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Related papers: Explicit Formulas for the p-adic Valuations of Fib…

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We state a general formula for the number of binomial coefficients $n$ choose $k$ that are divided by a fixed power of a prime $p$, i.e., the number of binomial coefficients divided by $p^j$ and not divided by $p^{j+1}$.

General Mathematics · Mathematics 2008-03-10 William B. Everett

We show that the $p$-adic valuation of the sequence of Fibonacci numbers is a $p$-regular sequence for every prime $p$. For $p \neq 2, 5$, we determine that the rank of this sequence is $\alpha(p) + 1$, where $\alpha(m)$ is the restricted…

Number Theory · Mathematics 2015-10-15 Luis A. Medina , Eric Rowland

If p is a prime and n a positive integer, let v(n) denote the exponent of p in n, and u(n)=n/p^{v(n)} the unit part of n. If k is a positive integer not divisible by p, we show that the p-adic limit of (-1)^{pke} u((kp^e)!) as e goes to…

Number Theory · Mathematics 2013-01-29 Donald M. Davis

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that $\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ and $\nu_p(\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{\binom{2k}k}{m^k})$ are at…

Number Theory · Mathematics 2011-04-14 Zhi-Wei Sun

Let p be a prime and let a be a positive integer. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=1}^{p-1}\binom{2k}{k+d}/(km^{k-1})$ modulo $p$ for all d=0,...,p^a, where m is any integer not divisible by p.…

Number Theory · Mathematics 2010-04-02 Zhi-Wei Sun , Roberto Tauraso

Let $p$ be a prime. In this paper, we present a detailed $p$-adic analysis to factorials and double factorials and their congruences. We give good bounds for the $p$-adic sizes of the coefficients of the divided universal Bernoulli number…

Number Theory · Mathematics 2013-08-23 Shaofang Hong , Jianrong Zhao , Wei Zhao

The study of prime divisibility plays a crucial role in number theory. The $p$-adic valuation of a number is the highest power of a prime, $p$, that divides that number. Using this valuation, we construct $p$-adic valuation trees to…

Number Theory · Mathematics 2023-08-24 Dillon Snyder

Let k>1 be an integer and let p be a prime. We show that if $p^a\le k<2p^a$ or $k=p^aq+1$ (with 2q<p) for some a=1,2,..., then the set {\binom{n}{k}: n=0,1,2,...} is dense in the ring Z_p of p-adic integers, i.e., it contains a complete…

Number Theory · Mathematics 2011-01-26 Zhi-Wei Sun , Wei Zhang

We attempt to quantify the exact proportion of monic $p$-adic polynomials of degree $n$ which are irreducible. We find an exact answer to this when $n$ is prime and $p \neq n$, and also when $n = 4$ and $p \neq 2$. Our answers are rational…

Number Theory · Mathematics 2025-03-19 Isaac Rajagopal

Let f(n)= Sum binomial(n,k)^(-1). First, we show that f:N to Q_p is nowhere continuous in the p-adic topology. If x is a p-adic integer, we say that f(x) is p-definable if lim f(x_j) exists in Q_p, where x_j denotes the jth partial sum for…

Number Theory · Mathematics 2012-08-02 Donald M. Davis

We propose higher-order generalizations of Jacobsthal's $p$-adic approximation for binomial coefficients. Our results imply explicit formulae for linear combinations of binomial coefficients $\binom{ip}{p}$ ($i=1,2,\dots$) that are…

Number Theory · Mathematics 2018-08-31 Rustem R. Aidagulov , Max A. Alekseyev

The coefficients c(n,k) defined by (1-k^2x)^(-1/k) = sum c(n,k) x^n reduce to the central binomial coefficients for k=2. Motivated by a question of H. Montgomery and H. Shapiro for the case k=3, we prove that c(n,k) are integers and study…

Number Theory · Mathematics 2015-05-13 Armin Straub , Tewodros Amdeberhan , Victor H. Moll

In this document will be proved a formula to compute the $p$-adic valuation of a hyperfactorial. We call a hyperfactorial the result of multiplying a given number of consecutive integers from 1 to the given number,each raised to its own…

Number Theory · Mathematics 2021-09-14 Luca Onnis

In number theory, we know Legendre's formula $ v_p(n!) = \sum_{k \ge 1} \lfloor \frac{n}{p^k} \rfloor $, which calculates the $p$-adic valuation of the factorial, i.e. the exponent of the greatest power of a prime $p$ that divides $n!$.…

Number Theory · Mathematics 2019-07-30 Gennady Eremin

This paper investigates the p-adic valuation trees of degree-2 and degree-3 polynomials in two variables over any prime p, building upon prior research outlined in [14].

General Mathematics · Mathematics 2024-07-16 Shubham

We prove an explicit formula for the $p$-adic valuation of the Legendre polynomials $P_n(x)$ evaluated at a prime $p$, and generalize an old conjecture of the third author. We also solve a problem proposed by Cigler in 2017.

Number Theory · Mathematics 2025-05-15 Max A. Alekseyev , Tewodros Amdeberhan , Jeffrey Shallit , Ingrid Vukusic

Let $p$ be an odd prime. It is well known that $F_{p-(\frac p5)}\equiv 0\pmod{p}$, where $\{F_n\}_{n\ge0}$ is the Fibonacci sequence and $(-)$ is the Jacobi symbol. In this paper we show that if $p\not=5$ then we may determine $F_{p-(\frac…

Number Theory · Mathematics 2013-11-01 Zhi-Wei Sun

We examine the behavior of the sequences of $p$-adic valuations of quadratic polynomials with integer coefficients for an odd prime $p$ through tree representations. Under this representation, a finite tree corresponds to a periodic…

Number Theory · Mathematics 2023-09-29 Will Boultinghouse , Emily Hammett , Stephen Hu , Olena Kozhushkina , Rachel Snyder , Justin Trulen

Let $p$ be a prime. In this paper, we give a complete classification of self-reciprocal polynomials arising from Fibonacci polynomials over $\mathbb{Z}$ and $\mathbb{Z}_p$, where $p=2$ and $p>5$. We also present some partial results when…

Number Theory · Mathematics 2019-01-01 Neranga Fernando , Mohammad Rashid

We prove an asymptotic formula for primes of the shape $f(a,b^2)$ with $a,b$ integers and of the shape $f(a,p^2)$ with $p$ prime. Here $f$ is a binary quadratic form with integer coefficients, irreducible over $\mathbb{Q}$ and has no local…

Number Theory · Mathematics 2024-09-25 Stanley Yao Xiao
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