A186430 Generalized Pascal triangle associated with the set of primes.

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, 2, 1, 1, 252, 252, 2520, 2520, 252, 252, 1, 1, 2, 252, 42, 2520, 42, 252, 2, 1, 1, 240, 240, 5040, 5040, 5040, 5040, 240, 240, 1, 1, 2, 240, 40, 5040, 84, 5040, 40, 240, 2, 1

Offset: 0

Author

Peter Bala, Feb 21 2011

Keywords

• nonn
• easy
• tabl

Comments

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.

Here we take S = {2,3,5,7,...} the set of primes.

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.

See A186432 for the generalized Pascal triangle associated with the set of squares.

Examples

			Triangle begins:
  n/k.|..0.....1.....2.....3.....4.....5.....6.....7
  ==================================================
  .0..|..1
  .1..|..1.....1
  .2..|..1.....2.....1
  .3..|..1....12....12.....1
  .4..|..1.....2....12.....2.....1
  .5..|..1...120...120...120...120.....1
  .6..|..1.....2...120....20...120.....2.....1
  .7..|..1...252...252..2520..2520...252...252.....1
		

Links

• M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly, 107(2000), 783-799.

Crossrefs

Cf. A053657, A186431, A186432.

Programs

Maple

#A186430
#Uses program for A053657 written by Peter Luschny
A053657 := proc(n) local P, p, q, s, r;
P := select(isprime, [$2..n]); r:=1;
for p in P do s := 0; q := p-1;
do if q > (n-1) then break fi;
s := s + iquo(n-1, q); q := q*p; od;
r := r * p^s; od; r end:
T := (n,k) -> A053657(n)/(A053657(k)*A053657(n-k)):
for n from 0 to 10 do
seq(T(n,k),k = 0..n)
end do;

Mathematica

b[n_] := Product[p^Sum[Floor[(n - 1)/((p - 1) p^k)], {k, 0, n}], {p, Prime[ Range[n]]}];
T[n_, k_] := b[n]/(b[k] b[n - k]);
Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jun 22 2019 *)

Formula

T(n,k) = A053657(n)/(A053657(k)*A053657(n-k)), for n,k >= 0, with the convention that A053657(0) = 1.

Row sums A186431.