A134941 - OEIS

%I #47 May 26 2024 18:59:00

%S 1,121,131,141,151,161,171,181,191,1231,1241,1251,1261,1271,1281,1291,

%T 1321,1341,1351,1361,1371,1381,1391,1421,1431,1451,1461,1471,1481,

%U 1491,1521,1531,1541,1561,1571,1581,1591,1621,1631,1641,1651,1671,1681,1691,1721

%N Mountain numbers.

%C For n > 1 the structure of digits represents a mountain. The first digit is 1. The last digit is 1. The first digits are in increasing order. The last digits are in decreasing order. The numbers only have one largest digit. This sequence is finite. The last term is 12345678987654321.

%C The total number of terms is 21846. - _Hans Havermann_, Nov 25 2007

%C A002450(8) + 1 = 21846. - _Reinhard Zumkeller_, May 17 2010

%C From _Reinhard Zumkeller_, May 25 2010: (Start)

%C A178333 is the characteristic function of mountain numbers: A178333(a(n)) = 1;

%C A178334(n) is the number of mountain numbers <= n;

%C A178052 and A178053 give sums of digits and digital roots of mountain numbers;

%C A178051(n) is the peak value of the n-th mountain number. (End)

%H Joshua Zucker, <a href="/A134941/b134941.txt">Table of n, a(n) for n = 1..21846</a> (shows all terms).

%e The A-number of this sequence (A134941) is itself a mountain number:

%e   . . . 9 . .

%e   . . . . . .

%e   . . . . . .

%e   . . . . . .

%e   . . . . . .

%e   . . 4 . 4 .

%e   . 3 . . . .

%e   . . . . . .

%e   1 . . . . 1

%t mountainQ[n_] := MatchQ[ IntegerDigits[n], {1, a___, b_, c___, 1} /; OrderedQ[{1, a, b}, Less] && OrderedQ[ Reverse[{b, c, 1}], Less]]; mountainQ[1] = True; Select[Range[2000], mountainQ] (* _Jean-François Alcover_, Jun 13 2012 *)

%t Prepend[Union @@ ((FromDigits@#&/@Flatten[Table[Join[(k=Prepend[#,1]&/@

%t Subsets[Range[2,#-1]])[[i]], {#}, (Reverse@# & /@k)[[j]]],

%t {i, 2^(# - 2)}, {j, 2^(# - 2)}], 1])&/@Range[9]), 1] (* _Hans Rudolf Widmer_, Apr 30 2024 *)

%o (Haskell)

%o import Data.List (elemIndices)

%o a134941 n = a134941_list !! (n-1)

%o a134941_list = elemIndices 1 a178333_list

%o -- _Reinhard Zumkeller_, Oct 28 2001

%o (Python)

%o from itertools import product

%o def ups():

%o     d = "23456789"

%o     for b in product([0, 1], repeat=8):

%o         yield "1" + "".join(d[i]*b[i] for i in range(8))

%o def downsfrom(apex):

%o     if apex < 3: yield "1"*int(apex==2); return

%o     d = "8765432"[-(apex-2):]

%o     for b in product([0, 1], repeat=len(d)):

%o         yield "".join(d[i]*b[i] for i in range(len(d))) + "1"

%o def A134941(): # return full sequence as a list

%o     mountain_strs = (u+d for u in ups() for d in downsfrom(int(u[-1])))

%o     return sorted(int(ms) for ms in mountain_strs)

%o print(A134941()[:45]) # _Michael S. Branicky_, Dec 31 2021

%Y Cf. A134853, A135417, A134951, A182721.

%Y Cf. A115300, A175044. - _Reinhard Zumkeller_, May 25 2010

%K base,fini,full,nonn

%O 1,2

%A _Omar E. Pol_, Nov 22 2007