Proof of the integrality of A211417 and A211418. Peter Bala, April 2012 (revised and corrected October 2015) For the convenience of users of the database we include a proof of a result of Chebyshev: A211417(n) = (30*n)!*n!/((15*n)!*(10*n)!*(6*n)!) is an integer for all n. We also prove the generalization: for real r >= 0 floor(30*r)!*floor(r)!/(floor(15*r)!*floor(10*r)!*floor(6*r)!) is an integer. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Let a(n) = A211417(n). Let c/d be a rational number and let ord_p(c/d) denote the power to which a prime p appears in the prime factorization of c/d. In order to show the rational number a(n) is an integer it will suffice to show that ord_p(a(n)) is nonnegative for each prime p. The proof of this assertion makes use of Legendre's theorem on the order in which a prime divides a factorial: ord_p(n!) = floor(n/p) + floor(n/p^2) + floor(n/p^3) + .... By Legendre's theorem we see that ord_p(a(n)) equals the sum over k = 1,2,... of the terms floor(30*n/p^k) + floor(n/p^k) - floor(15*n/p^k) - floor(10*n/p^k) - floor(6*n/p^k) (the sum of course has only a finite number of nonzero terms). The nonnegativity of this sum, and hence the integrality of a(n), is then a consequence of the following result: LEMMA For real x >= 0 we have floor(15*x) + floor(10*x) + floor(6*x) <= floor(30*x) + floor(x). PROOF We require the following simple property of the floor function: For real x and y floor(x) + floor(y) <= floor(x + y). ... (1) We can write the real number 30*x in the form 30*x = 30*n + m + e ... (2) where n and m are nonnegative integers with 0 <= m < 30, and where 0 <= e < 1. From (2) we find floor(30*x) = 30*n + m and floor(x) = n. Thus floor(30*x) + floor(x) = (30*n + m) + n = 31*n + m. ... (3) Again using (2) we find floor(15*x) + floor(10*x) + floor(6*x) = (15*n + floor(m/2)) + (10*n + floor(m/3)) + (6*n + floor(m/5)) = 31*n + floor(m/2) + floor(m/3) + floor(m/5) <= 31*n + floor(m/2 + m/3 + m/5) (by (1)) = 31*n + floor(31*m/30) = 31*n + m (since 0 <= m < 30) = floor(30*x) + floor(x). (by (3)). END PROOF. We now give a generalization of Chebyshev's result, which includes the integrality of A211418 as a particular case. It will be convenient in what follows to abbreviate notation and write f(x) for floor(x). We require the following standard result for the floor function: for positive integers m and n and real x f(f(x/m)/n) = f(x/(m*n)). . .. (4) PROPOSITION Let r >= 0 be a real number. The sequence of rational numbers a(r,n) defined by a(r,n) = f(30*r*n)!*f(r*n)!/(f(15*r*n)!*f(10*r*n)!*f(6*r*n)!) takes only integer values. PROOF The result is clearly true when r = 0, so from now on we assume r > 0. The approach to integrality is the same as above: we show that ord_p(a(r,n)), the power to which a prime p appears in a(r,n), is nonnegative for all p. We first establish this for the case when r is a rational number r = c/d, (c, d positive integers). By Legendre's theorem, ord_p(a(c/d,n)) is given by the sum over k of the expressions f(f(30*c*n/d)/p^k) + f(f(c*n/d)/p^k) - f(f(15*c*n/d)/p^k) - f(f(10*c*n/d)/p^k) - f(f(6*c*n/d)/p^k). By means of (4) this simplifies to the sum over k of f(30*c*n/(d*p^k)) + f(c*n/(d*p^k)) - f(15*c*n)/(d*p^k)) - f(10*c*n)/(d*p^k)) - f(6*c*n)/(d*p^k)) and for each k the expression is nonnegative by the above Lemma. Finally, to deal with the case when r is irrational let {r_k}k>=0 be a sequence of positive rational numbers tending to r lim {k -> inf} r_k = r. For a fixed nonnegative integer n, a(r,n), qua function of r, is a piecewise constant function on [0,inf) having step discontinuities at a discrete set of rational points. Hence for irrational values of r a(r,n) = lim {k -> inf} a(r_k,n) is a limit of a sequence of integers and so must also be an integer. END PROOF The particular case of the proposition when r = 1/30 is listed in the database as A211418.