User nilotpal sinha - MathOverflow

Answer by Nilotpal Sinha for Are sums of the inverses of prime siblings finite?

2013-05-21T06:03:50Z

The following heuristic study suggests that for any gap $d \ge 2$ the sum of the reciprocal of primes having a gap of $d$ should be convergent.

Let $L(r,x)$ be the number of conjectured occurrences of gaps of size $2r$ between successive prime $\le x$. We have

$$ L(r,x) = \int_{2}^{x} \sum_{}{}\frac{(-1)^k A(r,k)}{(\log t)^{k+1}} dt $$

The coefficients $A(r,k)$ as well the details of the above formula are explained in the following link:

http://mac6.ma.psu.edu/primes/

The above study shows that till the point $1.7427435732 * 10^{35}$ the gap of 6 is most frequent but after this point the gap of 2 and 4 are most frequent and they have the same asymptotic density. Since we sum of the reciprocal of twin primes are convergent and they are asymptotically more denser than primes with any other gaps, we expect that the sum of the reciprocal of primes with a gap $d > 2$ to be convergent as well.

Sequences equidistributed modulo 1

2013-05-18T03:15:03Z

Let $\alpha$ be any positive irrational and $\beta$ be any positive real. We have the following results.

H. Weyl (1909): The fractional part of the sequence $\alpha n$ is equidistributed modulo 1.

I. Vinogradov (1935): The fractional part of the sequence $\alpha p_n$ is equidistributed modulo 1 where $p_n$ is the $n$-th prime.

E. Hlawka (1975): The fractional part of the sequence $\beta \gamma_n$ is equidistributed modulo 1 where $\gamma_n$ is the imaginary part of the $n$-th zero the Riemann zeta function.

The common thing in each of the above three celebrated results is that the sequences are of the form $as_n$ where $a$ is a positive real and $s_n$ has the property that the sequence

$$ \frac{s_1}{s_n}, \frac{s_2}{s_n}, \ldots , \frac{s_{n-1}}{s_n} $$

approaches equidistribution modulo 1 as $n \to \infty$.

Question: I would like a nontrivial counterexample of a positive real $a$ and a sequence $s_n$ such that the fractional part of the sequence $as_n$ is equidistributed modulo 1 but the sequence of the ratios $s_i/s_n$ do not approach equidistribution modulo 1 as $n \to \infty$.

Edit: I am explaining what I mean by nontrivial because the example given by Noam indicates that it is necessary to explain it explicitly. The examples of Weyl, Vinogradov and Hlawka are nontrivial because there is no assumption on the normality constant. If we take the constant to be normal, we indirectly already assume what we want to prove and so we can construct many artificial examples.

Answer by Nilotpal Sinha for Square and reversed integer

2013-05-07T10:28:43Z

Way back when I was in school, I had investigated number with these properties. These numbers are a special case of a more general property of numbers satisfying $R(mn) =R(m)R(n)$ where $R(n)$ means the digits reversal of $n$.

Examples

12*13 = 156, 21*31 = 651

101*102 = 10302, 101*103 = 20301

12012*11212 = 134678544, 21021*21211 = 445876431

11013*10212 = 112464756, 31011*21201 = 657464211

Infinitely many such numbers can be constructed but as Gerry Myerson said, the digits will be 0's, 1's, 2's or 3's.

A more challenging problem would be to study numbers with the property that

$R(n_1 n_2 \ldots n_k) =R(n_1)R(n_2)\ldots R(n_k)$.

Some interesting question in this direction:

What is the largest $k$ for which we will find a solution?
What is the largest $k$ for which we will find a solution if all $n_i$'s are equal? i.e. what is the largest $k$ for which $R(n^k) =R(n)^k$ has a solution?

For some reason I stopped at $k=2$. May be someone might want to revisit the problem.

Mathematicians whose works were criticized by contemporaries but became widely accepted later

2013-02-12T07:25:53Z

Gauss famously discarded Abel's proof that an algebraic equation of degree five or more cannot have a general solution (Abel himself had rejected divergent series as the work of the devil). Cantor's theory of transfinite numbers was originally regarded as so counter-intuitive—even shocking—that it encountered resistance from mathematical contemporaries such as Leopold Kronecker and Henri Poincaré and later from Hermann Weyl and L. E. J. Brouwer, while Ludwig Wittgenstein raised philosophical objections. Ramanujan's work on divergent series was rejected by three leading English mathematicians of the time before he was discovered by Hardy.

