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Numbers having multiple representations as the product of non-overlapping ranges of consecutive numbers.
2

%I #9 Jul 05 2024 18:02:48

%S 210,720,175560,17297280

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

%C 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.

%H Art Kalb, <a href="https://www.youtube.com/watch?v=i8bn0US_k84">Why Number Theory is Hard</a>, YouTube video, 2024

%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_469.htm">Puzzle 469. 5040</a>, The Prime Puzzles and Problems Connection.

%e 210 = 5*6*7 = 14*15.

%e 720 = 2*3*4*5*6 = 8*9*10.

%e 175560 = 19*20*21*22 = 55*56*57.

%e 17297280 = 8*9*10*11*12*13*14 = 63*64*65*66.

%o (Python)

%o import heapq

%o def aupton(terms, verbose=False):

%o p = 2*3; h = [(p, 2, 3)]; nextcount = 4; alst = []; oldv = None

%o while len(alst) < terms:

%o (v, s, l) = heapq.heappop(h)

%o if v == oldv and ((s > oldl) or (olds > l)) and v not in alst:

%o alst.append(v)

%o if verbose: print(f"{v}, [= Prod_{{i = {s}..{l}}} i = Prod_{{i = {olds}..{oldl}}} i]")

%o if v >= p:

%o p *= nextcount

%o heapq.heappush(h, (p, 2, nextcount))

%o nextcount += 1

%o oldv, olds, oldl = v, s, l

%o v //= s; s += 1; l += 1; v *= l

%o heapq.heappush(h, (v, s, l))

%o return alst

%o print(aupton(4, verbose=True)) # _Michael S. Branicky_, Jun 24 2021

%Y Cf. A064224.

%K nonn

%O 1,1

%A _T. D. Noe_, Jul 29 2009