Valid or not valid: that is the question

3 + 4 = 7 11 + 12 = 23 \large {\begin{aligned} 3+4&=&7 \\ 11+12&=&23 \end{aligned} }

A certain sequence of consecutive numbers, with first term as a prime number, can have its sum equal to a prime number. The examples above demonstrate this property with two different sequences, each with two consecutive numbers, and the first ones are prime numbers. Is this property valid for sequences like these with (strictly) more than two terms?

Yes There is insufficient Information No

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7 solutions

Let n be a prime number. The sum of consecutive numbers, being n the first one, can be written as

n + ( n + 1 ) + ( n + 2 ) + ( n + 3 ) + . . . + ( n + y ) {n + (n+1) + (n+2) + (n+3) +...+ (n+y)}

If the sum has x terms, the value of y is (x-1), and n will be added to itself x times. Hence, the equation can be simplified to

n ( y + 1 ) + k = 1 y a k {n(y+1) + \sum_{k=1}^y a_k}

Which is

n x + k = 1 x 1 a k \LARGE{nx + \sum_{k=1}^{x-1} a_k}

a k a_k is the term that represent the sequence of the values of y: a 1 a_1 , a 2 a_2 , a 3 a_3 ,..., a y a_y , which is 1 + 2 + 3 + . . . + y {1 + 2 + 3 +...+ y} . For any y > 1 y > 1 , the equation above will be divisible by a certain positive integer, or more than one, which makes the result of it not a prime number. For y = 1 y = 1 , the result is

n + ( n + 1 ) = x n + k = 1 x 1 a 1 = 2 n + 1 {n + (n+1) = xn + \sum_{k=1}^{x-1} a_1 = 2n + 1}

For some values of n to ( 2 n + 1 ) (2n + 1) , the result is a prime. For instance, for n = 3, the result is 7, which is a prime number. So, to y > 1 y > 1 , the result can't be a prime number.

Note : I tried to simplify the main equation that I used to calculate the results of the sums, but I could just get n x + k = 1 x 1 a k \LARGE{nx + \sum_{k=1}^{x-1} a_k} because it matters if x is an odd or even number, what makes the calculations quite confusing and complex.

If you are able to simplify it even more, I would be glad if you could share your thoughts here. Thank you.

EDIT : I used Gauss' Sum Equation to expand my sigma notation and I got an interesting result. Here are the calculations:

z = n x + k = 1 x 1 a k z = n x + ( x 1 ) [ ( x 1 ) + 1 ] 2 z = n x + x ( x 1 ) 2 {z = nx + \sum_{k=1}^{x-1}a_k \Rightarrow z = nx + \frac{(x-1)[(x-1)+1]}{2} \Rightarrow z = nx + \frac{x(x-1)}{2}}

A s ( x 1 ) = y k = 1 x 1 a k = y ( y + 1 ) 2 z = n ( y + 1 ) + y ( y + 1 ) 2 {As (x-1) = y \therefore \sum_{k=1}^{x-1}a_k = \frac{y(y+1)}{2} \Rightarrow z = n(y+1) + \frac{y(y+1)}{2} \Rightarrow}

z = 2 n ( y + 1 ) + y ( y + 1 ) 2 z = ( y + 1 ) ( 2 n + y ) 2 {\Rightarrow z = \frac{2n(y+1) + y(y+1)}{2} \Rightarrow \LARGE{z = \frac{(y+1)(2n + y)}{2}}}

Cool Fact : for y > 1 y >1 , z z is divisible by, for instance, ( y + 1 ) (y+1) , so z z will not a prime number. For y = 1 y =1 (which means that it is a sum of n plus (n+1) ), z z will be represented as 2 n + 1 {2n + 1} , which can result on a prime number (as I had already stated before). Considering y > 1 y >1 , the way the equation that expresses z z works does not depend anymore on the fact that x x (or ( y + 1 ) (y + 1) ) is odd or even.

Now that it's proven that, for y > 1 y >1 and n as a prime, z z won't be a prime, we can say, for real, that the answer to the problem is no .

Given n n sequenced integers numbers, n 3 n \ge 3 , being the first one a prime number, p p , the sum will be:

p + ( p + 1 ) + ( p + 2 ) + + ( p + ( n 1 ) ) = n p + ( n 1 ) ( 1 + ( n 1 ) ) 2 \small p + ( p+1) + (p+2) + \dots + (p + (n-1)) = np + \frac{(n-1)( 1 + (n-1))}{2}

n ( 2 p + ( n 1 ) ) 2 \Rightarrow \frac{n(2p+(n-1))}{2}

n n or ( n 1 ) (n-1) it´s an even number. 2 p 2p it´s clearly an even number.

We always can assume n = 2 k n = 2k or ( 2 p + ( n 1 ) ) = 2 k (2p+(n-1)) = 2k , one of them an even number.

Then the sum represented by the fraction it will be, k . ( 2 p + ( n 1 ) ) k.(2p+(n-1)) or n . k n.k .

Therefore, the sum it´s a product of two numbers and so not a prime number.

