Mersenne numbers that are prime written as the sum of consecutive integers starting with 1 while skipping exactly one number.

[Martin Musatov]

The Mersenne numbers that are prime can be elegantly written as a sum of consecutive integers starting from 1 while skipping exactly one number.

I wanted to know if anyone sees anything that could make for an interesting series for OEIS.

Thanks,

Martin Musatov

### 1. M2 = 3 (Exponent p = 2)

* Consecutive sequence: 1 to 3

* Skipped number: 3

* Sum representation: 1 + 2 = 3

### 2. M3 = 7 (Exponent p = 3)

* Consecutive sequence: 1 to 4

* Skipped number: 3

* Sum representation: 1 + 2 + 4 = 7

### 3. M5 = 31 (Exponent p = 5)

* Consecutive sequence: 1 to 8

* Skipped number: 5

* Sum representation: 1 + 2 + 3 + 4 + 6 + 7 + 8 = 31

### 4. M7 = 127 (Exponent p = 7)

* Consecutive sequence: 1 to 16

* Skipped number: 9

* Sum representation: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 127

### 5. M13 = 8,191 (Exponent p = 13)

* Consecutive sequence: 1 to 128

* Skipped number: 65

* Sum representation: 1 + 2 + ... + 64 + 66 + ... + 128 = 8,191

### 6. M17 = 131,071 (Exponent p = 17)

* Consecutive sequence: 1 to 512

* Skipped number: 257

* Sum representation: 1 + 2 + ... + 256 + 258 + ... + 512 = 131,071

### 7. M19 = 524,287 (Exponent p = 19)

* Consecutive sequence: 1 to 1,024

* Skipped number: 513

* Sum representation: 1 + 2 + ... + 512 + 514 + ... + 1,024 = 524,287

### 8. M31 = 2,147,483,647 (Exponent p = 31)

* Consecutive sequence: 1 to 65,536

* Skipped number: 32,769

* Sum representation: 1 + 2 + ... + 32,768 + 32,770 + ... + 65,536 = 2,147,483,647

### 9. M61 = 2,305,843,009,213,693,951 (Exponent p = 61)

* Consecutive sequence: 1 to 2,147,483,648

* Skipped number: 1,073,741,825

* Sum representation: 1 + 2 + ... + 2^30 + (2^30 + 2) + ... + 2^31 = 2,305,843,009,213,693,951

### 10. M89 = 618,970,019,642,690,137,449,562,111 (Exponent p = 89)

* Consecutive sequence: 1 to 35,184,372,088,832

* Skipped number: 17,592,186,044,417

* Sum representation: 1 + 2 + ... + 2^44 + (2^44 + 2) + ... + 2^45 = 618,970,019,642,690,137,449,562,111

### 11. M107 = 1,622,592,768,292,133,633,915,780,102,881,27 (Exponent p = 107)

* Consecutive sequence: 1 to 18,014,398,059,481,984

* Skipped number: 9,007,199,254,740,993

* Sum representation: 1 + 2 + ... + 2^53 + (2^53 + 2) + ... + 2^54 = 1,622,592,768,292,133,633,915,780,102,881,27

### 12. M127 = 170,141,183,460,469,231,731,687,303,715,884,105,727 (Exponent p = 127)

* Consecutive sequence: 1 to 18,446,744,073,709,551,616

* Skipped number: 9,223,372,036,854,775,809

* Sum representation: 1 + 2 + ... + 2^63 + (2^63 + 2) + ... + 2^64 = 170,141,183,460,469,231,731,687,303,715,884,105,727

[Allan Wechsler]

I fear that all numbers can be written in this form. In my examples I will mark the addend I am omitting by multiplying it by 0.

10 = 1 + 2 + 3 + 4 + 5*0.

11 = 1 + 2 + 3 + 4*0 + 5

12 = 1 + 2 + 3*0 + 4 + 5

13 = 1 + 2*0 + 3 + 4 + 5

14 = 1*0 + 2 + 3 + 4 + 5

15 = 1 + 2 + 3 + 4 + 5 + 6*0

I trust it is obvious how to continue this pattern.

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[Geoffrey Caveney]

Any odd prime p can be expressed as p = 2n - 1.

So the Mersenne prime 2^p - 1 can be expressed as 2^(2n-1) - 1.

Then the following basic algebraic transformations can be applied:

2^(2n-1) - 1  =  2^2n / 2^1 - 1

2^2n / 2^1 - 1  =  (2^n)^2 / 2 - 1

(2^n)^2 / 2 - 1  =  (2^n)^2 / 2 + 2^n / 2 - 2^n / 2 - 1

(2^n)^2 / 2 + 2^n / 2 - 2^n / 2 - 1  =  ((2^n)^2 + 2^n) / 2  -  (2^n / 2^1 + 1)

((2^n)^2 + 2^n) / 2  -  (2^n / 2^1 + 1)  =  ((2^n)^2 + 2^n) / 2  -  (2^(n-1) + 1)

In this last expression, the first term ((2^n)^2 + 2^n) / 2 is the formula for the (2^n)-th triangular number, which is the sum of the positive integers from 1 to 2^n.

The subtracted second term is 2^(n-1) + 1.

This is precisely the pattern seen in your examples of Mersenne primes with odd prime exponents.

On Sat, Sep 26, 2026 at 4:36 PM Martin Musatov <martinm...@gmail.com> wrote:

[L. Edson Jeffery]

Or, for the sequence 

a = {3, 3, 5, 9, 65, ...}. 

of missing numbers, defined in terms of A-numbers, it appears that

a(n) = A000217(floor((A000043(n) + 1)/2)) - A000668(n).

I think it is interesting enough to at least propose to OEIS for review and see what the editors say.

Ed Jeffery 

[Tomasz Ordowski]

Hi Martin, 

I once did something similar from scratch, as far as I remember. 

Let us first examine all Mersenne numbers of the form 2^n - 1 for n > 0. 

 Let a(n) be the smallest k > 0 such that b(n) = k(k+1)/2 - (2^n - 1) > 0.

Question: For which n does b(n) < a(n) hold in this greedy sequence?

Maybe for almost all n. 

 

Unfortunately, I don't have time to check this, and if I misunderstood or missed other points raised in the discussion, I apologize.

Best, 

Thomas 

[M F Hasler]

On Monday, September 28, 2026 at 6:08:11 AM UTC-4 tomaszo...@gmail.com wrote:

 Let a(n) be the smallest k > 0 such that b(n) = k(k+1)/2 - (2^n - 1) > 0.

Question: For which n does b(n) < a(n) hold in this greedy sequence? 

Maybe for almost all n.

Yes, more precisely, for all n except the 5 terms of

A215795

Numbers n such that 2^n-1 is a triangular number (A000217).

+20
5

0, 1, 2, 4, 12 

[Daniel Mondot]

A006516 contains a link to "descartes numbers" which is broken.

A222263 and A174292 both contain a different link to an archived paper, which might be the same "descartes numbers" paper.

Could someone verify that it is indeed the same paper, and fix the first link?

Daniel.

[jpallouche.math]

Hi

You can find the paper cited in 1006516 at
https://web.archive.org/web/20210506121806/http://faculty.missouri.edu/~bankswd/papers/2008_Descartes_Final.pdf
best
jpa

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[jpallouche.math]

and yes this the same paper

jpa

Le 28/09/2026 à 17:42, Daniel Mondot a écrit :

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[Tomasz Ordowski]

Maximilian, thanks!

No comment.

Best,

Thomas

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[Sean A. Irvine]

Link updated. Note it is essentially going to the Internet Archive to get it.

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