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%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
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