A231115 Start with 1; then a(n+1) = concatenation of c(d) and d*c(d), where d is the last digit of a(n) and the counter c(d) is increased at each occurrence of d, and once more if d*c(d) would end in zero.

1, 11, 22, 12, 24, 14, 28, 18, 216, 16, 212, 36, 318, 324, 312, 48, 432, 612, 714, 416, 424, 624, 728, 648, 756, 636, 742, 816, 848, 864, 832, 918, 972, 1122, 1224, 936, 954, 1144, 1248, 1188, 1296, 1166, 1272, 1326, 1378, 13104, 1352, 1428

Offset

1,2

Comments

The resulting sequence shows the multiplications that have been done, where the "," becomes the multiplication operator. E.g., ...,312,48,... shows that at this point the digit '2' occurred for the 4th time and the calculation was 2*4=8. At the next occurrence one would get 2 x 5 = 10 which is forbidden, so the counter is set to 6, which yields the 432,612 <=> 2*6=12.

Links

Table of n, a(n) for n=1..48.

E. Angelini, Concatenating multiplications and successive results, post to SeqFan list, Nov 04 2013

Prog

(PARI) c=vector(9); a=1; for(n=1, 99, print1(a", "); d=a%10; until(P%10, P=d*c[d]++); a=eval(Str(c[d], P)))

Crossrefs

Sequence in context: A064263 A138838 A139337 * A040110 A140452 A048136

Adjacent sequences: A231112 A231113 A231114 * A231116 A231117 A231118

Keyword

nonn,base

Author

Eric Angelini and M. F. Hasler, Nov 04 2013

Status

approved