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A373391
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Integers whose American English names (minus punctuation) can be split into two parts wherein the product of the letter-ranks of one part is equal to that of the other.
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0
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3960, 13986, 15368, 80547, 85740, 111789, 111987, 386048, 408649, 408946, 410699, 410969, 410996, 486014, 519487, 519784, 609408, 609804, 615430, 619814, 629428, 629824, 639438, 639834, 649448, 649844, 659458, 659854, 669468, 669864, 679478, 679874
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OFFSET
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1,1
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COMMENTS
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By letter-rank we mean a=1, b=2, ..., z=26. If the qualifying split happens beside a letter "a" (as for 408649) there will be two solutions, as moving that "a" to the other side of the split will not affect either product.
There can be no terms involving "million" less than 10^60 ("novemdecillion") since "m" = 13 would otherwise have no counterpart to make the product of all valuations square. - Michael S. Branicky, Jun 03 2024
Similarly, there can be no terms involving "quadrillion" less than 10^18 because "q" = 17. a(84)=1000107588, a(10887)=1000000611455. - Hans Havermann, Jun 15 2024
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LINKS
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EXAMPLE
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3960 = threethousandni|nehundredsixty has the product of the letter-ranks of each side of the vertical pipe equal (to 486491443200000).
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PROG
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(Python)
from math import prod, isqrt
from num2words import num2words
def n2w(n): return "".join(c for c in num2words(n).replace(" and", "") if c.isalpha())
def ok(n):
d = [ord(c) - ord('A') + 1 for c in n2w(n).upper()]
if isqrt(s:=prod(d))**2 != s: return False
return any(prod(d[:i]) == prod(d[i:]) for i in range(1, len(d)))
(Python)
from math import prod, isqrt
from num2words import num2words
def n2w(n):
return "".join(c for c in num2words(n).replace(" and", "") if c.isalpha())
def ok(n):
d = [ord(c) - ord("A") + 1 for c in n2w(n).upper()]
pr = prod(d)
s = isqrt(pr)
if s**2 != pr:
return False
p = 1
for i in range(len(d)):
p *= d[i]
if p > s:
return False
if p == s:
return True
for n in range(1, 100000):
if ok(n):
print(n, end=", ")
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CROSSREFS
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KEYWORD
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nonn,base,word
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AUTHOR
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STATUS
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approved
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