The above stories have become mathematical folklore. I would like to know the examples of other mathematicians whose works were initially criticized or rejected by contemporaries but later became widely accepted famous. I am particularly interested in modern mathematicians or lesser known mathematicians of the classical era who stories may not be as popular as those of other mathematical giants.

Answer by Nilotpal Sinha for A divergent series related to the number of divisors of of p-1

2013-04-25T08:31:23Z

I did a heuristic analysis to study how this sum is growing. I calculated $d(p_n-1)$ and the sum and plotted the curve of the sum vs $n$. I obtained a very smooth curve which looked like a cumulative distribution curve. Next I did curve fitting to model this curve where the sum is the dependent variable and $n$ is the independent variable. The boundary condition I imposed was that the sum should be divergent. The following model was found to best fit the desired sum.

$$ \sum_{n=1}^{x}\frac{1}{d(p_n - 1)} \sim e^{a + b/x + c\ln x} $$

where $a,b$ and $c$ are suitable constants.

The true asymptotic formula could be different from the above but I believe this can give some hints in the direction of the true asymptotics.

Answer by Nilotpal Sinha for Integer dynamics hitting infinitely many primes

2013-04-26T05:21:47Z

In 1947, American William . H. Mills proved that there is a real number $A$, greater than 1 but not an integer, such that integer part of

$$ A^{3^n} $$

is prime for all $n =1, 2, 3, \ldots$. The numnber $A$ is known as the Mill's constant. Its value is unknown, but if the Riemann hypothesis is true then,

$$ A \approx 1.3063778838630806904686144926... $$

[1] Mills, W. H. (1947) A prime representing function. Bull. Amer. Math. Soc., 53: 604: MR 8, 567.

[2] [Mill's constant]1

Multivariate functions whose value is independent of the order of the arguments

2013-03-13T05:07:02Z

Let $r_1, r_2, \ldots, r_k$ be positive integers with or without repetition such that $1\le r_i \le n$ for $i = 1, 2, \ldots, k$. Let $f$ be a continuous multivariate function with the property that the value of $f(r_1, r_2, \ldots, r_k)$ is independent of the order the arguments $r_1, r_2, \ldots, r_k$. The trivial examples are:

$f(r_1, r_2, \ldots, r_k) = g(r_1) + g(r_2) + \ldots + g(r_k)$,

$f(r_1, r_2, \ldots, r_k) = g(r_1)g( r_2) \ldots g(r_k)$ and

$f(r_1, r_2, \ldots, r_k) = c$

where $g$ is a continuous univariate function and $c$ is a constant. I have two questions

Questions:

Are there other non trivial examples of such functions?
Are there infinitely many such non trivial functions $f$?

Answer by Nilotpal Sinha for How can an integer be factorized as n*m so that n^m has the highest value.

2013-02-27T05:14:07Z

If $n$ is divisible by three than the required split is $(3,n/3)$. If $n$ is not divisible by 3 then the required split is $(p,n/p)$ where $p$ is the smallest prime divisor of $n$.

Probability that randomly chosen integers from a restricted set of natural numbers are coprime

2013-01-30T06:55:03Z

We know that the probability $P(k)$ of $k$ randomly chosen integers $(k \ge 2)$ from the set of natural number are coprime is

$$ P(k) = \frac{1}{\zeta(k)}. $$

I am looking at a special case of this problem. Let $S_n$ be the set of all natural numbers which do not have a prime factor greater than $n$-th prime (i.e $S_n$ is the set of natural numbers that can be formed using only the first $n$ prime numbers). What is the probability $P(k, S_n)$ that $k$ randomly chosen integers $(k \ge 2)$ from the set $S_n$ are coprime?

I do not know the answer but I think it could be in a parametric form involving $n$ such that in the trivial case when $n\to \infty$, $P(k, S_{\infty}) = P(k) = 1/\zeta(k)$.

Edit: Explained the meaning of "all natural numbers."

