Let $p$ be an odd prime. We prove that
$$\sum_{k=0}^{p-1}\binom{2k}{k}/2^k=(-1)^{(p-1)/2}-p^2E_{p-3} (mod p^3),$$
where E_0,E_1,E_2,... are Euler numbers. We also determine
$\sum_{k=0}^{p-1}\binom{2k}{k}^2/16^k$ mod $p^4$ and show that
$$\sum_{k=0}^{(p-1)/2}(4k+1)\binom{2k}{k}^2/16^k=p^2(2^p-1) (mod p^4).$$ We
formulate many conjectures concerning such super congruences and relate most of
them to Euler numbers.
In 1997 Ken Ono and K. Soundararajan [Invent. Math. 130(1997)] proved that
under the generalized Riemann hypothesis any positive odd integer greater than
2719 can be represented by the famous Ramanujan form $x^2+y^2+10z^2$,
equivalently the form $2x^2+5y^2+4T_z$ represents all integers greater than
1359, where $T_z$ denotes the triangular number $z(z+1)/2$.
Harmonic numbers $H_k=\sum_{0<j\le k}1/j (k=0,1,2,...)$ arise naturally in
many fields of mathematics. In this paper we initiate the study of congruences
involving both harmonic numbers and Lucas sequences. One of our three theorems
is as follows: Let u_0=0, u_1=1, and u_{n+1}=u_n-4u_{n-1} for n=1,2,3,....
Then, for any prime p>5 we have $$\sum_{k=0}^{p-1}u_{k+\delta}H_k/2^k=0 (mod
p),$$ where $\delta=0$ if p=1,2,4,8 (mod 15), and $\delta=1$ otherwise.
In this paper we establish some sophisticated congruences involving central
binomial coefficients and Fibonacci numbers. For example, we show that if
$p\not=2,5$ is a prime then
$$\sum_{k=0}^{p-1}F_{2k}\binom{2k}{k}=(-1)^{[p/5]}(1-(p/5)) (mod p^2)$$ and
$$\sum_{k=0}^{p-1}F_{2k+1}\binom{2k}k=(-1)^{[p/5]}(p/5) (mod p^2).$$ We also
obtain similar results for some other second-order recurrences.
In this paper we obtain some congruences involving central binomial
coefficients and Lucas sequences.We also raise several conjectures.
We collect here various conjectures on congruences made by the author in a
series of papers, some of which involve binary quadratic forms and other
advanced theories. Part A consists of unsolved conjectures of the author while
conjectures in Part B have been recently confirmed. We hope that this material
will interest number theorists and stimulate further research. Number theorists
are welcome to work on those open conjectures.
Harmonic numbers $H_k=\sum_{0<j<=k}1/j (k=0,1,2,...)$ play important roles in
mathematics. In this paper we investigate their arithmetic properties and
obtain various basic congruences. Let p>3 be a prime. We show that
$$\sum_{k=1}^{p-1}H_k/(k2^k)=0 (mod p), \sum_{k=1}^{p-1}H_k^2=2p-2 (mod p^2),
\sum_{k=1}^{p-1}H_k^3=6 (mod p),$$ and $$\sum_{k=1}^{p-1}H_k^2/k^2=0 (mod p)
provided p>5.$$ Our tools include some sophisticated combinatorial identities
and properties of Bernoulli numbers.
In this paper we obtain several congruences modulo an odd prime p which are
related to central binomial coefficients. For example,
$$\sum_{k=0}^{p-1}\binom{2k}{k}^3/64^k=\cases4x^2 (mod p)&if p=x^2+y^2 with x
odd and y even, \\0 (mod p)& if p=3 (mod 4);$$ and
$$\sum_{k=0}^{p-1}C_k^2}/16^k=-3 (mod p) and \sum_{k=0}^{p-1}C_k^3/64^k=7 (mod
p),$$ where $C_k$ denotes the Catalan number $\binom{2k}{k}/(k+1)$. We also
pose several challenging conjectures one of which states that if p=3,5,6 (mod
7)$ then $$\sum_{k=0}^{p-1}\binom{2k}{k}^3=0 (mod p^2).$$
Let p be an odd prime and let a be a positive integer. In this paper we
investigate the sum $\sum_{k=0}^{p^a-1}\binom{hp^a-1}{k}\binom{2k}{k}/m^k$ mod
p^2, where h,m are p-adic integers with m\not=0 (mod p). For example, we show
that if h(2h-1)\not=0 (mod p) then $$
sum_{k=0}^{p^a-1}\binom{hp^a-1}{k}\binom{2k}{k}(-h/2)^k
=(\frac{1-2h}{p^a})(1+h((4h-2)^{p-1}/h^{p-1}-1)) (mod p^2),$$ where (-) denotes
the Jacobi symbol.
