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%I #30 Apr 22 2023 16:46:35
%S 8,0,1,3,3,12,13,17,21,23,23,73,101,103,103,112,113,117,121,123,123,
%T 173,323,373,1103,1103,1112,1113,1117,1121,1123,1123,1173,1323,1373,
%U 3323,3373,11373,13323,13373,17373,23323,23373,73373,101373,103323,103373,111373
%N a(n) is the smallest nonnegative number whose American English name has the letter "e" in the n-th position.
%D GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 70.
%e a(1)=8 ("Eight"), a(2)=0 ("zEro"), a(3)=1 ("onE"), a(4)=3 ("thrEe").
%o (Python)
%o from num2words import num2words
%o from itertools import count, islice
%o def n2w(n):
%o return "".join(c for c in num2words(n).replace(" and ", "") if c.isalpha())
%o def A164790(n, t="e", i0=0): # t is target letter, i0 is start
%o return next(i for i in count(i0) if len(w:=n2w(i))>=n and w[n-1]==t)
%o print([A164790(n) for n in range(1, 38)]) # _Michael S. Branicky_, Apr 21 2023
%o (Python) # faster for initial segment of sequence; uses n2w/imports above
%o def A164790gen(t="e", i0=0, offset=1): # generator of terms w
%o adict, n = dict(), offset
%o for i in count(i0):
%o w = n2w(i)
%o if t in w:
%o locs = [i+1 for i, c in enumerate(w) if w[i] == t]
%o for v in locs:
%o if v not in adict: adict[v] = i
%o while n in adict: yield adict[n]; n += 1
%o print(list(islice(A164790gen(), 50))) # _Michael S. Branicky_, Apr 21 2023
%Y See A362120-A362122 for other versions.
%Y Cf. A164789 ("o"), A164791 ("n"), A164792 ("t"), A164793 ("i"), A164794 ("f"), A164795 ("h"), A164796 ("r"), A164797 ("u").
%K nonn,word
%O 1,1
%A _Claudio Meller_, Aug 26 2009
%E a(25) and beyond from _Michael S. Branicky_, Mar 25 2021
%E Name edited by _N. J. A. Sloane_, Apr 20 2023
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