A350519 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.

3, 13, 63, 325, 1719, 9237, 50199, 275149, 1518263, 8422961, 46935819, 262512929, 1472854451, 8285893713, 46723439019, 264009961733, 1494486641911, 8473508472009, 48112827862527, 273541139290857, 1557023508876891, 8872219429659729, 50605041681538595, 288897992799897481

Offset: 1

Author

Yigit Oktar, Jan 02 2022

Keywords

• nonn

Comments

Replacing prime(n+1) by other functions f(n) we can get many other sequences. For example, with f(n) = 1 we get A001850.

Examples

			The two-dimensional recurrence A(m,n) can be depicted in matrix form as
   3   5   7   11   13    17    19 ...
   5  13  25   43   67    97   133 ...
   7  25  63  131  241   405   635 ...
  11  43 131  325  697  1343  2383 ...
  13  67 241  697 1719  3759  7485 ...
  17  97 405 1343 3759  9237 20481 ...
  19 133 635 2383 7485 20481 50199 ...
  ...
and then a(n) is the main diagonal of this matrix, A(n,n).
		

Crossrefs

Cf. A000040, A001850, A002002, A050151, A344576 (see comments).

Programs

MATLAB

clear all
close all
sz = 14
f = zeros(sz,sz);
pp = primes(50);
f(1,:) = pp(2:end);
f(:,1) = pp(2:end);
for m=2:sz
    for  n=2:sz
        f(m,n) = f(m-1,n-1)+f(m,n-1)+f(m-1,n);
    end
end
an = []
for n=1:sz
    an = [an f(n,n)];
end
S = sprintf('%i,',an);
S = S(1:end-1)

Mathematica

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 *)