# Happy mathematical new year: 2011 is the sum of 11 consecutive prime numbers

Posted by: Gary Ernest Davis on: January 1, 2011

xOn December 31, 2010 @mathematicsprof posted two interesting tweets:

@mathematicsprof FINALLY, a prime number year 2011 …. first one since 2003.
x

@mathematicsprof 2011 is also the sum of 11 CONSECUTIVE prime numbers: 2011=157+163+167+173+179+181+191+193+197+199+211

x

Two prime numbers are “consecutive” if they follow one upon the other, in the collection of prime numbers. So, for example, 3 and 5 are consecutive prime numbers, as are 7 and 11.

These tweets, re-tweeted, attracted a fair amount of attention, deservedly so, because the second observation is both funny and thought provoking.

_____________________________________________________________________________________________________

Many people said to me they wished they had the “2011=157+163+167+173+179+181+191+193+197+199+211” on a T-shirt, so …. by popular demand I am happy to say that you can now purchase just such a T-shirt:

_____________________________________________________________________________________________________

I asked the question “Which prime numbers are sums of consecutive prime numbers” and David Radcliffe (@daveinstpaul) tweeted:

Primes expressible as the sum of consecutive primes: http://bit.ly/dKuh9S (2011 is an example!)

The link leads us to an entry in the On-Line Encyclopedia of Integer Sequences™ in which there is some data on prime numbers that are sums of two or more consecutive prime numbers.

This On-Line Encyclopedia of Integer Sequences™ entry includes Mathematica code to find such prime numbers.

The problem of finding prime numbers that are sums of two or more non-consecutive primes is a non-trivial computational exercise for school students and, it seems to me, one that many of them would find fun.

Here’s a variant of that problem, stemming from @mathematicsprof’s observation:

2011 is the sum of a prime number (11) of consecutive prime numbers.

Which prime numbers are a sum of a prime number of consecutive prime numbers?

Enjoy, and Happy New Year!

#### Postscript

The next year after 2010 that is a prime number and a sum of consecutive prime numbers is 2027 = 29+31+37+41+43+47+53+59+61+67+71+73+

79+83+89+97+101+103+107+109+113+127+131+137+139

but that year is not a sum of a prime number of consecutive prime numbers.

However, 2027 does have the curious and interesting property that it is prime and the sum of its digits 2+0+2+7=11 is also a prime number. It is the first year after 2003 that has this property.

The next year that is a prime number and a sum of a prime number of consecutive prime numbers is 2081 =401+409+419+421+431.

The following modification of Vladimir Orlovsky’s Mathematica code produces prime numbers that are a sum of a prime number of consecutive primes:

p = {};

Do[a = Table[Prime[i], {i, n, 100}];

l = Length[a];

k = 1;

While[Prime[k] < l + 1, b = Plus @@@ Partition[a, Prime[k]];

k++;

p = Append[p, Select[b, PrimeQ[#] &]]], {n, 1, 99}];

S =Union[Flatten[p]]

The first prime number that is a sum of two or more consecutive primes but is not a sum of a prime number of consecutive primes is 17 = 2+3+5+7

@mathematicsprof further tweeted on January 1, 2011:

2011 is also the sum of THREE consecutive primes 2011 = 661+673+677 . Can you write 2011 as the sum of 5, 7, 9, 11, 13, … primes?

in how many ways can a prime number be written as a sum of consecutive primes?

(allowing, for convenience, a sum to have only 1 term, so that, for example 7 =7 can be written as sum of consecutive primes in only 1 way, whereas 5= 5 = 2+3 can be written as a sum of consecutive primes in 2 ways).

#### R packages that might be useful

People familiar with R might want to recode the above-mentioned Mathematica code in R.

Here are details of two packages that should prove useful in doing this:

Partitions: partitions

Primality checking: schoolmath

### 39 Responses to "Happy mathematical new year: 2011 is the sum of 11 consecutive prime numbers"

I don’t have Mathematica, and my R skills aren’t up to this, but I wonder something else ….

Are there an infinite number of prime numbers that are the sum of consecutive prime numbers?

Excellent question Peter. It looks like it from the growth of the primes that are sums of consecutive prime numbers, but right now I do not know a proof.

My R skills are definitely not (yet) up to writing code to do this in R, but it should not be too difficult, following the idea of the Mathematica code. I put summaries of two R packages to find partitions of integers, and to check if an integer is prime, at the bottom of the blog post.

of course there are; a variant of Euclid’s proof of the infinity of primes will work, unless you have a prime number that doesn’t begin a chain of sequential primes that eventually sum to a prime. What’s the standard of proof?