Most inconsistent ranking

2013-02-19T09:46:46Z

A matrix of $k$ rows and $n$ columns is filled with the numbers $1,2,\ldots,k$ such that the following conditions are satisfied:

Every column contain all the numbers form 1 to $k$ without repetition.
The variance of the elements of each row is calculated. The matrix is filled in such a way that the total sum $S_{kn}$ of the variance of each row is maximized.
$k < n < \infty$.

Questions:

What is the representation of the maximum value of $S_{kn}$ in a closed form in terms of $k$ and $n$? If exact representaion is not possible, can we have the upper and lower bound.
Is there an algorithm to fill the matrix such that $S_{kn}$ is maximized?
If repetition is allowed, what would be the answers for the above two questions.

Motivation: I am doing a worst case scenario analysis of the theoretically most inconsistent ranking where I need the solution of the following problem. In my problem, I have a ranking system which gave me the rank matrix. In the best case when the ranking system is completely consistent, the variance of each row will be zero and hence the total sum will be zero. In my case I have a finite total sum of variance say $S$ and I want to compare it against the worst or the maximum possible total sum $S_{kn}$ in order to quantify how consistent the rank matrix is.

Mathematical techniques to reduce the amount of storage memory

2013-02-05T17:09:40Z

Apologies for the length question. Those acquainted with the analytics industry will know that the next big thing in the information technology world will be the Big Data revolution where huge volumes of data will be processed. Big Data revolution will imply huge requirement of storage space/memory hence it is critical to store data as efficiently as possible. We want to store data in the smartest way that requires the least amount of storage. The following question is on the application of mathematical ideas to reduce the amount of storage memory required to store information of a particular kind.

Problem: This is a consumer information storage problem. There are $k$ customers and $n$ books. Each customer can buy one or more books (without repetition). We want to identify the book purchased by a given customer using the minimum number of memory required for storage. We do not seek to improve the time complexity, we only seek to reduce storage space required.

Method 1 - Traditional approach: This is the common and default approach. Give each book a $d$ digit code and create a field in the data base where the code of the books purchased by the customer in entered, separated by commas. So in the worst case when a customer has brought all $n$ books, we will need $D = nd+d-1$ characters to store this information (including $d-1$ commas) about the customer.

Method 2. Using prime numbers: Assign a unique prime number to each of the books, 2 denotes the first book, 3 the second, $\ldots$ and $p_n$ denotes the $n$-th. Every customer $C_i$ assigned a number $N_i$ which is equal to the product of the prime numbers corresponding to books purchased. By factoring $N_i$, unique factorization theorem ensures that we can identify the exactly the books purchased by the customer. More over two customers will have a common book if and only if they have a common factor. The greater the number of common factors, the greater is the similarity between the customers. This this method in principle carries more business information. Let us do the heuristics for the number of characters needed. In the worst case when the customer $C_i$ has purchased all the $n$ books, $$ N_i = p_1 p_2 \ldots p_n < p_n^n. $$ (This inequality can be strengthened using the estimates of the Chevyshev function of the first kind but right now, that is not the objective as I only want to demonstrate the underlying idea). Hence the number of characters required is $$ D_i = \log_{10} N_i < n\log_{10}p_n < nd + d - 1 $$

for $p_n < 10^d$ which is a safe assumption since most of the UPC code numbers given to products sold in the market have multiple digits. Hence with a small number of books say $<100$ we expect method 2 to use lesser number of character (hence memory) to store the same information. For example if I have 10 books them to store the information about a customer who purchased all the 10 books, method one with a 2 digit code for each book will require 21 characters where as method 2 will require only 10 characters. Unfortunately when there are a large number of books, the products of primes grows very fast and we may end up requiring more memory space than in method 1. Hence this method is not scale-able.

Method 3: We can do better than method 2. Let the $k$-th book be given the number $2^{k}$. Every time a customer buys a unique book, we add its corresponding number. Since a number can be decomposed as the sum of non-repeating powers of two in only one way, we can identify the exact books purchased by the customer $C_i$ by decomposing his/her total sum $S_i$. For example if the sum of the book numbers for a customer is 154, since $154=2^7 + 2^4 + 2^3 + 2^1$, we know that the customer has purchased the first, third, fourth and the seventh book. So instead of storing the code of these four books or the product of three primes, I can just store the three digit number 154 which will give me the same information. In the worst case when the customer has purchased all the n books,

$$ S_i = 2^1 + 2^2 +\ldots + 2^n = 2^{n+1} - 2 $$

$$ D_i < \log_{10}(2^{n+1}-2) < n\log_{10}2 + 1 $$

Thus with method 3, we can store the same information using the least number of characters thus far. Also two customers will have a book in common if and only if the decomposition of their sum contains an identical power of 2.

