from 1 to 99 then skip 100: undulating numbers in base 10 and cents you can have in US coins without having change for a dollar.

The number of divisions of $\mathbb{R}^3$ by $k \ge 0$ planes in general position starts 1,2,4,8, then 15, etc. For $\mathbb{R}^6$ it is 1,2,4,8,16,32,64 then 127. In general for $\mathbb{R}^N$ it is the sum of the binomial coefficients from $\binom{k}{0}$ up to $\binom{k}{N}$ and hence it agrees with $2^k$ for terms 0,1,2, up to N before starting to fall off.

other answers Of course for prime p, $2^{p-1}=1 \mod{p}$ but there are only 2 known cases $p=1093$ and $3511$ where $2^{p-1}=1 \mod{p^2}$. SO primes and primes with $2^{p-1} \ne 1 \mod{ p^2}$ agree for the first 182 primes.

For "listed in the OEIS" there are a couple which go from 1 to 99 then skip 100: undulating numbers in base 10 and cents you can have in US coins without having change for a dollar (the latter being 1-99 along with $105, 106, 107, 108, 109, 115, 116, 117, 118, 119$.)

The number of divisions of $\mathbb{R}^3$ by $k \ge 0$ planes in general position starts 1,2,4,8, then 15, etc. For $\mathbb{R}^6$ it is 1,2,4,8,16,32,64 then 127. In general for $\mathbb{R}^N$ it is the sum of the binomial coefficients from $\binom{k}{0}$ up to $\binom{k}{N}$ and hence it agrees with $2^k$ for terms 0,1,2, up to N before starting to fall off.

other answers Of course for prime p, $2^{p-1}=1 \mod p$ but there are only 2 known cases $p=1093$ and $3511$ where $2^{p-1}=1 \mod p^2$. SO primes and primes with $2^{p-1} \ne 1 mod p^2$ agree for the first 182 primes. For "listed in the OEIS there are a couple which go from 1 to 99 then skip 100.

The number of divisions of $\mathbb{R}^3$ by $k \ge 0$ planes in general position starts 1,2,4,8, then 15, etc. For $\mathbb{R}^6$ it is 1,2,4,8,16,32,64 then 127. In general for $\mathbb{R}^N$ it is the sum of the binomial coefficients from $\binom{k}{0}$ up to $\binom{k}{N}$ and hence it agrees with $2^k$ for terms 0,1,2, up to N before starting to fall off.

other answers Of course for prime p, $2^{p-1}=1 \mod p$ but there are only 2 known cases $p=1093$ and $3511$ where $2^{p-1}=1 \mod p^2$. SO primes and primes with $2^{p-1} \ne 1 mod p^2$ agree for the first 182 primes. 

For "listed in the OEIS" there are a couple which go from 1 to 99 then skip 100: undulating numbers in base 10 and cents you can have in US coins without having change for a dollar.

The number of divisions of $\mathbb{R}^3$ by $k \ge 0$ planes in general position starts 1,2,4,8, then 15, etc. For $\mathbb{R}^6$ it is 1,2,4,8,16,32,64 then 127. In general for $\mathbb{R}^N$ it is the sum of the binomial coefficients from $\binom{k}{0}$ up to $\binom{k}{N}$ and hence it agrees with $2^k$ for terms 0,1,2, up to N before starting to fall off.

The number of divisions of $\mathbb{R}^3$ by $k \ge 0$ planes in general position starts 1,2,4,8, then 15, etc. For $\mathbb{R}^6$ it is 1,2,4,8,16,32,64 then 127. In general for $\mathbb{R}^N$ it is the sum of the binomial coefficients from $\binom{k}{0}$ up to $\binom{k}{N}$ and hence it agrees with $2^k$ for terms 0,1,2, up to N before starting to fall off.

other answers Of course for prime p, $2^{p-1}=1 \mod p$ but there are only 2 known cases $p=1093$ and $3511$ where $2^{p-1}=1 \mod p^2$. SO primes and primes with $2^{p-1} \ne 1 mod p^2$ agree for the first 182 primes. For "listed in the OEIS there are a couple which go from 1 to 99 then skip 100.