Hi David – I don’t see how you would apply Euclid’s proof here.

Let’s call all the primes P and all the primes that are sums of consecutive primes S

1. Assume there is a largest S. Let’s call it X
2. Write down …. what did you have in mind here? All P smaller than X, or all S smaller than X?
3. Multiply them and add 1
4. Either the new number is a member of S or there is a member of S that is larger than X and smaller than the result in 3.

The first few number in S are 5, 17, 23. Let’s say we think 23 is the largest one. So, 5*17*23 + 1 = 1956. Not prime, obviously, but = 2*2*3*489, and 489 is not a member of S.

or, perhaps all P less than X?

2*3*5*7*11*13*17*19 + 1 = 9699691, which isn’t prime, but I don’t have a program that factors things quickly … perhaps someone else does.

In any case, I don’t understand how this proof would work

Peter

the sense of that proof, that there are always more. You can start with any prime number and add consecutive primes to it until you get a prime. There are an infinity of primes, so an infinity of starting points, all resulting in a prime that is a sum of consecutive primes.

9699691 is divisible by 347.
David has failed to prove his conjecture that “all prime numbers begin series of consecutive prime numbers which will sum to a prime” without presuming “there is no prime number that does not begin such a series” which is a restatement, therefore circular. How would one search for such a number? It’s nonexistence seems obvious, but obvious is often meaningless in math theory.

Hi.

Euclid’s proof does not state a way of finding new primes. It just just produces a number that’s either a prime, or not.
If it is a prime, it adds to the “complete” list, and proves that there is no complete list of primes.
If it is not a prime, that means that there is missing a prime number amongst the list of the product: who can factor the new number/product….

That’s amazing! Not just prime, but the sum of so many consecutive primes. 2011, 157, and 11 are my new fav. primes. Actually commenting just to get follow-up comments, this has made my year! (In number only of course.)

On average, roughly 70% of the natural numbers can be represented as a sum of consecutive primes in at least one way. This was proven by Moser in 1963.

So, there is about a 7 in 10 chance that any year from now on will be the sum of consecutive primes.

Also, there may be more than one way to write 2011 as a sum of consecutive primes.

Of course, New Year’s Day was 1/1/11, and 2011 is prime. Maybe the end of times is near.

I tried to reproduce your results with R; my code (see below) does seem to produce the right results for 2011

2011 : 661 673 677
2011 : 157 163 167 173 179 181 191 193 197 199 211

Just as in the Mathematica example, I did create a list of consecutive prime numbers, and for each possible set of consecutive prime numbers from this list, I checked whether or not the sum is a prime number.

Checking all possible combinations doesn’t seem the most efficient way to solve this problem. Can this be done somehow more efficiently?

===================

#from stackoverflow.com/questions/3789968/generate-a-list-of-primes-in-r-up-to-a-certain-number
primes <- function(n){
primesR <- function(p, i = 1){
f <- p %% p[i] == 0 & p != p[i]
if (any(f)){
p <- primesR(p[!f], i+1)
}
p
}
primesR(2:n)
}

#from stackoverflow.com/questions/3789968/generate-a-list-of-primes-in-r-up-to-a-certain-number
isPrime <- function(x){
div <- 2:floor(sqrt(x))
!any(x %% div == 0)
}

#find all partitions of length n in x
partition<-function(x,n) {
N<-length(x);
lapply(seq(1,N-n+1,1),function(i) x[i:min(i+n-1,N)])
}

#find all partitions in x
partitions<-function(x) {
N<-length(x);
lapply(seq(1,N,1),function(i) partition(x,i))
}

#display only partitions where sum is prime
printPrimeSumOfPrimes<-function(x) {
N<-length(x);
for(i in 1:N) {
M<-length(x[[i]]);
for(j in 1:M) {
s <- sum(x[[i]][[j]]);
if(isPrime(s)) {
print(c(s,x[[i]][[j]]));
}
}
}
}

a <- primes(1000)
b <- partitions(a)
printPrimeSumOfPrimes(b)

4603769 is the sum of 159 consecutive primes starting with 28151, the 3069th prime number. What do I win? Also, did I work that right? 28151 is the first prime that is the first in a sequence of more than a hundred to find a prime sum. Here’s my code:

# plot primes, by sequence, against how many additional ones must be
# added to it to get another.
#
# p1 is 2, and by adding p2, which is 3, to it, you get 5, so f(1) == 1.
# p2 is 3, 3+5 is 8, 8+7 is 15, 15+11 is 26, 26+13 is 39, …? how many
# will it take?