Questions: Is there are better method of identifying the books uniquely using less than $n\log_{10}2$ characters? I think that if there is indeed a better method, it could possible be using some of the property of integers.

Answer by Nilotpal Sinha for Mathematicians whose works were criticized by contemporaries but became widely accepted later

2013-02-12T09:10:50Z

Not in pure mathematics but in applied mathematics we have the case Ludwig Boltzman, the Austrian physicist (also the founder of the Austrian Mathematical Society) whose greatest achievement was in the development of statistical mechanics. He spent a life time trying to defend the now famous equation

$$ S = k\log W $$

Boltzmann's mentor and colleague Josef Loschmidt criticized Boltzmann's demonstration of entropy increase on the grounds that dynamical laws are reversible. If all the particles could be turned around exactly (or if time could be reversed), Boltzmann's work indicated the entropy should decrease, violating the second law. Eventually he committed suicide out of depression.

Today the above equation is one of the most important and fundamental equations of science.

On a sum involving prime numbers

2012-06-19T05:20:54Z

I find myself needing the asymtotics of the following summation for my work. Let $a$ be a positive real number and $p_n$ be the $n$-th prime.

$$ \sum_{k=1}^{n} [k^a - (k-1)^a]p_k $$

At $a=1$, this becomes the sum of the first $n$ primes and the asymptotics of this is well known. Moreover it is easy to prove that

$$ \sum_{k=1}^{n} [k^a - (k-1)^a]p_k = n^a p_n - \int_{2}^{p_n} \pi(x)^a dx. $$

I want an asymptotic expansion in terms of either $n$ or $p_n$ or a combination of both and get rid of the integration. (Don't ask me how many terms of the asymptotic expansion you want, do your best.)

Answer by Nilotpal Sinha for Sum involving binomial coefficients

2013-02-04T11:41:22Z

The closed form of the integral is

$$ \int_{-1}^{0}\frac{(1+x)^k - 1}{x} = \frac{s(k+1,2)}{k!} $$

where $s(k+1,2)$ denote the Stirling number of the first kind.

Answer by Nilotpal Sinha for Not especially famous, long-open problems which anyone can understand

2013-02-03T09:58:43Z

Waring's problem inequality

One of the oldest (Since 1770) and famous open problem in number theory is Waring's problem. It has been conjectured that if

$$ Frac\bigg[\bigg(\frac{3}{2}\bigg)^n\bigg] \le 1 - \bigg[\bigg(\frac{3}{4}\bigg)^n\bigg]. $$

(where $Frac$ denotes the fractional part) true then, the general solution of Waring's problem is

$$ g(n) = 2^n + Int\bigg[\bigg(\frac{3}{2}\bigg)^n\bigg] - 2. $$

Why do primes dislike dividing the sum of all the preceding primes?

2013-02-01T12:29:33Z

I was investigating primes with the property that the sum of the first $n$ primes is divisible by $p_n$. It turns out that these primes are extremely extremely rare. For primes less than $10^9$, I have found that there are only five primes with this property:

$$ p_1 = 2 $$

$$ p_3 = 5 $$

$$ p_{20} = 71 $$

$$ p_{31464} = 369,119 $$

$$ p_{22096548} = 415,074,643 $$

This raises the curious and equivalent questions:

Q1. Are there infinitely many primes which divide the sum of all the preceding primes?

Q2. Even if we assume that there are infinitely many such primes, why are they so rare? In other words, why do primes dislike dividing the sum of all the preceding primes? Is there any heuristic argument to show that such primes will indeed be extremely rare?