Cleres Cupertino - 5 years, 10 months ago

If you sum a prime number n, which is odd, to the second term of the sucesion (n+1), wich is even, to the third term (n+2, therefore odd), you get an even number, that cannot be prime. More simply, n+(n+1)+(n+2) = 3n+3= even number.

Lupita Vásquez Favela - 5 years, 10 months ago

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Correct. That's a simple way to prove that the property is not valid for y = 2 (a sequence of three consecutive numbers, with first term as a prime). But the point is: how would you prove that, for a sequence of 50 consecutive numbers, with the first one as a prime, the result is not a prime? Trying it manually would take a while :/

Bruno Oggioni Moura - 5 years, 10 months ago

Let us assume that n n is a prime number and there k k consecutive numbers in the sequence.

Then the sum would be i = 1 n + k 1 i i = 1 n 1 i \sum_{i=1}^{n+k-1} i - \sum_{i=1}^{n-1} i

= ( n + k 1 ) ( n + k ) n ( n 1 ) 2 = k ( k + 2 n 1 ) 2 =\frac{(n+k-1)(n+k)-n(n-1)}{2} = \frac{k(k+2n-1)}{2}

There is a chance of the sum to be a prime number only if either k / 2 = 1 k/2=1 and ( k + 2 n 1 ) (k+2n-1) is prime OR ( k + 2 n 1 ) / 2 = 1 (k+2n-1)/2=1 and k k is prime.

k / 2 = 1 k/2=1 would mean k = 2 k=2 which is given in the question and which we are not interested in.

The minimum value of n n is n = 2 n=2 and we need k 3 k \ge 3 . Hence, k + 2 n 1 6 k+2n-1 \ge 6 . Therefore, the second case would not arise.

Hence, there are no sequences of length greater than two.

Brilliant! The sigma notations' expansion makes it easier to calculate the results for high values of k. Thank you!

Bruno Oggioni Moura - 5 years, 10 months ago

Let the first number be 'n'

The sum will be n + n + 1 + n + 2 = 3n + 3

3n +3 is always even, and aside, from 2, there are no even primes, therefore the answer is NO.

Moderator note:

You only considered the case of 3 terms. Why can't we have 4 or more terms?

Let n be the number of consecutive numbers in a summation, n>1:

If n is odd: Let k = n + 1 2 k = \frac{n+1} {2} so that x = a k x = a_{k} is the middle term in the summation. We can write the summation as:

. . . + ( x 2 ) + ( x 1 ) + x + ( x + 1 ) + ( x + 2 ) + . . . ...+ (x-2) + (x-1) + x + (x +1) + (x+2) +... =

= x + ( x + 1 ) + ( x 1 ) + ( x + 2 ) + ( x 2 ) + . . . x + (x+1) +(x-1) +(x+2)+(x-2) + ... =

= x n = a k n x*n = a_k*n

So, the summation is composite .

If n is even: Let k = n 2 k = \frac{n}{2} and x = a k + 1 2 x = a_{k}+\frac{1}{2} . So, we can write the summation as:

. . . + ( x 3 2 ) + ( x 1 2 ) + ( x + 1 2 ) + ( x + 3 2 ) ...+(x-\frac{3}{2}) +(x-\frac{1}{2})+(x+\frac{1}{2}) + (x+\frac{3}{2}) =

= ( x + 1 2 ) + ( x 1 2 ) + ( x + 3 2 ) + ( x 3 2 ) + . . . (x+\frac{1}{2})+(x-\frac{1}{2})+(x+\frac{3}{2})+(x-\frac{3}{2})+... =

= n x = 2 k x = 2 k ( a k + 1 2 ) = 2 k ( 2 a k + 1 2 ) = k ( 2 a k + 1 ) n*x = 2k *x = 2k*(a_k+\frac{1}{2} ) = 2k*(\frac{2a_k+1}{2} ) = k*(2a_k +1)

and, for k > 1 k>1 or, rather, n > 2 n>2 , it is composite.

Moderator note:

Good analysis. This shows us why it doesn't work for > 2 > 2 terms.

Do you know how these primes are related to each other?

Peter Sumner
Feb 20, 2017

For a sequence of length k > 2, starting from prime n, the sum will be: n + (n+1) + (n+2) + ... + (n+k-1) = kn + SUM(consecutive integers from 1 to k-1) = kn + k(k-1) / 2 = k(n + (k-1)/2)

Since this number is always divisible by integer k, it is never prime.

Edit: I feel like this is close, but it fails when k is even, as the term inside the brackets is then a fraction..?

Manuj Dagar
Aug 9, 2015

Its very simple 3terms- let the middle term be x Therefore : x-1+x+x+1=3x which means that it has 3 as it's factor . 4-terms-let the first term be x Therefore : x+x+1+x+2+x+3=4x+6=2(2x+3) and so on.. ..HOWZ IT ? PLS TELL..

Amjad Ali
Jul 30, 2015

There is only of the sequence to be resulted in odd number(Prime number), i.e., 2+3+4 but it is not a prime one. After that all the such sequences have two odd and one even number which always results in even number and it can never be a prime number.

Why can't it be three odd and two even, which gives us an odd number?

Calvin Lin Staff - 5 years, 10 months ago

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