%I #8 Nov 03 2012 13:05:13

%S 0,1,2,9,10,12,14,16,17,18,20,22,23,24,25,248,250,251,254,257,258,259,

%T 262,263,264,265,267,269,272,275,276,277,281,285,287,288,289,291,293,

%U 295,296,298,299,300,301,303,305,306,307,309,311,313,314,315,317,319,320,321,322,326,329,330,331,335

%N Numbers with d distinct ternary digits (d=1,2,3) such that for each k=1,...,d, some digit occurs exactly k times.

%C For each of the terms, the number of ternary (= base 3) digits is a triangular number A000217.

%C The base 2 analog would have only the 5 terms 0,1,4,5,6. See A218556 for the base 10 analog.

%C The sequence A167819 is a subsequence containing exactly all terms >= 9.

%C The sequence is finite, with 255=3+12+240 (= 1 + sum of the 3rd row of A218566) terms.

%H M. F. Hasler, <a href="/A218560/b218560.txt">Table of n, a(n) for n = 1..255</a> (full sequence).

%e The terms a(1)=0 through a(3)=2 have exactly 1 digit occurring exactly once.

%e The terms a(4)=9=100[3] through a(15)=25=221[3], have one ternary digit occurring once and a second, different digit occurring exactly twice.

%e The terms a(16)=248=100012[3] through a(255)=714=222110[3] contain each ternary digit at least once. There are no other terms in this sequence.

%o (PARI) {my(T(n)=n*(n+1)\2); print1(0); for(i=1,3, s=vector(i+1,j,j-1); for(n=3^(T(i)-1),3^T(i)-1,i !=#Set(digits(n,3)) & next; c=vector(4); for(j=1,#d=digits(n,3),c[d[j]+1]++); vecsort(c,,8)==s & print1(","n)))}

%o (PARI) is_A218560(n,b=3)={ my(c=vector(b+1)); for(i=1,#n=digits(n,b),c[n[i]+1]++); #(c=vecsort(c,,8))==1+c[#c] && 2*#n==c[#c]*#c }

%Y Cf. A218559, A182040, A218556.

%K nonn,easy,base,fini,full

%O 1,3

%A _M. F. Hasler_, Nov 02 2012