A problem in group theory (indices of imprimitive groups) gives rise to the following conjectures in number theory. Suppose a positive integer $n$ has binary and ternary expansions $n=\sum_{k\geqslant0}b_k2^k=\sum_{k\geqslant0}t_k3^k$. For example, $6=2^2+2=2\cdot 3$; $81=2^6+2^4+1=3^4$.

Conjecture 1. If $\sum_{k\geqslant0}b_k=\sum_{k\geqslant0}t_k=2$, then $n\in\{6,10,12,18,36\}$.

Conjecture 2. If $\sum_{k\geqslant0}(b_k+t_k)=4$, then $n\in\{6,10,12,18,36,81\}$.

Terry Tao discusses the separation of powers of 2 and powers of 3; see his blog:

http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/

Conjectures 1 and 2 are related to separation problems. For example, the four $n$ with $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,2)$ or $(2,1)$ are related to the solutions to $|3^p-2^q|=1$, namely $(p,q)=(1,1),(2,3),(0,1),(1,2)$. More trivially, $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,1)$ is related to the solution to $|3^p-2^q|=0$, namely $(p,q)=(0,0)$. I wonder whether one needs the values for the `effective constants' in Tao's blog, or whether elementary arguments suffice to prove these conjectures.

Answers to the question " $3^n - 2^m = \pm 41$ is not possible. How to prove it? " may help.

A problem in group theory (indices of imprimitive groups) gives rise to the following conjectures in number theory. Suppose a positive integer $n$ has binary and ternary expansions $n=\sum_{k\geqslant0}b_k2^k=\sum_{k\geqslant0}t_k3^k$. For example, $6=2^2+2=2\cdot 3$; $81=2^6+2^4+1=3^4$.

Conjecture 1. If $\sum_{k\geqslant0}b_k=\sum_{k\geqslant0}t_k=2$, then $n\in\{6,10,12,18,36\}$.

Conjecture 2. If $\sum_{k\geqslant0}(b_k+t_k)=4$, then $n\in\{6,10,12,18,36,81\}$.

Terry Tao discusses the separation of powers of 2 and powers of 3; see his blog:

http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/

Conjectures 1 and 2 are related to separation problems. For example, the four $n$ with $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,2)$ or $(2,1)$ are related to the solutions to $|3^p-2^q|=1$, namely $(p,q)=(1,1),(2,3),(0,1),(1,2)$. More trivially, $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,1)$ is related to the solution to $|3^p-2^q|=0$, namely $(p,q)=(0,0)$. I wonder whether one needs the values for the `effective constants' in Tao's blog, or whether elementary arguments suffice to prove these conjectures.

Answers to the question " $3^n - 2^m = \pm 41$ is not possible. How to prove it? " may help.

A problem in group theory (indices of imprimitive groups) gives rise to the following conjectures in number theory. Suppose a positive integer $n$ has binary and ternary expansions $n=\sum_{k\geqslant0}b_k2^k=\sum_{k\geqslant0}t_k3^k$. For example, $6=2^2+2=2\cdot 3$; $81=2^6+2^4+1=3^4$.

Conjecture 1. If $\sum_{k\geqslant0}b_k=\sum_{k\geqslant0}t_k=2$, then $n\in\{6,10,12,18,36\}$.

Conjecture 2. If $\sum_{k\geqslant0}(b_k+t_k)=4$, then $n\in\{1,2,3,4,6,9,10,12,18,36,81\}$.

Terry Tao discusses the separation of powers of 2 and powers of 3; see his blog:

http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/

Conjectures 1 and 2 are related to separation problems. For example, the four $n$ with $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,2)$ or $(2,1)$ are related to the solutions to $|3^p-2^q|=1$, namely $(p,q)=(1,1),(2,3),(0,1),(1,2)$. More trivially, $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,1)$ is related to the solution to $|3^p-2^q|=0$, namely $(p,q)=(0,0)$. I wonder whether one needs the values for the `effective constants' in Tao's blog, or whether elementary arguments suffice to prove these conjectures.

Answers to the question " $3^n - 2^m = \pm 41$ is not possible. How to prove it? " may help.

A problem in group theory (indices of imprimitive groups) gives rise to the following conjectures in number theory. Suppose a positive integer $n$ has binary and ternary expansions $n=\sum_{k\geqslant0}b_k2^k=\sum_{k\geqslant0}t_k3^k$. For example, $6=2^2+2=2\cdot 3$; $81=2^6+2^4+1=3^4$.

Conjecture 1. If $\sum_{k\geqslant0}b_k=\sum_{k\geqslant0}t_k=2$, then $n\in\{6,10,12,18,36\}$.

Conjecture 2. If $\sum_{k\geqslant0}(b_k+t_k)=4$, then $n\in\{6,10,12,18,36,81\}$.

Terry Tao discusses the separation of powers of 2 and powers of 3; see his blog:

http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/

Conjectures 1 and 2 are related to separation problems. For example, the four $n$ with $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,2)$ or $(2,1)$ are related to the solutions to $|3^p-2^q|=1$, namely $(p,q)=(1,1),(2,3),(0,1),(1,2)$. More trivially, $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,1)$ is related to the solution to $|3^p-2^q|=0$, namely $(p,q)=(0,0)$. I wonder whether one needs the values for the `effective constants' in Tao's blog, or whether elementary arguments suffice to prove these conjectures.

Answers to the question " $3^n - 2^m = \pm 41$ is not possible. How to prove it? " may help.