use strict; # good perl practice; default past 5.12

# here’s the Concise But Dumb Perl Prime Tester:
sub CBDisPrime(\$){
(‘A’x\$_[0]) !~ m/^(AA+)\1+\$/
}

#here’s a proper seive:
my @primes = (2,3,5); my \$TooBig = 6;
sub isPrime{
my \$x = shift;
\$x % \$_ or return 0 for @primes;
if(\$x > \$TooBig){
my \$top = 1+ int sqrt \$x;
while (\$TooBig++ 25;
1
};

# make it faster
use Memoize;
memoize(‘isPrime’);

# no partial lines in output
\$| = 1;

# stop when we find one that takes a sequence of 100

# loop over starter primes
my (\$nn,\$sum,@peas);
my \$p = 1; my \$n;
\$sum = \$n;
@peas = (\$n);
# loop over sequential primes
findseq: {
push @peas, \$_;
\$sum += \$_;
isPrime(\$sum) and last findseq;
};
while(1){
isPrime(++\$nn) or next;
push @peas, \$nn;
\$sum += \$nn;
isPrime(\$sum) and last findseq;
};
};
BEGIN { \$” = ‘, ‘ } # make a nice list
print @peas.”. Sum (@peas) == \$sum\n”;
};

[…] Happy mathematical new year: 2011 is the sum of 11 consecutive prime numbers x […]

2+0+2+7=13 ???????

Wow!

LOL! Thanks Alexander. I didn’t spot that, and made the correction.

As 13 is my lucky number, I’ll note this about the new year day:

01 + 01 + 11 = 13

It’s going to be a very* good year.

I am enjoying discussions

I love mathematics

November 11th, 2011, sometime before lunch, which will be

11/11/11 11:11:11

Maybe we should use this to redefine “prime time”.

Nice, but you better ask Neon Deion first…

LOL. http://en.wikipedia.org/wiki/Deion_Sanders for those who didn’t get the reference.

Also worth noting is that this would be the last binary datetime until 1 AM January 1, 2100 (1/1/00 1:00:00).

Another interesting question might be how quantum computing methods could be applied to these searches.

This would be a good topic for my MathCounts team this week.

4603769 is the sum of 159 consecutive primes starting with 28151, the 3069th prime number. What do I win? Also, did I work that right? 28151 is the first prime that is the first in a sequence of more than a hundred to find a prime sum. Here’s my code:

# plot primes, by sequence, against how many additional ones must be
# added to it to get another.
#
# p1 is 2, and by adding p2, which is 3, to it, you get 5, so f(1) == 1.
# p2 is 3, 3+5 is 8, 8+7 is 15, 15+11 is 26, 26+13 is 39, …? how many
# will it take?

use strict; # good perl practice; default past 5.12

# here's the Concise But Dumb Perl Prime Tester:
sub CBDisPrime(\$){
('A'x\$_[0]) !~ m/^(AA+)\1+\$/
}

#here's a proper seive:
my @primes = (2,3,5); my \$TooBig = 6;
sub isPrime{
my \$x = shift;
\$x % \$_ or return 0 for @primes;
if(\$x > \$TooBig){
my \$top = 1+ int sqrt \$x;
while (\$TooBig++ <= \$top ){
isPrime(\$TooBig) or next;
\$x % \$TooBig or return 0;
};
};
warn "\$x is prime";
push @primes, \$x;
# die "@primes" if @primes > 25;
1
};

# make it faster
use Memoize;
memoize('isPrime');

# no partial lines in output
\$| = 1;

# stop when we find one that takes a sequence of 100

# loop over starter primes
my (\$nn,\$sum,@peas);
my \$p = 1; my \$n;
\$n = shift @lookahead; # isPrime(++\$n) or next;

print \$p++," ==> ";
\$sum = \$n;
@peas = (\$n);
# loop over sequential primes
findseq: {
push @peas, \$_;
\$sum += \$_;
isPrime(\$sum) and last findseq;
};
while(1){
isPrime(++\$nn) or next;
push @peas, \$nn;
\$sum += \$nn;
isPrime(\$sum) and last findseq;
};
};
BEGIN { \$" = ', ' } # make a nice list
print @peas.". Sum (@peas) == \$sum\n";
};