Answer by Nilotpal Sinha for Upper bounds for the sum of primes <= n

2013-01-30T11:12:53Z

The following paper gives the asymptotic expansion of the sum of the first $n$ prime numbers. Hence for sufficiently large $n$, the first few positive and negative terms of the asymptotic expansion will give best upper and lower bound on the sum of primes.

http://arxiv.org/pdf/1011.1667.pdf

$$ \sum_{r \le n}p_r = \frac{n^2}{2}\Bigg[\ln n + \ln\ln n - \frac{3}{2} + \frac{\ln\ln n}{\ln n} - \frac{3}{\ln n}- \frac{\ln^2 \ln n}{2\ln^2 n} $$

$$ + \frac{7 \ln \ln n}{2\ln^2 n} - \frac{27}{4\ln^2 n} + o\Bigg(\frac{1}{\ln^2 n}\Bigg) \Bigg]. $$

Asymptotics of the n-th prime using the gamma function

2012-06-01T11:47:53Z

In the paper http://rgmia.org/papers/v8n2/eepnt.pdf, the author proves that proves an explicit inequality on prime numbers using the gamma function and as a corollary, he showed that.

$$ p_n = n \frac{\Gamma'(n)}{\Gamma(n)} + o(n \ln n). $$

I obtained a stronger form of this result namely

$$ p_n = n \ln \frac{\Gamma'(n)}{\Gamma(n-1)} + O\Big(\frac{n\ln\ln n}{\ln n}\Big). $$

The gamma function seems to beautifully approximate $p_n$. To get the same error term using the regular Cipolla's asymptotic expansion of the $p_n$ we would need three terms.

Can someone explain why the gamma function approximated the n-th prime so nicely? Is this a coincidence or is there some underlying phenomenon governing this result that can shed some new light distribution of prime numbers.

A simple looking problem in partitions that became increasingly complex

2012-05-07T11:27:46Z

I began with problem which looked simple in the beginning but became increasingly complex as I dug deeper.

Main questions: Find the number of solutions $s(n)$ of the equation $$ n = \frac{k_1}{1} + \frac{k_2}{2} + \ldots + \frac{k_n}{n} $$ where $k_i \ge 0$ is a non-negative integer. This is my main questions. After tying different approaches, the one that I found most optimistic is as follows. But soon even this turned out to be devil (as we shall see why).

Let $l_n$ be the LCM of the first $n$ natural numbers We know that $\log l_n =\psi(n)$. Multiplying both sides by $l_n$ we obtain $$ n l_n = \frac{k_1 l_n}{1} + \frac{k_2 l_n}{2} + \ldots + \frac{k_n l_n}{n} $$

Each term on the RHS is a positive integer thus our question is equivalent to finding the number of partitions of $nl_n$ in which each part satisfy some criteria.

Criteria 1: How small can a part be? Assume that there is a solution with $k_n = 1$ then the smallest term in the above sum will be the $n$-th term which is $l_n / n$. Hence each term in our partition is $\ge l_n/n$.

Criteria 2: How many prime factors can each part contain? If my calculation is correct then for $n \ge 2, 2 \le r \le n$, the minimum number of prime factors that $l_n /r$ can contain is $\pi(n)-1$. With these two selection criterion we have:

$s(n) \le $ No. of partitions of $n l_n$ into at most $n$ parts such that each part is greater than $l_n / n$ and has at least $\pi(n) - 1$ different prime factors.

May be we can narrow down further by adding sharper selection criterions but I thought it was already complicated enough for the time being. The asymptotics of the number of partitions of $n$ into $k$ parts $p(n,k)$ is well known, but I have not found in literature any asymptotics for the number of partitions of $n$ into $k$ parts such that each part is at least $m$, let alone the case when each part has a certain minimum number of prime factors. I am looking for any suggestions, reference materials that would help in these intermediate questions that would ultimately help in answering the main question.

Answer by Nilotpal Sinha for Why Mertens could not prove the prime number theorem?

2012-05-04T06:27:57Z

Thanks to David very interesting remark and GH for the calculation of R(x). I like to present my reasoning why I thought on the Merten's theorem could imply PNT. I would have put it as a comment but due to the length, I am writing as a reply. In all likely hood, I must have committed an error somewhere hence I am reaching the ridiculous conclusion that Merten's theorem implies PNT. I would appreciate if someone tell me where I went wrong.

