16
votes
Accepted
simple conjecture on palindromes in base 10
Using
$$\frac{10^c-1}{9}=\sum_{m=0}^{c-1} 10^m,$$
the product in question equals
$$\sum_{n=0}^{2a+2b+c-1}r(n)10^n,$$
where $r(n)$ is the number of times $n$ occurs among the numbers (counted with ...
- 105k
10
votes
Probability of sequence of coin flips palindrome
Let's distinguish some cases. If the first two flips result in HT or TH then with probability 1 you will flip a palindrome. Indeed, this will happen at the time $n$ when the $n$-th flip coincides with ...
- 796
8
votes
Accepted
Is it true that there are infinite palindromic primes that when squared give palindromic number?
There are in fact more than those 4, and they have their own page on OEIS. Two conjectures, then, would be:
are they infinitely many? I suspect so.
are $2$ and $3$ the only ones formed with decimal ...
- 5,407
8
votes
Accepted
Conjecture on palindromic numbers
The conjecture is true.
Let $b=a(n)=2^n+1$.
First, notice that the number in question is
$$N=(b^c-1)\frac{b^{m_1}+1}2\cdots \frac{b^{m_n}+1}2 = \frac{b^c-1}{b-1}(b^{m_1}+1)\cdots (b^{m_n}+1).$$
...
- 34.3k
2
votes
Probability of sequence of coin flips palindrome
Long comment: I think the expected number of tosses in $\infty$ and you can show it along the following lines: The probability of a palindrome of even length is the the same as the probablility ...
- 1,172
2
votes
Combinatorics of palindromic decompositions
To get started you can use oeis.org to investigate this. For instance, from plugging in the numerators corresponding to some of your data it seems that
$$\#P_n^{(2)}(1)=n(n-1)$$
("the oblong numbers")
...
- 24.8k
Only top scored, non community-wiki answers of a minimum length are eligible