here is a more efficient sieve. The memoization is intrinsic, caching the found primes that are too large to append to the sieve list yet.

``` #here's a proper sieve: my @primes = (2,3,5); my \$TooBig = 5;my \$counter; sub isPrime{ my \$x = shift; # warn "q: \$x p:[@primes] tb:\$TooBig"; # die "debugging?" if ++\$counter > 50; my \$top = 1+ int sqrt \$x; if (\$top < \$primes[-1]){ for (@primes){ \$x % \$_ or return 0; \$_ > \$top and last }; # warn "\$x is prime, found no factors < \$top in [@primes]"; return 1 }; # warn "\$top is > \$primes[-1]"; for (@primes){ \$x % \$_ or return 0; }; # warn "will add to list now."; if(\$x > \$TooBig){ while (\$TooBig <= \$top ){ # warn "tb ( \$TooBig ) LE \$top"; isPrime(++\$TooBig) or next; push @primes, \$TooBig; warn "\$TooBig is prime"; # warn "list now [@primes]"; \$x % \$TooBig or return 0; }; warn "\$x is too big for our sieve"; return 1; }; die "IMPOSSIBLE"; }; # make it faster use Memoize; memoize('isPrime'); ```

2011 prime sexiness

Hi,
I’m Gaurav Garg From http://www.Hackersfind.com

This post IS amazing. I like the one in 2011.

#from stackoverflow.com/questions/3789968/generate-a-list-of-primes-in-r-up-to-a-certain-number
primes <- function(n){
primesR <- function(p, i = 1){
f <- p %% p[i] == 0 & p != p[i]
if (any(f)){
p <- primesR(p[!f], i+1)
}
p
}
primesR(2:n)
}

#from stackoverflow.com/questions/3789968/generate-a-list-of-primes-in-r-up-to-a-certain-number
isPrime <- function(x){
div <- 2:floor(sqrt(x))
!any(x %% div == 0)
}

#find all partitions of length n in x
partition<-function(x,n) {
N<-length(x);
lapply(seq(1,N-n+1,1),function(i) x[i:min(i+n-1,N)])
}

#find all partitions in x
partitions<-function(x) {
N<-length(x);
lapply(seq(1,N,1),function(i) partition(x,i))
}

#display only partitions where sum is prime
printPrimeSumOfPrimes<-function(x) {
N<-length(x);
for(i in 1:N) {
M<-length(x[[i]]);
for(j in 1:M) {
s <- sum(x[[i]][[j]]);
if(isPrime(s)) {
print(c(s,x[[i]][[j]]));
}
}
}
}

a <- primes(1000)
b <- partitions(a)
printPrimeSumOfPrimes(b)

Just another Nice Post. I will keep up with this one. Nice Post mate and Happy 2011

This whole conversation is way over my basic math head, but I love primes!

Hey Cassandra,

let us know if there’s something you want to know, or that we could explain better. Great to hear from you!

Great observations! You may be interested in some other general numerical observations at numbersknack.wordpress.com

[…] Republic of Math blog follows the consecutive-prime inquiry further, with the observation that 2011 can also be […]

Beautiful, isn’t it? I must link it. Don’t worry, I will attribute.

[…] Loving and HatingMathematics.Gary Davis at the Republic of Mathematics has a fun article about the number 2011. Not only is 2011 prime but it is also the sum of 11consecutive prime numbers. […]

Checking all possible combinations doesn’t seem the most efficient way to solve this problem. Can this be done somehow more efficiently?

===================

#from stackoverflow.com/questions/3789968/generate-a-list-of-primes-in-r-up-to-a-certain-number
primes <- function(n){
primesR <- function(p, i = 1){
f <- p %% p[i] == 0 & p != p[i]
if (any(f)){
p <- primesR(p[!f], i+1)
}
p
}
primesR(2:n)
}

#from stackoverflow.com/questions/3789968/generate-a-list-of-primes-in-r-up-to-a-certain-number
isPrime <- function(x){
div <- 2:floor(sqrt(x))
!any(x %% div == 0)
}

#find all partitions of length n in x
partition<-function(x,n) {
N<-length(x);
lapply(seq(1,N-n+1,1),function(i) x[i:min(i+n-1,N)])
}

#find all partitions in x
partitions<-function(x) {
N<-length(x);
lapply(seq(1,N,1),function(i) partition(x,i))
}

#display only partitions where sum is prime
printPrimeSumOfPrimes<-function(x) {
N<-length(x);
for(i in 1:N) {
M<-length(x[[i]]);
for(j in 1:M) {
s <- sum(x[[i]][[j]]);
if(isPrime(s)) {
print(c(s,x[[i]][[j]]));
}
}
}
}

a <- primes(1000)
b <- partitions(a)
printPrimeSumOfPrimes(b)

[…] If we take the 11 consecutive prime numbers from 157 to 211 & sum them up; we get 2011. Image via RepulicOfMath.WordPress.com […]