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A183086
Generalized canyon numbers.
2
101, 102, 103, 104, 105, 106, 107, 108, 109, 201, 202, 203, 204, 205, 206, 207, 208, 209, 212, 213, 214, 215, 216, 217, 218, 219, 301, 302, 303, 304, 305, 306, 307, 308, 309, 312, 313, 314, 315, 316, 317, 318, 319, 323, 324, 325, 326, 327, 328, 329, 401, 402, 403, 404, 405, 406, 407, 408, 409, 412
OFFSET
1,1
COMMENTS
Supersequence of A134970. The structure of digits represents a canyon (a deep valley between cliffs). The first digits are in decreasing order. The last digits are in increasing order. There is only one smaller digit which represents the bottom of the canyon. But the restriction that both cliffs are at same level (first digit equal to the final digit) is dropped here.
This sequence is finite. The final term is 9876543210123456789.
Question: How many terms are there in this sequence?
There are 347489 terms in the sequence. They may be generated in seconds using the posted Python program. - Michael S. Branicky, Aug 02 2022
LINKS
EXAMPLE
Illustration of 751378 as a generalized canyon number:
. . . . . .
. . . . . 8
7 . . . 7 .
. . . . . .
. 5 . . . .
. . . . . .
. . . 3 . .
. . . . . .
. . 1 . . .
. . . . . .
PROG
(Python)
from itertools import chain, combinations as combs
ups = list(chain.from_iterable(combs(range(10), r) for r in range(2, 11)))
s = set(L[::-1] + R[1:] for L in ups for R in ups if L[0] == R[0])
afull = sorted(int("".join(map(str, t))) for t in s)
print(afull[:60]) # Michael S. Branicky, Aug 02 2022
CROSSREFS
KEYWORD
nonn,base,fini
AUTHOR
Omar E. Pol, Jan 19 2011
STATUS
approved