A218560 Numbers with d distinct ternary digits (d=1,2,3) such that for each k=1,...,d, some digit occurs exactly k times.

0, 1, 2, 9, 10, 12, 14, 16, 17, 18, 20, 22, 23, 24, 25, 248, 250, 251, 254, 257, 258, 259, 262, 263, 264, 265, 267, 269, 272, 275, 276, 277, 281, 285, 287, 288, 289, 291, 293, 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

Offset: 1

Author

M. F. Hasler, Nov 02 2012

Keywords

• nonn
• easy
• base
• fini
• full

Comments

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

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

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

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

Examples

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

Links

• M. F. Hasler, Table of n, a(n) for n = 1..255 (full sequence).

Crossrefs

Cf. A218559, A182040, A218556.

Programs

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)))}

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 }