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%I #25 Jan 24 2022 15:50:57
%S 3,13,63,325,1719,9237,50199,275149,1518263,8422961,46935819,
%T 262512929,1472854451,8285893713,46723439019,264009961733,
%U 1494486641911,8473508472009,48112827862527,273541139290857,1557023508876891,8872219429659729,50605041681538595,288897992799897481
%N a(n) = A(n,n) where A(1,n) = A(n,1) = prime(n+1) and A(m,n) = A(m-1,n) + A(m,n-1) + A(m-1,n-1) for m > 1 and n > 1.
%C Replacing prime(n+1) by other functions f(n) we get:
%C A001850 (with f(n) = 1),
%C A002002 (with f(n) = n+1),
%C A050151 (with f(n) = n/2), and
%C A344576 (with f(n) = Fibonacci(n)).
%e The two-dimensional recurrence A(m,n) can be depicted in matrix form as
%e 3 5 7 11 13 17 19 ...
%e 5 13 25 43 67 97 133 ...
%e 7 25 63 131 241 405 635 ...
%e 11 43 131 325 697 1343 2383 ...
%e 13 67 241 697 1719 3759 7485 ...
%e 17 97 405 1343 3759 9237 20481 ...
%e 19 133 635 2383 7485 20481 50199 ...
%e ...
%e and then a(n) is the main diagonal of this matrix, A(n,n).
%t f[1,1]=3;f[m_,1]:=Prime[m+1];f[1,n_]:=Prime[n+1];f[m_,n_]:=f[m,n]=f[m-1,n]+f[m,n-1]+f[m-1,n-1];Table[f[n,n],{n,25}] (* _Giorgos Kalogeropoulos_, Jan 03 2022 *)
%o (MATLAB)
%o clear all
%o close all
%o sz = 14
%o f = zeros(sz,sz);
%o pp = primes(50);
%o f(1,:) = pp(2:end);
%o f(:,1) = pp(2:end);
%o for m=2:sz
%o for n=2:sz
%o f(m,n) = f(m-1,n-1)+f(m,n-1)+f(m-1,n);
%o end
%o end
%o an = []
%o for n=1:sz
%o an = [an f(n,n)];
%o end
%o S = sprintf('%i,',an);
%o S = S(1:end-1)
%Y Cf. A000040, A001850, A002002, A050151, A344576 (see comments).
%K nonn
%O 1,1
%A _Yigit Oktar_, Jan 02 2022
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