You want a $\neq$ at the end there.

I must be misreading something. Take $v_1 = (a_1 + a_2, a_1a_2)$ and $v_2 = a_1a_2, a_1 + a_2)$. Then $\pi_2 = 1$, but $\pi_3 = 1/2$.

Try checking Ribenboim's book "My Numbers, My Friends". It's very much a survey, but if there's something to be said (i.e. if these are not open questions), it has the source. Your sequences are $A_0 = 0$, $A_1 = p-1$, and $A_n = (p+1)A_{n-1} - pA_{n-2}$, so covered by general linear recurrences.

It seems the congruences you want are $p \equiv \pm 11 \mod 30$, and $p \equiv \pm 1 \mod 30$. The other cases $(7,13,17,23)$ are taken care of.

$2$ is a root mod $29$: $16 + 96 + 56 - 24 + 1 = 88 + 57 = 87 + 58 = 3(29) + 2(29)$, so $h$ has a linear factor.

You can restrict to $m<n$, otherwise $A$ is invertible, and the bracketed expression is $0$.

@Anthony: Consider $x=0$, so that we want $\frac{p}{q} < \frac{1}{q^r}$. This has no solution for any $r \geq 1$ when $p \neq 0$. The statement of the question should say that $\frac{p}{q} \neq x$, then it follows that all rationals have $\mu(x) = 1$.

@Geoff: For that we have $(1,3,4), (2,9,13), (5,12,15), (6,8,10), (7,11,14)$. This comes from considering how many of each triple (either $(0,0,0)$ or $(0,1,1)$) we must use for $\mathbb{Z}/2 \mathbb{Z}$. There's only one option, and the rest of it falls into place. You're right, I wasn't considering even values, and I also don't know $\mathbb Z / \left(10\right) \times \mathbb Z / \left(4\right)$ for $n=3$ or $\mathbb Z / \left(18\right) \times \mathbb Z / \left(2\right)$ for $n=5$. They seem to boil down to finding positive solutions to underdetermined linear systems.

I assume you mean the shape can be rotated, but can it be reflected?

Can you solve $n=3$ when $G = \mathbb Z / \left(5\right) \times \mathbb Z / \left(11\right)$? Cases $\mathbb Z / \left(p\right) \times \mathbb Z / \left(q\right)$ where $n$ divides both of $p-1, q-1$ are solved by patching together the partitions given by each prime. $\left|G\right| = 55$ is the smallest case for $n=3$ where this doesn't apply. I haven't made it work, but I also can't say that it doesn't.

For G odd, no element has order 2, hence we can pair each non-identity element with its inverse to get zero sum sets of size 2. For multiples of 2, combine the correct number of these pairs.

How many such orderings do you plan to use in the same section? It's probably best for the reader if you just use $\preceq$ and explicitly state what you mean. That way there's no confusion, and no unwieldy notation (like $X_2^2 \preceq_{\overrightarrow{231}} X_2X_3$)

See this question: mathoverflow.net/questions/19840/… @Guillaume: Here is a link to Poonen's paper lifted directly from above www-math.mit.edu/~poonen/papers/ae.pdf

"Since $∗+∗=0$ even as misere games" I think you mean $\\{0\\} + \\{0\\} = \\{0|0\\}$, i.e. it is a first player win as a misere game. In misere Nim, the one matchstick game is a first player loss, so one usually does not denote it by $*$.

Ricky Demer's argument works for any grid with a playable line of symmetry, so it also includes all rectangles in Lipp's version. But the same strategy does not work on asymmetric boards, like an L. Or starting with any rectangle, remove an interior point and say that lines cannot cross the missing point. Splitting the game into two disjoint, equal subgames would require passing through the missing point, so you will want to compute the nimbers of smaller subgames instead to find the winning strategy.

The gap can be resolved as follows: If $p$ is an odd prime dividing a Fibonacci number with odd index (it need not be a primitive factor), then it does not divide any Lucas number. This follows from $\gcd(F_a, F_b) = F_{\gcd(a,b)}$. We can always find such a prime with odd exponent in the factorization of $F_q$ for q >3 and odd, since otherwise we get $F_q = x^2$ or $F_q = 2x^2$.