|
|
A173070
|
|
Palindromic mountain numbers.
|
|
2
|
|
|
1, 121, 131, 141, 151, 161, 171, 181, 191, 12321, 12421, 12521, 12621, 12721, 12821, 12921, 13431, 13531, 13631, 13731, 13831, 13931, 14541, 14641, 14741, 14841, 14941, 15651, 15751, 15851, 15951, 16761, 16861, 16961, 17871, 17971, 18981
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
There are 256 terms, the last of which is 12345678987654321. - Michael S. Branicky, Aug 04 2022
|
|
LINKS
|
|
|
EXAMPLE
|
13731 is in the sequence because it is a palindrome (A002113) and it is also a mountain number (A134941).
. . . . .
. . . . .
. . 7 . .
. . . . .
. . . . .
. . . . .
. 3 . 3 .
. . . . .
1 . . . 1
|
|
PROG
|
(Python)
from itertools import chain, combinations as combs
def c(s): return s[0] == s[-1] == 1 and s == s[::-1]
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[-1] == R[-1])
afull = [1] + sorted(int("".join(map(str, t))) for t in s if c(t))
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base,fini,full
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|