This sounds like the h-index. The one-line matlab code to calculate it is

N=max(find(sort(A,2,'descend')-(1:length(A))>=0))

clearly the order of the elements in the set does not matter, so we can assume that we have a set of k non-increasing positive integers which we write as p_1 >= ... >= p_k

for each position j, we have at least j elements at least as big as p_j

as j increases from 1 to k, p_j decreases from p_1 to p_k -

if those sequences do not cross, then the value is k (consider the integer 20, repeated 6 times, or the 6 integers from 20 down to 15; both have at least 6 elements larger than 6) - or maybe the answer is 1 - the definition is ambiguous

if those sequences cross (as they do in the example), then i think the answer is where they cross

but the statement of the property in question needs to be refined to be much clearer about the quantifiers and how they scope the minimum - you might want to ask instead what is the largest N for which there are at least N elements of value at least N, because as long as you use positive integers, the minimum value for N is always 1

in either case, i think that there are sets with any positive integer answer you want (as long as it is no more than the size k of the set)

I am looking for the maximum value of N. I need an equation to express it. Like set that belongs to this variable which has these properties.

The matlab code i used in my experiment was this:

N=0

[m,n]=size(SET);

for i=1:n

a=max(SET(:,i));

a=uint32(a)+1;

s=0;

while s == 0

count=0;

a=a-1;

for j=1:m

if uint32(SET(j,i)) >= a

count=count+1;

end

end

if count >= a

s=1;

else

s=0;

end

end

N=a;

end

I do not think that it is possible to find a formula for solving the problem. It is , however, possible to define an algorithm as follows:

1) The sequence must be arranged in decreasing order;

2) Each element of the ordered sequence should be indexed starting from 1 so that A(1) is the maximum value;

3) Solve the following equation : A(i)=i with the condition that A(i+1)≠i

For instance the original sequence was: 3, 4, 5, 2, 1, 4, 1, 3, 2, 6

The sequence with numbers in decreasing order is

6, 5, 4, 4, 3, 3, 2, 2, 1, 1

The sequence should be indexed as follows A(1)=6, A(2)=5, A(3)=4, A(4)=4, A(5)=3, A(6)=3, A(7)=2, A(8)=2, A(9)=1, A(10)=10

We have A(4)=4 and A(5)≠4. Thus the solution is 4.

Counterexample:

Assume that the sequence is 3, 3, 3, 3, 7. The indexed sequence is

A(1)=7, A(2)=3, A(3)=3, A(4)=3, A(5)=3

We have A(3)=3 but A(4)=3 therefore the problem has no solution as there are 4 elements that are ≥3.

This sounds like the h-index. The one-line matlab code to calculate it is

N=max(find(sort(A,2,'descend')-(1:length(A))>=0))

It does look like jo:rg has written a program that computes what i wrote (though the program is, of course, more precise)

Yes, I agree that h can be the correct function by interpreting the first description of the problem posed by Afaz. However, the example he showed seems requiring that the number of elements (4) should be equal to the minimum value (4). It would be interesting if Afaz can better clarify is question. However, I think we have answered to both kinds of problems.