There are at least “a” primes within the range of n!(a) and n!(a+1), (n,a) being any positive integer greater than 2.

I have to introduce the concept of “degree of Factorial” as I have not been successful in finding any notation for representing a factorial of a factorial, ad. inf.

I define n!(a) as n! of degree a>0.

n!(1) indicates the factorial of degree 1 which is the conventional factorial.

Ex. 5!(1) indicates simply 5! i.e 5*4*3*2*1, i.e 120.

Ex. 5!(2) indicates factorial of 5! i.e (5*4*3*2*1)! or (5!)! or 120!

However, I have not extended this for non positive integer arguments. I welcome our readership to help me extend this via the Gamma function. As always, I invite readers to prove or disprove this conjecture. Thank you for reading this blog.

Nunghead

There exist at least (a-1) primes in the interval n^a to (n+1)^a, n and a being integers greater than 1.

If this sounds familiar, yes, it was inspired by (and is an extension of) Legendre’s conjecture. which states;

There exists at least 1 prime in the interval n^2 to ( n+1)^2, n being an integer.

It was not possible to verify this using Excel, as I did with the earlier conjecture. Excel does not have a “isPrime” function and the workarounds are too complex. However, I have done the verification for quite a number of cases using TI Nspire CX.

Once again, I invite the readers to prove or disprove this conjecture.

Thank you for taking the time to the read this blog and the support you have given.

Nunghead

A few months ago, I came up with this conjecture that I am going to post today. This conjecture may seem elementary but I assure you it is a very difficult question to resolve. It goes like this.

x^a + x^b + x^c is divisible by 12, for all positive even integer values of (x, a, b, c) > 0.

I have not found a way to prove it formally. But I have tested it for various combinations of values for x, a, b and c. Here is an interactive excel sheet where you can test it out yourself.

https://numeracy.car.blog/wp-content/uploads/2019/07/akash-conjecture-01a-1.xlsx

The worksheet has a column and a row of random even integers. The random even integer greater than zero is generated with the formula =INT(RAND()$B$1)2+2

Cell B1 is the multiplier for the column of random numbers and Cell B2 is the multiplier for the row.

You can plug different values for A (cell B1) and B (Cell B2).

Each cell in the evaluated area calculates B$3^$A4+B$3^$A5+B$3^$A6)/12

In order to check if the division results in an integer, this formula is further enhanced to =IF((B$3^$A4+B$3^$A5+B$3^$A6)/12=INT((B$3^$A4+B$3^$A5+B$3^$A6)/12),1,0), so that the final result in the cell is a 1 or 0. A ‘1’ indicates that the conjecture is true for those set of values. Unfortunately, Excel’s capabilities to handle large numbers are limited. These are the times when you would find #NUM!, instead of ‘1’ (0r ‘0’; although I have never come across a ‘0’ so far). In such cases, I have verified it with a TI Nspire CX. You may want to do the same.

Instead of this elementary ad-hoc method of verification, I welcome the readers to provide a more generic proof. Although it sounds trivial, it does not appear so.

My sincere thanks to my grandfather, for helping me out with the Excel sheet.

Welcome to my blog! This blog of mine is primarily meant to present to the world my conjectures in mathematics. However,you can also find me occasionally blogging on other subjects as well. As they say, “Only time will tell”.