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By induction, $a_i=(p+u)\{i\frac{u}{p+u}\}$, where $\{x\}$ is a fractional part of $x$. Thus, by Weyl equidistribution theorem, if $p/u$ is irrational, the average equals $(p+u)/2$, otherwise $u/(p+u)=... • 105k 14 votes Accepted ### Closed form for$\sum_{n=0}^\infty \frac1{2^{2^n}}$? If you allow for a named number to be a closed form representation, the answer is "yes".$\sum_{n=0}^\infty (1/2)^{2^n}$is known as the Kempner number [1], a transcendental number [2]. More ... • 184k 12 votes ### Closed form expression for$\sum_{n=0}^{\infty} J_n^2(x) \cos(ny)$, where$J_n(x)$is the Bessel function of order$n$Just solve for the sum in the addition formula (31) of Neumann 1867, p. 65 (also in Watson p. 128, or more conceptually Vilenkin 1968, formula (4) p. 209): $$J_0\left(2r\sqrt{\frac{1-\cos\theta}2}\... • 30.3k 10 votes Accepted ### Permutations of the natural numbers with a common conditionally convergent series Let me rephrase your Question 2: How many permutations of the natural numbers are needed so that every conditionally convergent series of real numbers can be rearranged to no longer converge? This ... • 18k 9 votes Accepted ### Summation of binomial coefficients with alternating signs It's not hard to obtain explicit formulas:$$S_1(n,\alpha) = \alpha^{-1}\binom{n+1-\frac1{\alpha}}n^{-1}$$and$$S_2(n,\alpha) = (1-2\alpha)\alpha^{-2}\binom{n+2-\frac1{\alpha}}n^{-1}.$$• 32.6k 6 votes ### Possible new series for \pi A related, and perhaps easier, question is whether there are other known series for 𝜋 that involve a complex parameter 𝜆 in the summand, but where the sum of the series is independent of the value ... • 2,549 5 votes ### Is it known that the Collatz-like sequence with 7n+1 diverges to infinity starting with 7? Since others are contributing knowledge about other factors than 7, I figured that I would chip in. It's been known since 1987 that the orbit of t^3 + 1 diverges in the ring \mathbb{F}_2[t] ... • 355 5 votes Accepted ### Do disjunctive sequences eventually get palindromic at some point? Any disjunctive sequence of 0's and 1's will always have a palindromic initial segment: say it starts with 1, the next digit must be 0 to avoid an immediate palindrome, and then the next 1 ... 4 votes Accepted ### 5 different ways to define the same family of integer sequences Let's prove that a_3=a_1. Note that the recurrence for R translates to the following PDE for the generating F(x,y):=\sum_{n,m\geq0} R(n,m,p,q) \frac{x^n}{n!}\frac{y^m}{m!}:$$\frac{\partial}{\... • 32.6k 4 votes ### 5 different ways to define the same family of integer sequences Here is a proof that$a_1(n, p, q) = a_2(n,p,q)$. The generating function for the Stirling numbers of the second kind is $$\sum_{n=i}^\infty {n\brace i}\frac{x^n}{n!}= \frac{(e^x-1)^i}{i!}.$$ Also $$\... • 16.8k 4 votes Accepted ### An inequality about binomial distribution Another try, now I claim that the inequality holds. For a random variable X, denote \|X\|=(\mathbb E |X|^\sigma)^{1/\sigma}. It is indeed a norm. Then$$ \|X_1+\ldots+X_n\|\leqslant \|X_1+\ldots+... • 105k 3 votes ### Is there another representation for the summation:$\sum_{j=1}^{N}\frac{a_j}{(c+a_j)(c+a_j+1)} $, how to reformulate that to keep$c$out of the sum A few words on existence and approximation. Write your equation as $$\sum_{n=1}^N \frac{c^2a_n}{(c+b_n)(c+b_n+1)}=1.$$ There is a nice partial fractions decomposition, $$\frac{c^2 }{(c+b )(c+b +1)}=1+... • 58.5k 3 votes Accepted ### R-recursion for Fibonacci numbers using signed Catalan numbers Let R(x,y) = \sum_{n,q} R(n,q)x^ny^q. Then, you can show that \begin{eqnarray*} R(x,y)\left(x - y - xyC(-y)\right) &=& x R(x,0) - yR(0,y) \\ &=& xR(x,0) - \frac{y}{1-y} \end{eqnarray*... • 4,162 2 votes ### Possible new series for \pi This answers the final question about other parametric similar series for \pi. Working on the Saha & Sinha paper, it is possible to get a bi-parametric generalization of a related formula. If s\... • 2,720 2 votes ### Conjectured Somos-like closed form of recurrences with polynomial coefficients This is just an extended comment. There is no need to invoke the algebraic dependency search. The recurrence can be found by directly constructing Groebner basis B under any term order, in which n ... • 32.6k 2 votes ### Lower bound in recurrence relation From the definition$$v_{k-1}=\frac1{15}\prod_{j=1}^k(2^j+1)== \frac1 {15}{2^{ {\frac12k(k+1)}}}\prod_{j=1}^k(1+2^{-j}),$$and from the bound of the infinite product$$\frac1 {15}\prod_{j=1}^\... • 58.5k 1 vote Accepted ### Lower bound in recurrence relation This recurrence looks fairly random, so let's try tackling it in parts. First, we will show that$\log v_k$and$\log n_k$grow like$k^2$. Indeed, it can not grow slower since even if we leave the ... • 5,736 1 vote Accepted ###$R$-recursion for unsigned Genocchi numbers (of first kind) of even index From the recurrence it follows that the generating function$F(x,y) := \sum_{n,k\geq 0} R(n,k) \frac{x^{2n}}{(2n)!} \frac{y^k}{k!}\$ satisfies a neat PDE: $$\frac{\partial^2}{\partial^2 x} F(x,y) = \... • 32.6k 1 vote ### Simplest way to generate integer coefficients with row sums equal to the terms of an arbitrary given sequence I don't have any answer, but here are a few terms to be sure it is clear and to show some regularity in the constant or linear terms. Writing f_n := f(n) and taking f_0=1, it looks like$$\begin{...

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