cp's OEIS Frontend

A107322 English name for number and its reverse have the same number of letters.

%I A107322 #20 Feb 25 2024 11:00:35
%S A107322 0,1,2,3,4,5,6,7,8,9,10,11,14,22,33,34,35,38,41,43,44,45,48,53,54,55,
%T A107322 58,66,67,69,76,77,79,83,84,85,88,96,97,99,101,102,103,104,105,106,
%U A107322 107,108,109,111,112,113,115,118,121,122,123,124,125,126,127,128,129,131,132
%N A107322 English name for number and its reverse have the same number of letters.
%C A107322 Obviously includes all palindromes (A002113).
%H A107322 Michael S. Branicky, <a href="/A107322/b107322.txt">Table of n, a(n) for n = 0..10000</a>
%e A107322 35 is in sequence because 35 ("thirty-five") and 53 ("fifty-three") each have 10 letters in English (dashes not counted).
%t A107322 Select[Range[0,132],Length[Select[Characters[IntegerName[#,"Words"]],LetterQ]]==Length[Select[Characters[IntegerName[FromDigits[Reverse[IntegerDigits[#]]],"Words"]],LetterQ]]&] (* _James C. McMahon_, Feb 12 2024 *)
%o A107322 (Python)
%o A107322 from num2words import num2words
%o A107322 def n2w(n):
%o A107322   map = {ord(c): None for c in "-, "}
%o A107322   return num2words(n).replace(" and", "").translate(map)
%o A107322 def ok(n): return len(n2w(n)) == len(n2w(int(str(n)[::-1])))
%o A107322 print([k for k in range(133) if ok(k)]) # _Michael S. Branicky_, Feb 12 2024
%Y A107322 Cf. A002113, A004086, A005589.
%K A107322 base,nonn,word
%O A107322 0,3
%A A107322 _David W. Wilson_, May 21 2005
%E A107322 10 inserted by _James C. McMahon_, Feb 12 2024