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A100934
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Numbers having more than one representation as the product of consecutive integers.
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3
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6, 24, 120, 210, 720, 5040, 40320, 175560, 362880, 3628800, 17297280, 19958400, 39916800, 259459200, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 20274183401472000, 121645100408832000
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OFFSET
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1,1
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COMMENTS
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All the factorials occur because we allow products to start with 1. See A064224 for a more restrictive case.
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LINKS
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EXAMPLE
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120 is a term since 120 = 1*2*3*4*5 = 2*3*4*5 = 4*5*6.
210 is a term since 210 = 14*15 = 5*6*7.
Other non-factorial terms are:
175560 = Product_{i=55..57} i = Product_{i=19..22} i,
17297280 = Product_{i=63..66} i = Product_{i= 8..14} i,
19958400 = Product_{i= 5..12} i = Product_{i= 3..11} i,
259459200 = Product_{i= 8..15} i = Product_{i= 5..13} i,
20274183401472000 = Product_{i=6..20} i = Product_{i=4..19} i.
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MATHEMATICA
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nn=10^10; t3={}; Do[m=0; p=n; While[m++; p=p(n+m); p<=nn, t3={t3, p}], {n, Sqrt[nn]}]; t3=Sort[Flatten[t3]]; lst={}; Do[If[t3[[i]]==t3[[i+1]], AppendTo[lst, t3[[i]]]], {i, Length[t3]-1}]; Union[lst]
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PROG
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(Python)
import heapq
def aupton(terms, verbose=False):
p = 1*2; h = [(p, 1, 2)]; nextcount = 3; alst = []; oldv = None
while len(alst) < terms:
(v, s, l) = heapq.heappop(h)
if v == oldv 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, 1, 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
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CROSSREFS
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Cf. A064224, A003015 (numbers occurring 5 or more times in Pascal's triangle).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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