A045619 Numbers that are the products of 2 or more consecutive integers.

0, 2, 6, 12, 20, 24, 30, 42, 56, 60, 72, 90, 110, 120, 132, 156, 182, 210, 240, 272, 306, 336, 342, 360, 380, 420, 462, 504, 506, 552, 600, 650, 702, 720, 756, 812, 840, 870, 930, 990, 992, 1056, 1122, 1190, 1260, 1320, 1332, 1406, 1482, 1560, 1640, 1680

Offset: 1

Author

Erich Friedman

Keywords

• easy
• nonn
• nice

Comments

Erdős and Selfridge proved that, apart from the first term, these are never perfect powers (A001597). - T. D. Noe, Oct 13 2002

Numbers of the form x!/y! with y+1 < x. - Reinhard Zumkeller, Feb 20 2008

Examples

			30 is in the sequence as 30 = 5*6 = 5*(5+1). - _David A. Corneth_, Oct 19 2021
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T.D. Noe)
• P. Erdős and J. L. Selfridge, The product of consecutive integers is never a power, Illinois Jour. Math. 19 (1975), 292-301.

Crossrefs

Union of A002378, A007531, A052762, A052787, A053625, etc. - R. J. Mathar, Oct 19 2021

Cf. A001597, A000142, A137895, A093449, A064224, A084720, A137899, A137900, A120436, A097889.

Programs

Mathematica

maxNum = 1700; lst = {}; For[i = 1, i <= Sqrt[maxNum], i++, j = i + 1; prod = i*j; While[prod < maxNum, AppendTo[lst, prod]; j++; prod *= j]]; lst = Union[lst]

PARI

list(lim)=my(v=List([0]),P,k=1,t); while(1, k++; P=binomial('n+k-1,k)*k!; if(subst(P,'n,1)>lim, break); for(n=1,lim, t=eval(P); if(t>lim, next(2)); listput(v,t))); Set(v) \\ Charles R Greathouse IV, Nov 16 2021

Python

import heapq
from sympy import sieve
def aupton(terms, verbose=False):
    p = 6; h = [(p, 2, 3)]; nextcount = 4; aset = {0, 2}
    while len(aset) < terms:
        (v, s, l) = heapq.heappop(h)
        aset.add(v)
        if verbose: print(f"{v}, [= Prod_{{i = {s}..{l}}} i]")
        if v >= p:
            p *= nextcount
            heapq.heappush(h, (p, 2, nextcount))
            nextcount += 1
        v //= s; s += 1; l += 1; v *= l
        heapq.heappush(h, (v, s, l))
    return sorted(aset)
print(aupton(52)) # Michael S. Branicky, Oct 19 2021

Formula

a(n) = A000142(A137911(n))/A000142(A137912(n)-1) for n>1. - Reinhard Zumkeller, Feb 27 2008

Since the oblong numbers (A002378) have relative density of 100%, we have a(n) ~ (n-1) n ~ n^2. - Daniel Forgues, Mar 26 2012

a(n) = n^2 - 2*n^(5/3) + O(n^(4/3)). - Charles R Greathouse IV, Aug 27 2013

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jul 20 2000

More terms from Reinhard Zumkeller, Feb 27 2008

Incorrect program removed by David A. Corneth, Oct 19 2021