MATLAB code for finding certain coefficients
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I am trying to input 
In the summation, 
s are distinct odd primes such that 
s are distinct odd primes such that How can it be written in MATLAB? Here 
values can be 0 also. All the other symbols involved in the expression are natural numbers and r can be any suitable value. I couldn't get any idea.
values can be 0 also. All the other symbols involved in the expression are natural numbers and r can be any suitable value. I couldn't get any idea. 9 个评论
Walter Roberson
2020-10-6
I do not understand what you mean by that "such that" ?
Walter Roberson
2020-10-6
The summation is over m, but m is not used in the terms ??
John D'Errico
2020-10-6
I found it immensely confusing. I even considered trying to answer the question until I tried to read it.
I assume that maybe the sum is over all partitions of the number m, so all possible sums of integers that sum to m? So perhaps for a fixed m, we compute the sum?
Is the denominator in that sum composed of binomial coefficients? Or are they Legendre symbols? (That would seem to make little sense, but the syntax could work.) Is that a factorial symbol in there?
A mathematical formula written completely out of context is sometimes just a set of meaningless characters.
Walter Roberson
2020-10-6
Integer partions is an interesting idea -- I would not have thought of that!
Bruno Luong
2020-10-6
编辑：Bruno Luong
2020-10-6
I guess (pi|li) meaning pi is one of the (odd) prime factor of li. So li/pi is just standard division.
The "distinct" part I don't know, what if there is no such distincts r-tuplets odd primes? Remove the term from the sum?
This formula begs for some explanation.
Sehra Sahu
2020-10-10
Yes, @Bruno, you are right. By 'distinct' I mean that if 
then we do not consider the same prime 
to divide any other 
. For example, if we are able to split 
, the summand concerning this partition will be 
. Even if you consider other primes in this figure, they will be like 
in the denominator, which is equal to 1. But we cannot split 
. If you consider like that, only 3 can divide the first two partitioned terms 3 and 6, since 2 is not involved. But here 3 should not simultaneously divide 3 and 6, or as a matter of fact, none of the primes should divide as such. So, when I split 
, then naturally we should be taking 
and 
, and hence the summand would be 
. I hope this is understandable.
then we do not consider the same prime to divide any other . For example, if we are able to split , the summand concerning this partition will be . Even if you consider other primes in this figure, they will be like in the denominator, which is equal to 1. But we cannot split . If you consider like that, only 3 can divide the first two partitioned terms 3 and 6, since 2 is not involved. But here 3 should not simultaneously divide 3 and 6, or as a matter of fact, none of the primes should divide as such. So, when I split , then naturally we should be taking and , and hence the summand would be . I hope this is understandable.
Sehra Sahu
2020-10-10
Bruno Luong
2020-10-10
One more question, does the order of (l1, l2, ... lr) matter? Meaning do you take all the permutations of {li} in the sum or they are selected said once, in the increasing order?
Sehra Sahu
2020-10-10
采纳的回答
Bruno Luong
2020-10-10
编辑：Bruno Luong
2020-10-11
Some detail of the formula is not totally clear to me (do we consider li>=0 or li>=1), but it goes like this
(EDIT, assume now the constraints li>=1)
m = 20;
r = 2;
verbose = true;
l = Partition(m, r);
% list all the odd primes
ptab = arrayfun(@(x) oddprime(x), 1:m, 'unif', 0);
% do the sum
s = 0;
for i=1:size(l,1)
[p, li] = alldistinctp(l(i,:), ptab);
if ~isempty(p) && size(p,2)==r
q = li./p;
if verbose
T = array2table([li; p; q]);
T.Properties.RowNames = {'l' 'p' 'l/p'};
disp(T)
end
s = s + 1/prod(factorial(q));
end
end
fprintf('s(m=%d,r=%d) = %f\n', m, r, s)
function [p, l] = alldistinctp(l, ptab)
l(l==0) = [];
n = size(l,2);
p = cell(1,n);
[p{:}] = ndgrid(ptab{l});
p = cat(n+1,p{:});
p = reshape(p, [], n);
% discard n-tuplets with repeated p
b = any(diff(sort(p,2),1,2)==0,2);
p(b,:) = [];
end
function p = oddprime(x)
p = unique(factor(x));
p(p==2) = [];
end
function v = Partition(n, lgt)
% v = Partition(n)
% INPUT
% n: non negative integer
% lgt: optional non negative integer
% OUTPUT:
% v: (m x lgt) non-negative integer array such as sum(v,2)==n
% each row of v is descending sorted
% v contains all possible combinations
% m = p(n) in case lgt == n, where p is the partition function
% v is (dictionnary) sorted
% Algorithm:
% Recursive
% Example:
% >> Partition(5)
%
% ans =
%
% 5 0 0 0 0
% 4 1 0 0 0
% 3 2 0 0 0
% 3 1 1 0 0
% 2 2 1 0 0
% 2 1 1 1 0
% 1 1 1 1 1
if nargin < 2
lgt = n;
end
v = PartR(lgt+1, n, Inf);
end % Partition
%% Recursive engine of integer partition
function v = PartR(n, L1, head)
rcall = isfinite(head);
if rcall
L1 = L1-head;
end
if n <= 2
if ~rcall
v = L1;
elseif head >= L1
v = [head, L1];
else
v = zeros(0, n, class(L1));
end
else % recursive call
j = min(L1,head):-1:0;
v = arrayfun(@(j) PartR(n-1, L1, j), j(:), 'UniformOutput', false);
v = cat(1,v{:});
if rcall
v = [head+zeros([size(v,1),1], class(head)), v];
end
end
end % PartR
2 个评论
Sehra Sahu
2020-10-11
You may consider 
for time being.
for time being.Apparently the code shows some error in line 19, says function definition not permitted.
Bruno Luong
2020-10-11
编辑：Bruno Luong
2020-10-11
Easy to fix: If you run on older version MATLAB, save the three functions in separate mfiles from the script.
It works for me it give out this for m=20, r=2
Var1 Var2
____ ____
l 19 1
p 19 1
l/p 1 1
Var1 Var2
____ ____
l 17 3
p 17 3
l/p 1 1
Var1 Var2
____ ____
l 15 5
p 3 5
l/p 5 1
Var1 Var2
____ ____
l 14 6
p 7 3
l/p 2 2
Var1 Var2
____ ____
l 13 7
p 13 7
l/p 1 1
Var1 Var2
____ ____
l 11 9
p 11 3
l/p 1 3
s(m=20,r=2) = 3.425000
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