The triangle is all zeroes if 0R0; reflexive relations are boring. Making every other entry 0, i.e. xRy when x(x+1)/2+y is odd, leaves a 1 at y=ceil(x/2) and 0s elsewhere. The triangle gets exciting when every third entry is 0, i.e. xRy when x(x+1)/2+y ==2mod3. For example, if y=3k, then the kth top-right-to-bottom-left diagonal is the sequence 2n^k with each member repeated k+1 times, unless k=0 and the diagonal is all 1s.
________________________________ From: Alin Soare <email@example.com> To: firstname.lastname@example.org Sent: Sunday, 30 June 2013, 12:52 Subject: Singularities in Pascal Triangle.
I have this problem:
Compute the numbers in the Pascal Triangle, but on some positions (X,Y), the pascal triangle has 0, instead of the sum of P(X, Y-1) + P(X-1, Y-1) , each time when X is in relation R with Y.
Obviously, if I am asked to compute the Pascal Triangle, I do not recursivelly compute the sums for each pair, but I compute the combinations(X,Y).
I wish to ask you whether the methods of analytic combinatorics can help to solve this problem -- to find the values using some analytical formula instead of recurrence.