%I #21 Jun 22 2019 11:29:48
%S 1,1,1,1,2,1,1,12,12,1,1,2,12,2,1,1,120,120,120,120,1,1,2,120,20,120,
%T 2,1,1,252,252,2520,2520,252,252,1,1,2,252,42,2520,42,252,2,1,1,240,
%U 240,5040,5040,5040,5040,240,240,1,1,2,240,40,5040,84,5040,40,240,2,1
%N Generalized Pascal triangle associated with the set of primes.
%C Given a subset S of the integers Z, Bhargava has shown how to associate with S a generalized factorial function, denoted n!_S, sharing many properties of the classical factorial function n! (which corresponds to the choice S = Z). In particular, he shows that the generalized binomial coefficients n!_S/(k!_S*(n-k)!_S) are always integral for any choice of S.
%C Here we take S = {2,3,5,7,...} the set of primes.
%C The generalized factorial n!_S is given by the formula n!_S = product {primes p} (p^(floor(n/(p-1)) + floor(n/(p^2-p)) + floor(n/(p^3-p^2)) + ...), and appears in the database as n!_S = A053657(n) for n>=1. We make the convention that 0!_S = 1.
%C See A186432 for the generalized Pascal triangle associated with the set
%C of squares.
%H M. Bhargava, <a href="http://mathdl.maa.org/images/upload_library/22/Hasse/00029890.di021346.02p0064l.pdf">The factorial function and generalizations</a>, Amer. Math. Monthly, 107(2000), 783-799.
%F T(n,k) = A053657(n)/(A053657(k)*A053657(n-k)), for n,k >= 0, with the convention that A053657(0) = 1.
%F Row sums A186431.
%e Triangle begins
%e n/k.|..0.....1.....2.....3.....4.....5.....6.....7
%e ==================================================
%e .0..|..1
%e .1..|..1.....1
%e .2..|..1.....2.....1
%e .3..|..1....12....12.....1
%e .4..|..1.....2....12.....2.....1
%e .5..|..1...120...120...120...120.....1
%e .6..|..1.....2...120....20...120.....2.....1
%e .7..|..1...252...252..2520..2520...252...252.....1
%e .8..|
%p #A186430
%p #Uses program for A053657 written by _Peter Luschny_
%p A053657 := proc(n) local P, p, q, s, r;
%p P := select(isprime, [$2..n]); r:=1;
%p for p in P do s := 0; q := p-1;
%p do if q > (n-1) then break fi;
%p s := s + iquo(n-1, q); q := q*p; od;
%p r := r * p^s; od; r end:
%p T := (n,k) -> A053657(n)/(A053657(k)*A053657(n-k)):
%p for n from 0 to 10 do
%p seq(T(n,k),k = 0..n)
%p end do;
%t b[n_] := Product[p^Sum[Floor[(n - 1)/((p - 1) p^k)], {k, 0, n}], {p, Prime[ Range[n]]}];
%t T[n_, k_] := b[n]/(b[k] b[n - k]);
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* _Jean-François Alcover_, Jun 22 2019 *)
%Y Cf. A053657, A186431, A186432.
%K nonn,easy,tabl
%O 0,5
%A _Peter Bala_, Feb 21 2011