Related papers: Curious congruences for Fibonacci numbers

In this paper we obtain some congruences involving central binomial coefficients and Lucas sequences. For example, we show that if p>5 is a prime then $\sum_{k=0}^{p-1}F_k*binom(2k,k)/12^k$ is congruent to 0,1,-1 modulo p according as p=1,4…

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.…

It is known that $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)4^k)=\pi/2$ and $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)16^k)=\pi/3$. In this paper we obtain their p-adic analogues such as…

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…

We prove congruences, modulo a power of a prime p, for certain finite sums involving central binomial coefficients $\binom{2k}{k}$.

In this paper we prove three conjectures on congruences involving central binomial coefficients or Lucas sequences. Let $p$ be an odd prime and let $a$ be a positive integer. We show that if $p\equiv 1\pmod{4}$ or $a>1$ then $$…

In this paper we establish some new congruences involving central binomial coefficients as well as Catalan numbers. Let $p$ be a prime and let $a$ be any positive integer. We determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}$ mod $p^2$ for…

Suppose that $p$ is an odd prime and $m$ is an integer not divisible by $p$. Sun and Tauraso [Adv. in Appl. Math., 45(2010), 125--148] gave $\sum_{k=0}^{n-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=0}^{n-1}\binom{2k}{k+d}/(km^k)$ modulo $p$ for…

We present several congruences modulo a power of prime $p$ concerning sums of the following type $\sum_{k=1}^{p-1}{m^k\over k^r}{2k\choose k}^{-1}$ which reveal some interesting connections with the analogous infinite series.

Let $p>5$ be a prime. We prove congruences modulo $p^{3-d}$ for sums of the general form $\sum_{k=0}^{(p-3)/2}\binom{2k}{k}t^k/(2k+1)^{d+1}$ and $\sum_{k=1}^{(p-1)/2}\binom{2k}{k}t^k/k^d$ with $d=0,1$. We also consider the special case…

Let $p>3$ be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{2k}{k}/2^k=(-1)^{(p-1)/2}-p^2E_{p-3} (mod p^3),$$ $$\sum_{k=1}^{(p-1)/2}\binom{2k}{k}/k=(-1)^{(p+1)/2}8/3*pE_{p-3} (mod p^2),$$…

Let $p$ be a prime and let $a$ be a positive integer. In this paper we investigate $\sum_{k=0}^{p^a-1}\binom[(h+1)k,k+d]/m^k$ modulo a prime $p$, where $d$ and $m$ are integers with $-h<d<=p^a$ and $m\not=0 (mod p)$. We also study…

Let $p>3$ be a prime. In this paper, we obtain the congruences for $$\sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^3}{(-8)^k},\ \sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^2\binom{3k}k}{(-192)^k},\…

For $n=0,1,2,\ldots$ let $W_n=\sum_{k=0}^{[n/3]}\binom{2k}k \binom{3k}k\binom n{3k}(-3)^{n-3k}$, where $[x]$ is the greatest integer not exceeding $x$. Then $\{W_n\}$ is an Ap\'ery-like sequence. In this paper we deduce many congruences…

Recently the first author proved a congruence proposed in 2006 by Adamchuk: $\sum_{k=1}^{\lfloor\frac{2p}{3}\rfloor}\binom{2k}{k}\equiv 0\pmod{p^2}$ for any prime $p=1 \pmod{3}$. In this paper, we provide more examples (with proofs) of…

Let $\{f_n\}$ be the Franel numbers given by $f_n=\sum_{k=0}^n\binom nk^3$, and let $p>5$ be a prime. In this paper we mainly determine $\sum_{k=0}^{p-1} \binom{2k}k\frac{f_k}{m^k}\pmod p$ for $m=5,-16,16,32,-49,50,96$. Let…

We present some congruences modulo $p^{6-d}$ for sums of the type $\sum_{k=0}^{(p-3)/2}x^k{2k\choose k}/(2k+1)^d$, for $d=1,2,3$ where $p>5$ is a prime.

For $k=1,2,\ldots$ let $H_k$ denote the harmonic number $\sum_{j=1}^k 1/j$. In this paper we establish some new congruences involving harmonic numbers. For example, we show that for any prime $p>3$ we have…

Let $p>3$ be a prime. For any $p$-adic integer $a$, we determine $$\sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k,\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k^{(2)},\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}k\frac{H_k^{(2)}}{2k+1}$$ modulo…

Let p be any odd prime. We mainly show that $$\sum_{k=1}^{p-1}binomial(3k,k)*2^k/k=0 (mod p)$$ and $$\sum_{k=1}^{p-1}2^{k-1}C_k^{(2)}=(-1)^{(p-1)/2}-1 (mod p),$$ where $C_k^{(2)}=binomial(3k,k)/(2k+1)$ is the $k$th Catalan number of order…