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A354523 Number of distinct letters in the English word for n that can also be found in the English word for n+1.

%I A354523 #24 Oct 01 2022 19:45:30
%S A354523 2,1,1,1,1,1,1,1,2,2,2,3,2,4,4,4,4,3,4,3,5,6,5,6,6,6,6,5,6,3,5,6,5,5,
%T A354523 6,5,6,6,6,3,5,5,5,5,5,6,6,6,7,4,4,5,4,5,4,4,5,5,5,3,5,6,5,6,6,5,5,6,
%U A354523 6,5,6,7,6,7,7,7,6,6,7,4,6,7,6,7,7,6,7,6,6,5,5,6,5,6,6,5,6,5,5,2,7
%N A354523 Number of distinct letters in the English word for n that can also be found in the English word for n+1.
%C A354523 US English is assumed (i.e., 101 = 'one hundred one' instead of 'one hundred and one').
%C A354523 There are no zero values since min{a(k) | 0 <= k < 1000} = 1 and "nine" and "one" share common letters whenever the initial power name changes. - _Michael S. Branicky_, Aug 19 2022
%e A354523 a(0) = 2 since the letters 'e' and 'o' in 'zero' can also be found in 'one'.
%e A354523 a(11)= 3 since the letters 'e', 'l' and 'v' in 'eleven' can also be found in 'twelve'.
%t A354523 a[n_]:= Length[Intersection[Characters[IntegerName[n]], Characters[IntegerName[n+1]], CharacterRange["a","z"]]]; Array[a,101,0] (* _Stefano Spezia_, Aug 19 2022 *)
%o A354523 (Python)
%o A354523 from num2words import num2words as n2w
%o A354523 def b(n): return set(c for c in n2w(n).replace(" and", "") if c.isalpha())
%o A354523 def a(n): return len(b(n) & b(n+1))
%o A354523 print([a(n) for n in range(101)]) # _Michael S. Branicky_, Aug 19 2022
%K A354523 nonn,easy,word
%O A354523 0,1
%A A354523 _Ray G. Opao_, Aug 16 2022