A373391 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.

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

Offset: 1

Author

Hans Havermann, Jun 03 2024

Keywords

• nonn
• base
• word

Comments

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

Examples

			3960 = threethousandni|nehundredsixty has the product of the letter-ranks of each side of the vertical pipe equal (to 486491443200000).
		

Links

• Hans Havermann, Equal products in split English integer names

Crossrefs

Cf. A372222.

Programs

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)))
print([k for k in range(10**5) if ok(k)]) # Michael S. Branicky, Jun 03 2024

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=", ")
# David A. Corneth, Jun 03 2024, adapted from Michael S. Branicky, Jun 03 2024