cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A036448 Smallest positive number containing n e's when spelled out in US English.

Original entry on oeis.org

2, 1, 3, 11, 17, 111, 117, 317, 1317, 3317, 11317, 17317, 111317, 117317, 317317, 1317317, 3317317, 11317317, 17317317, 111317317, 117317317, 317317317, 1317317317, 3317317317, 11317317317, 17317317317, 111317317317, 117317317317, 317317317317, 1317317317317, 3317317317317
Offset: 0

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From Michael S. Branicky, Oct 24 2020: (Start)
"US English" connotes that no "and" is used ("one hundred one") and, importantly here, that the names of large numbers follow the "American system" (Weisstein link), also known as the short scale (Wikipedia link). The previous a(8) and a(9) were based on "eleven hundred and seventeen" and "seventeen hundred and seventeen", which are less common written forms (Wikipedia English numbers link). To make the sequence precise, the common written form is adopted ("one thousand one hundred seventeen"; Wilson link; A000052 Example). Thus, a(n) is the least m such that A085513(m)=n.
The sequence follows the pattern of 1(317)^n, 3(317)^n, 11(317)^n, 17(317)^n, 111(317)^n, 117(317)^n, 317(317)^n for n = 0 through 7 and whenever the largest named power has no "e". a(50) > 10^21 = "one sextillion" which is the first power name that has an "e", breaking the pattern. In that case, a(50) = 1117(317)^6 and a(51) = 1(317)^7. Whenever the largest power has 1 "e" it follows this pattern. If it has m>1 "e"'s, the first block of three is shifted lower to a(7-m). See Wikipedia link for Names of large numbers for power names.
(End)

Examples

			One has 1 e.
Three has 2 e's.
		

References

  • Rodolfo Marcelo Kurchan, Problem 1882, Another Number Sequence, Journal of Recreational Mathematics, vol. 23, number 2, p. 141.

Crossrefs

Programs

  • Python
    from num2words import num2words
    def A036448(n):
        i = 1
        while num2words(i).count("e")!=n:
            i += 1
        return i
    print([A036448(n) for n in range(1,12)]) # Michael S. Branicky, Oct 23 2020

Extensions

a(8)-a(9) changed and a(11)-a(30) added by Michael S. Branicky, Oct 23 2020
a(0)=2 inserted by Sean A. Irvine, Nov 02 2020