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%I #21 Jul 20 2021 20:59:46
%S 0,1,3,2,4,5,7,8,13,14,11,17,6,12,23,15,9,18,16,21,19,24,22,27,25,28,
%T 26,29,33,10,20,37,34,31,38,35,32,39,36,43,47,48,44,30,45,41,42,53,49,
%U 57,46,58,123,54,127,73,128,55,117,124,125,113,63,114,133
%N Nonnegative integers ordered by lowest number value per letter (in American English) and then, in case of ties, numerically.
%C Uses the convention of omitting a trailing 'and', so 101 is 'one hundred one' rather than 'one hundred and one.'
%e 17 has a value per letter of 17/9, which is more than 11's value of 11/6 but less than 6's value of 2.
%e From _Jon E. Schoenfield_, Jun 27 2021: (Start)
%e Table begins:
%e letter count
%e n a(n) name A005589(a(n)) a(n)/A005589(a(n))
%e -- ---- ------------ ------------- ------------------
%e 1 0 zero 4 0/4 = 0
%e 2 1 one 3 1/3 = 0.33333...
%e 3 3 three 5 3/5 = 0.6
%e 4 2 two 3 2/3 = 0.66666...
%e 5 4 four 4 4/4 = 1
%e 6 5 five 4 5/4 = 1.25
%e 7 7 seven 5 7/5 = 1.4
%e 8 8 eight 5 8/5 = 1.6
%e 9 13 thirteen 8 13/8 = 1.625
%e 10 14 fourteen 8 14/8 = 1.75
%e 11 11 eleven 6 11/6 = 1.83333...
%e 12 17 seventeen 9 17/9 = 1.88888...
%e 13 6 six 3 6/3 = 2
%e 14 12 twelve 6 12/6 = 2
%e 15 23 twenty-three 11 23/11 = 2.09090...
%e (End)
%o (Python)
%o from num2words import num2words as n2w
%o def A005589(n): return sum(c.isalpha() for c in n2w(n).replace(" and", ""))
%o def aseq(N): return sorted(range(10*N), key=lambda x: (x/A005589(x), x))[:N]
%o print(aseq(65)) # _Michael S. Branicky_, Jun 27 2021
%Y Cf. A005589.
%K nonn,word
%O 1,3
%A _Jack Zilinskas_, Jun 26 2021