A257299 Numbers n for which each of the digits 0-9 appears exactly once as first digit in the orbit of n under iterations of n -> (first digit of n)*(n with first digit removed) until a single digit is reached; no leading zeros allowed.

9848, 51948, 56648, 68648, 77712, 84157, 87207, 98142, 98642, 249217, 298242, 325803, 328957, 381082, 383003, 423027, 461992, 516957, 549492, 721712, 796523, 812157, 879707, 925492, 945992, 948742, 950742, 960492, 1248242, 1957313, 2211992, 2259492, 2282707

Offset: 1

Author

Eric Angelini and M. F. Hasler, May 08 2015

Keywords

• nonn
• base
• fini
• full

Comments

Numbers for which a leading zero appears in "n with first digit removed" are excluded from this sequence. One could consider the variant where this is allowed in case of a "multi digit zero", i.e., if the last step is x0...0 -> x*0...0 -> 0, see the example of 79855.

The sequence is necessarily finite, because the considered iterations must end in 0 and reach one of the 9 values {10, 20, ..., 90} just before this last iteration, and there must be exactly 9 iterations. This leaves only a finite number of possible starting values n.

Examples

			a(1) = 9848 is in the sequence because if we consider 9848 -> 9 * 848 = 7632 -> 7 * 632 = 4424 -> 4 * 424 = 1696 -> 1 * 696 = 696 -> 6 * 96 = 576 -> 5 * 76 = 380 -> 3 * 80 = 240 -> 2 * 40 = 80 -> 8 * 0 = 0, each of the digits 0-9 appears exactly once as first digit.
For a(2) = 51948, the sequence is 51948 -> 9740 -> 6660 -> 3960 -> 2880 -> 1760 -> 760 -> 420 -> 80 -> 0.
For 79855 -> 68985 -> 53910 -> 19550 -> 9550 -> 4950 -> 3800 -> 2400 -> 800 -> 0, there appears a "leading zero", but only in front of zero.
a(54) = 24578492 is in the sequence because it yields the sequence 24578492 -> 9156984 -> 1412856 -> 412856 -> 51424 -> 7120 -> 840 -> 320 -> 60 -> 0.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..55 (a(1)-a(54) from M. F. Hasler)
• L. Blomberg, in reply to E. Angelini, 10-line tables ?, SeqFan list, Apr 28 2015

Programs

PARI

is(n,d=0)={while(n,bittest(d,(n=divrem(n,10^L=#Str(n\10)))[1])&&return;#Str(n[2])==L||return;d+=1<
				

PARI

A257299(v=0,d=vector(9,i,i))={Set(concat(vector(#d,i,if(v%d[i],[],if(#d>1, A257299(eval(Str(d[i],v/d[i])),vecextract(d,Str("^"i))),[eval(Str(d[i],v/d[i]))])))))} \\ Use just A257299() for the complete list. - M. F. Hasler, May 11 2015

Python

from itertools import permutations
A257299_list = []
for n in permutations('123456789',9):
    x = 0
    for d in n:
        q, r = divmod(x,int(d))
        if r:
            break
        x = int(d + str(q))
    else:
        A257299_list.append(x)
A257299_list = sorted(A257299_list) # Chai Wah Wu, May 11 2015