Let $s(n)$ be a strictly increasing sequence, $s(1) \ge 2$, such that $$ S_s(x) = \sum_{s(r) \le x}\frac{1}{s(r)} = \ln\ln x + C_s + R_s(x). $$ where $C_s$ is a constant that depends only on $s$ and $R_s(x) = o(1)$ is the error term which depends on $s$ and $x$. Let $A={[s(1)],[s(2)], [s(3)], \ldots }$ where [.] denotes the greatest integer function. Let $a(r) = 1$ if $r \in A$ and $a(r) = 0$ if $r \notin A$. Then $N_s(x) = \sum_{r \le x} a(r)$ is the number of elements of $A$ in the interval $(0,x)$. Using Abel's summation formula we obtain $$ \sum_{s(r) \le x}\frac{1}{s(r)} = \frac{N_s(x)}{x} + \int_{s(1)}^{x} \frac{N_s(t)}{t^2}dt = \ln\ln x + C_s + R_s(x) $$ Differentiating under the integral sign with respect to $x$, we obtain $$ \frac{N_s'(x)}{x} - \frac{N_s(x)}{x^2} + \frac{N_s(x)}{x^2} = \frac{1}{x\ln x} + R_s'(x) $$ Simplifying the above equation and integrating both sides, we obtain $$ N_s(x) = \int_{s(1)}^{x}\frac{dt}{\ln t} + \int_{s(1)}^{x} tR_s'(t)dt $$ Without loss of generality, we can define $s(1) =2$ as this will only effect the constant term $C_s$ and not $R_s(x)$. Hence we have $$ N_s(x) = \int_{2}^{x}\frac{dt}{\ln t} + E(x) = Li(x) + E(x). $$ Thus any sequence $s(n)$ satisfying $S_s(x) = \ln\ln x + C_s + R_s(x)$ must have a density function which is asymptotic to the logarithmic integral. We can verify this with the sequences $n\ln n, n\ln n + n\ln\ln n, nH_n$ etc; where $H_n$ is the harmonic number. Also we have found the explicit relationship between the error term in $S_s(x)$ and that of $N_s(x)$ which is $$ E(x) = \int_{2}^{x} tR_s'(t)dt $$

In case of prime numbers, $N_s(x) = \pi (x)$ and Merten's theorem shows that primes satisfy the condition on $S_s(x)$. This gives $$ \pi(x) = \int_{2}^{x}\frac{dt}{\ln t} + E(x). $$

This shows that the fact that $\pi(x) \sim Li(x)$ is a consequence of $S_p(x) \sim \ln\ln x$ and it is not a unique property of primes. In fact it is a common attribute of a general family of sequences which grow at the same asymptotic rate in their dominant term. Primes happen to be just one sequence in this family. What differentiates primes and other members of this family of sequences form each other is difference in their respective error terms $E(x)$ or in terms of $S_s(x)$ the difference in their respective constant terms $C_s$ and their error terms $R_s(x)$. Therefore the correct way to interpret Merten's theorem is that for prime numbers, the constant term in $S_p(x)$ is $$ C_p = M = \gamma + \sum_{k=2}^{\infty} \frac{\mu (k)\ln \zeta(k)}{k} \simeq 0.2614972128 $$ because if it were any other constant, then we know that we are dealing with some other sequence and not the sequence of primes even though the dominant term in asymptotic expansion of the sum of the reciprocal of the sequence would still be $\ln\ln x$ and its density would still be asymptotic to $Li(x)$.

Why Mertens could not prove the prime number theorem?

2012-05-02T10:03:24Z

We know that $$ \sum_{n \le x}\frac{1}{n\ln n} = \ln\ln x + c_1 + O(1/x) $$

where $c_1$ is a constant. Again Mertens' theorem says that the primes $p$ satisfy $$ \sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/\ln x). $$

Thus both these divergent series grow at the same rate. Mertens' theorem was proved without using the prime number theorem, some 25 years before PNT was proved. However from these two examples, we cannot conclude that

$$ \lim_{n \to \infty} \frac{p_n}{n\ln n} = 1 $$ otherwise Mertens' would have been the first to prove PNT. My question is - based on the above two series, what are the technical difficulties that prevent us from reaching the conclusion that $p_n/n\ln n = 1$. There may be counter examples with other series, so such conclusions may not be true in general. However I am not interested in the general case. Instead I am asking only in case of the sequence $1/n\ln n$ and $1/p_n$.

