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%I #27 Jul 25 2018 04:28:58
%S 5,7,29,15,36,3,95,11,1,5,2,19,2,0,1,0,1,1,1,1,2,1,1,1,1,0,1,1,1,1,3,
%T 0,2,0,2,0,2,0,0,0,4,1,4,1,4,0,1,0,0,0,12,0,5,0,2,0,6,0,0,0,12,0,1,0,
%U 1,0,12,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,1,1
%N Number of numbers whose American English name has no letter in common with that of n.
%C The sequence is well defined and zero for n sufficiently large (> 6.1*10^30 ?) because "million", "billion" etc. have a letter in common with all small numbers except for three, which is letter-disjoint with six. Therefore, 3 is letter-disjoint with six, six million, six billion, six nonillion (10^30) and any nonempty sum of two or more of these. See also example of a(5).
%H E. Angelini, <a href="http://www.cetteadressecomportecinquantesignes.com/ChaineFR.htm">Chaîne de noms de nombres</a>
%H E. Angelini, <a href="/A227857/a227857.pdf">Chaîne de noms de nombres</a> [Cached copy, with permission]
%H <a href="/index/El#English">OEIS Index entries for sequences related to the English name of numbers</a>
%e a(0) = 5 = # {6, 50, 56, 60, 66} because "zero" has no letter in common with: six, fifty, fifty-six, sixty, sixty-six.
%e a(5) = 3 = # {2, 2000, 2002} because "five" has no letter in common with: two, two thousand, two thousand two. ("thousand" is not considered; "one thousand" is excluded.)
%e a(3) = 15 = 2^4-1 because any nonzero sum_{i=0,6,9,30} e_i*10^i with e_i in {0, 6} is "letter-disjoint" with three.
%o (PARI) A227857(n,lang=English/*see A052360*/,L=999,o=0)={n==5 && L+=2000; n==3 && return(15)/*can't be computed explicitely*/; n=setminus(Set(Vec(lang(n))),Set([" ","-"])); sum(k=o,L,!setintersect( Set(Vec(lang(k))), n))}
%K nonn,word
%O 0,1
%A _M. F. Hasler_, Nov 04 2013