Sun and Tauraso conjectured that for any positive integer $a$ we have
$$\sum_{k=0}^{3^a-1}\binom{2k}{k}=0 (mod 3^{2a})$$ and furthermore
$$3^{-2a}}\sum_{k=0}^{3^a-1}\binom{2k}k=1 (mod 3).$$ Recently a $q$-analogue of
the first congruence was conjectured by Guo and Zeng. In this paper we prove
both conjectures.
Sun and Tauraso conjectured that for any positive integer $a$ we have
$$\sum_{k=0}^{3^a-1}\binom{2k}{k}=0 (mod 3^{2a})$$ and furthermore
$$3^{-2a}}\sum_{k=0}^{3^a-1}\binom{2k}k=1 (mod 3).$$ Recently a $q$-analogue of
the first congruence was conjectured by Guo and Zeng. In this paper we prove
both conjectures.
Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but
not dividing $m$. We show that
$$\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ is at least $\nu_p(n)$, where
$\nu_p(x)$ denotes the $p$-adic valuation of $x$ at $p$. Furthermore, if
$p\not=3$ or $3\nmid n$ then
$$n^{-1}\sum_{k=0}^{n-1}\frac{\bi{2k}k}{m^k}=\frac{\binom{2n-1}{n-1}}{m^{n-1}}
(mod p^{\nu_p(m-4)}).$$ This implies several conjectures of Guo and Zeng [GZ].
Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but
not dividing $m$. We show that
$$\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ is at least $\nu_p(n)$, where
$\nu_p(x)$ denotes the $p$-adic valuation of $x$ at $p$. Furthermore, if
$p\not=3$ or $3\nmid n$ then
$$n^{-1}\sum_{k=0}^{n-1}\frac{\bi{2k}k}{m^k}=\frac{\binom{2n-1}{n-1}}{m^{n-1}}
(mod p^{\nu_p(m-4)}).$$ This implies several conjectures of Guo and Zeng [GZ].
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. For example, we show that if
$p\not=2,5$ then
$$\sum_{k=1}^{p-1}(-1)^k\frac{\binom{2k}k}k=-5\frac{F_{p-(\frac p5)}}p (mod
p),$$ where F_n is the n-th Fibonacci number and (-) is the Jacobi symbol. We
also prove that if p>3 then $$\sum_{k=1}^{p-1}\frac{\binom{2k}k}k={8/9}
p^2B_{p-3} (mod p^3),$$ where B_n denotes the n-th Bernoulli number.
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. For example, we show that if
$p\not=2,5$ then
$$\sum_{k=1}^{p-1}(-1)^k\frac{\binom{2k}k}k=-5\frac{F_{p-(\frac p5)}}p (mod
p),$$ where F_n is the n-th Fibonacci number and (-) is the Jacobi symbol. We
also prove that if p>3 then $$\sum_{k=1}^{p-1}\frac{\binom{2k}k}k={8/9}
p^2B_{p-3} (mod p^3),$$ where B_n denotes the n-th Bernoulli number.
We prove that for any nonnegative integers $n$ and $r$ the binomial sum $$
\sum_{k=-n}^n\binom{2n}{n-k}k^{2r} $$ is divisible by
$2^{2n-\min\{\alpha(n),\alpha(r)\}}$, where $\alpha(n)$ denotes the number of
1's in the binary expansion of $n$. This confirms a recent conjecture of Guo
and Zeng.
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 2.
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 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 congruences involving higher-order Catalan numbers
$C_k^{(h)}=\binom[(h+1)k,k]/(hk+1)$. Our tools include linear recurrences and
the theory of cubic residues. Here are some typical results in the paper.
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 congruences involving higher-order Catalan numbers
$C_k^{(h)}=\binom[(h+1)k,k]/(hk+1)$. Our tools include linear recurrences and
the theory of cubic residues. Here are some typical results in the paper.
For m=3,4,..., the polygonal numbers of order m are given by
$p_m(n)=(m-2)n(n-1)/2+n (n=0,1,2,...)$. For positive integers $a,b,c$ and
$i,j,k>2$ with max{i,j,k}>4, we call the triple $(ap_i,bp_j,cp_k)$ universal if
for any n=0,1,2,... there are nonnegative integers $x,y,z$ such that
$n=ap_i(x)+bp_j(y)+cp_k(z)$. We show that there are only 95 candidates for
universal triples (two of which are $(p_4,p_5,p_6)$ and $(p_3,p_4,p_{27})$),
and conjecture that they are indeed universal triples.