A173070 - OEIS

%I #6 Aug 04 2022 11:00:46

%S 1,121,131,141,151,161,171,181,191,12321,12421,12521,12621,12721,

%T 12821,12921,13431,13531,13631,13731,13831,13931,14541,14641,14741,

%U 14841,14941,15651,15751,15851,15951,16761,16861,16961,17871,17971,18981

%N Palindromic mountain numbers.

%C There are 256 terms, the last of which is 12345678987654321. - _Michael S. Branicky_, Aug 04 2022

%H Michael S. Branicky, <a href="/A173070/b173070.txt">Table of n, a(n) for n = 1..256</a> (full sequence)

%e 13731 is in the sequence because it is a palindrome (A002113) and it is also a mountain number (A134941).

%e . . . . .

%e . . . . .

%e . . 7 . .

%e . . . . .

%e . . . . .

%e . . . . .

%e . 3 . 3 .

%e . . . . .

%e 1 . . . 1

%o (Python)

%o from itertools import chain, combinations as combs

%o def c(s): return s[0] == s[-1] == 1 and s == s[::-1]

%o ups = list(chain.from_iterable(combs(range(10), r) for r in range(2, 11)))

%o s = set(L[:-1] + R[::-1] for L in ups for R in ups if L[-1] == R[-1])

%o afull = [1] + sorted(int("".join(map(str, t))) for t in s if c(t))

%o print(afull[:40]) # _Michael S. Branicky_, Aug 04 2022

%Y Cf. A002113, A134941, A134951, A134810, A134853, A173071

%K nonn,base,fini,full

%O 1,2

%A _Omar E. Pol_, Feb 09 2010