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%I #16 Feb 02 2023 16:52:32
%S 1,1,1,1,1,1,1,10,10,1,1,21,140,21,1,1,3960,55440,55440,3960,1,1,78,
%T 205920,432432,205920,78,1,1,28560,1485120,588107520,588107520,
%U 1485120,28560,1,1,171,3255840,25395552,4788875520,25395552,3255840,171,1
%N Triangular array T(n,k) read by antidiagonals: T(2,1) = 1; otherwise T(n,k) = p(n)!/(p(k)!*p(n-k)!), where p(0)=1 and p(m)=prime(m) for m > 0.
%C Essentially analogous to Pascal's triangle, A007318.
%F T(2,1) = 1; otherwise T(n,k) = p(n)!/(p(k)!*p(n-k)!), where p(0)=1 and p(m)=prime(m) for m > 0.
%e First six rows:
%e 1
%e 1 1
%e 1 1 1
%e 1 10 10 1
%e 1 21 140 21 1
%e 1 3960 55440 55440 3960 1
%e ...
%p p:= n-> `if`(n=0, 1, ithprime(n)):
%p T:= (n, k)-> floor(p(n)!/(p(k)!*p(n-k)!)):
%p seq(seq(T(n, k), k=0..n), n=0..10); # _Alois P. Heinz_, Jan 30 2023
%t p[0] = 1; p[n_] := Prime[n];
%t t = Table[p[n]!/(p[k]!*p[n - k]!), {n, 0, 10}, {k, 0, n}]
%t t[[3, 2]] = 1;
%t TableForm[t] (* A360207 array *)
%t Flatten[t] (* A360207 sequence *)
%Y Cf. A007318, A008578, A360208.
%K nonn,tabl
%O 0,8
%A _Clark Kimberling_, Jan 30 2023