Comment by Nilotpal Sinha

2013-05-21T05:30:38Z

If I understand your question correctly, you are asking the following. The sum of the reciprocal of primes is divergent but the sum of the reciprocal of twin primes is convergent (as shown by Brun). Your question is there any gap $d$ greater than 2 (2 is for twin primes) between primes for which the sum of the reciprocal primes differing by $d$ is convergent. Is this understanding correct?

Comment by Nilotpal Sinha

2013-05-20T07:35:03Z

@Gerhard Paseman: I would rather NOT ask you about system design but I would definitely ask this question in this forum because if it is more than perfectly fitted here :)

Comment by Nilotpal Sinha

2013-05-18T11:17:15Z

But this is nothing but the extension of the Equidistributuion Theorem (first result) to polynomials, which was done my Weyl himself. So not just $n^2$, but any polynomial $s_n=f(n)$ is trivial.

Comment by Nilotpal Sinha

2013-05-18T08:31:38Z

@Noam: While your example is correct, it is highly trivial because if we take $a$ to be normal in base $b$, then by the very definition of normality, we can construct similar example in that base $b$. Please note that in my examples of Weyl, Vonogradov had Hlawka, there is no such assumption on the normality constant which makes them non trivial. My question is can we have such a non trivial example such that $s_i/s_n$ does approach equidistribution modulo 1.

Comment by Nilotpal Sinha

2013-05-03T05:25:52Z

"There are a lot more connections known between π and e and other numbers than between γ and other numbers." This may not entirely true. The numbers $e^{\gamma}$ pops up every now and then in the theory of primes. For example Merten's Theorems, Cramer-Granville's conjecture etc to name a few. But I do agree that the connection between $\gamma$ and other numbers that has nothing to do with primes directly or indirectly is far less common.

Comment by Nilotpal Sinha

2013-04-23T09:44:08Z

Any inputs/comments?

Comment by Nilotpal Sinha

2013-03-13T06:18:19Z

@Survit: That practical application for which I need such a function has discrete arguments (positive integers) but we want to keep it flexible enough to accommodate all positive real values if required in future.

Comment by Nilotpal Sinha

2013-03-13T06:17:05Z

@Qiaochu: That is because the current application for which I need such a function has discrete arguments (positive integers) but we want to keep it flexible enough to accommodate all positive real values if required in future.

Comment by Nilotpal Sinha

2013-02-19T18:14:24Z

@Kevin: Thanks for the insight and the lower bound. In my problem, I have a ranking system which gave me the rank matrix. In the best case when the ranking system is completely consistent, the variance of each row will be zero and hence the total sum will be zero. In my case I have a finite total sum of variance say $S$ and I want to compare it against the worst or the maximum possible total sum $S_{kn}$ in order to quantify how consistent the rank matrix is. I am updating this comment in the question.

Comment by Nilotpal Sinha

2013-02-19T17:36:07Z

@Kevin: Yes this is the right understanding

Comment by Nilotpal Sinha

2013-02-19T11:42:49Z

@ Gerry, I have now defined the sum of variance of the rows in condition number 2.

Comment by Nilotpal Sinha

2013-02-13T14:16:02Z

No conditions are known yet.

Comment by Nilotpal Sinha

2013-02-12T07:45:54Z

@Yemon. No I had no such intention to distinguish between the two. All as wanted to know the lesser known mathematicians or stories that need to be heard. How do I make it community Wiki?

Comment by Nilotpal Sinha

2013-02-04T16:10:32Z

@Pietro Yes there is such an integral which is in fact a contour integral. $$ s(n,r) = \frac{n!}{2\pi r!}\int_{|z|=1} z^{-n-1} \log^r (z+1)dz $$ @Danne, Yes I am sure.

Comment by Nilotpal Sinha

2013-01-31T09:19:40Z

@Quid: Yes, I assumed that anyone attempting the solution already knows what you have mentioned. More specifically, the probability $P(k)$ of $k$ randomly chosen integers from the set $[1,N]$ being coprime is $$ P(k)=\frac{1}{\zeta(k)} + O(\ln N/N). $$ Making $N \to \infty$, the error term vanishes and we say in a simplified manner that the probability of two randomly chosen integers being coprime is $1/\zeta(k)$. However since this does not seem trivial and apparent to many readers, I am mentioning it explicitly in the question so that it is easier to understand.