A163263 Numbers having multiple representations as the product of non-overlapping ranges of consecutive numbers.

210, 720, 175560, 17297280

Offset: 1

Author

T. D. Noe, Jul 29 2009

Keywords

• nonn

Comments

A subsequence of A064224. This sequence gives solutions P to the equation P = (x+1)...(x+m) = (y+1)...(y+n) with x>0, y>0 and x+m < y+1. So far, no numbers P with more than two representations have been discovered. Note that the only the lowest range of consecutive numbers (x+1 to x+m) can contain prime numbers; the other ranges are in a gap between consecutive primes. Gaps between the first 45000 primes were searched for additional terms, but none were found.

Examples

			210 = 5*6*7 = 14*15.
720 = 2*3*4*5*6 = 8*9*10.
175560 = 19*20*21*22 = 55*56*57.
17297280 = 8*9*10*11*12*13*14 = 63*64*65*66.
		

Links

• Art Kalb, Why Number Theory is Hard, YouTube video, 2024
• Carlos Rivera, Puzzle 469. 5040, The Prime Puzzles and Problems Connection.

Crossrefs

Cf. A064224.

Programs

Python

import heapq
def aupton(terms, verbose=False):
    p = 2*3; h = [(p, 2, 3)]; nextcount = 4; alst = []; oldv = None
    while len(alst) < terms:
        (v, s, l) = heapq.heappop(h)
        if v == oldv and ((s > oldl) or (olds > l)) and v not in alst:
            alst.append(v)
            if verbose: print(f"{v}, [= Prod_{{i = {s}..{l}}} i = Prod_{{i = {olds}..{oldl}}} i]")
        if v >= p:
            p *= nextcount
            heapq.heappush(h, (p, 2, nextcount))
            nextcount += 1
        oldv, olds, oldl = v, s, l
        v //= s; s += 1; l += 1; v *= l
        heapq.heappush(h, (v, s, l))
    return alst
print(aupton(4, verbose=True)) # Michael S. Branicky, Jun 24 2021