This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(18) = 1 because there is exactly one generalized mountain number with 18 digits: 123456789876543210
The A-number of this sequence (A134941) is itself a mountain number: . . . 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 . 4 . . 3 . . . . . . . . . . 1 . . . . 1
import Data.List (elemIndices) a134941 n = a134941_list !! (n-1) a134941_list = elemIndices 1 a178333_list -- Reinhard Zumkeller, Oct 28 2001
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 *)
Prepend[Union @@ ((FromDigits@#&/@Flatten[Table[Join[(k=Prepend[#,1]&/@
Subsets[Range[2,#-1]])[[i]], {#}, (Reverse@# & /@k)[[j]]],
{i, 2^(# - 2)}, {j, 2^(# - 2)}], 1])&/@Range[9]), 1] (* Hans Rudolf Widmer, Apr 30 2024 *)
from itertools import product
def ups():
d = "23456789"
for b in product([0, 1], repeat=8):
yield "1" + "".join(d[i]*b[i] for i in range(8))
def downsfrom(apex):
if apex < 3: yield "1"*int(apex==2); return
d = "8765432"[-(apex-2):]
for b in product([0, 1], repeat=len(d)):
yield "".join(d[i]*b[i] for i in range(len(d))) + "1"
def A134941(): # return full sequence as a list
mountain_strs = (u+d for u in ups() for d in downsfrom(int(u[-1])))
return sorted(int(ms) for ms in mountain_strs)
print(A134941()[:45]) # Michael S. Branicky, Dec 31 2021
Illustration using the final term of this sequence: 9 . . . . . . . . . . . . . . . . . 9 . 8 . . . . . . . . . . . . . . . 8 . . . 7 . . . . . . . . . . . . . 7 . . . . . 6 . . . . . . . . . . . 6 . . . . . . . 5 . . . . . . . . . 5 . . . . . . . . . 4 . . . . . . . 4 . . . . . . . . . . . 3 . . . . . 3 . . . . . . . . . . . . . 2 . . . 2 . . . . . . . . . . . . . . . 1 . 1 . . . . . . . . . . . . . . . . . 0 . . . . . . . . .
ups = list(tuple(range(i, j)) for i in range(9) for j in range(i+2, 11))
s = set(L[::-1] + R[1:] for L in ups for R in ups if L[0] == R[0])
s |= set(L[:-1] + R[::-1] for L in ups for R in ups if L[-1] == R[-1])
afull = sorted(int("".join(map(str, t))) for t in s if t[0] != 0)
print(afull[:47]) # Michael S. Branicky, Aug 02 2022
Illustration using the final term of this sequence: 9 . . . . . . . . . . . . . . 8 . . . . . . . . . . . 8 . . 7 . . . . . . . . . . . . . . 6 . . . . . . . . 6 . . . . . 5 . . . . . . . . . . . . . . 4 . . . . . 4 . . . . . . . . 3 . . . . . . . . . . . . . . 2 . . 2 . . . . . . . . . . . 1 . . . . . . . . . . . . . . 0 . . . .
progressions = set(tuple(range(i, j+1, d)) for i in range(10) for d in range(1, 10-i) for j in range(i+d, 10))
s = set()
for left in progressions:
for right in progressions:
dl, dr = left[1] - left[0], right[1] - right[0]
if dl + dr > 2:
if left[-1] == right[-1]: s.add(left[:-1] + right[::-1])
if left[0] == right[0]: s.add(left[::-1] + right[1:])
afull = sorted(int("".join(map(str, t))) for t in s if t[0] != 0)
print(afull[:53]) # Michael S. Branicky, Aug 02 2022
Illustration of the number 9753111: 9 . . . . . . . . . . . . . . 7 . . . . . . . . . . . . . . 5 . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . 1 1 1 . . . . . . .
from itertools import count, islice
def ok3(n):
if n < 100: return False
d = list(map(int, str(n)))
m1, m2 = (d[1]-d[0], d[-1]-d[-2])
return len({m1, m2}) == 2 and m1*m2 >= 0
def agen():
seeds = [k for k in range(100, 1000) if ok3(k)]
for digits in count(4):
yield from sorted(seeds)
new, pow10 = set(), 10**(digits-1)
for q in seeds:
d = list(map(int, str(q)))
e1, e2 = d[0] - (d[1]-d[0]), d[-1] + (d[-1]-d[-2])
if 9 >= e1 > 0: new.add(e1*pow10 + q)
if 9 >= e2 >= 0: new.add(10*q + e2)
seeds = new
print(list(islice(agen(), 54))) # Michael S. Branicky, Aug 03 2022
The acute-angled number 12342 (see A135601): . . . . . . . . 4 . . . 3 . . . 2 . . 2 1 . . . . The right-angled number 12343 (see A135602): . . . . . . . . 4 . . . 3 . 3 . 2 . . . 1 . . . . The obtuse-angled number 12344 (see A135603): . . . . . . . . 4 4 . . 3 . . . 2 . . . 1 . . . . The straight-angled (or straight-line) number 12345 (see A135643): . . . . 5 . . . 4 . . . 3 . . . 2 . . . 1 . . . .
\\ See PARI link. David A. Corneth, Aug 02 2022
from itertools import count, islice
def agen():
seeds = [k for k in range(100, 1000)]
for digits in count(4):
yield from sorted(seeds)
new, pow10 = set(), 10**(digits-1)
for q in seeds:
d = list(map(int, str(q)))
e1, e2 = d[0] - (d[1]-d[0]), d[-1] + (d[-1]-d[-2])
if 9 >= e1 > 0: new.add(e1*pow10 + q)
if 9 >= e2 >= 0: new.add(10*q + e2)
seeds = new
print(list(islice(agen(), 50))) # Michael S. Branicky, Aug 03 2022
Illustration of 136973 as a generalized mountain prime: . . . 9 . . . . . . . . . . . . 7 . . . 6 . . . . . . . . . . . . . . . . 3 . . . 3 . . . . . . 1 . . . . .
13731 is in the sequence because it is a palindrome (A002113) and it is also a mountain number (A134941). . . . . . . . . . . . . 7 . . . . . . . . . . . . . . . . . . 3 . 3 . . . . . . 1 . . . 1
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))
print(afull[:40]) # Michael S. Branicky, Aug 04 2022
Illustration of 751378 as a generalized canyon number: . . . . . . . . . . . 8 7 . . . 7 . . . . . . . . 5 . . . . . . . . . . . . . 3 . . . . . . . . . . 1 . . . . . . . . .
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
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