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.
Contribution from _Reinhard Zumkeller_, May 25 2010: (Start) a(1) = A178334(9) = 1; a(2) = A178334(99) - A178334(9) = 1 - 1 = 0; a(3) = A178334(999) - A178334(99) = 9 - 1 = 8; a(4) = A178334(9999) - A178334(999) = 65 - 9 = 56. (End)
n=552, A134941(552)=134941 (example from A134941): . . level 9: . . . . . . . . X . . . . . . . . level 8: . . . . . . . . X . . . . . . . . level 7: . . . . . . . . X . . . . . . . . level 6: . . . . . . . . X . . . . . . . . level 5: . . . . . . . . X . . . . . . . . level 4: . . . . . . X . X . X . . . . . . level 3: . . . . X . X . X . X . . . . . . level 2: . . . . X . X . X . X . . . . . . level 1: . . X . X . X . X . X . X . . ---------- -------------------------------- . . . . a(552) = 1 + 3 + 4 + 9 + 4 + 1 = 22.
A134941(552)=134941 --> 1+3+4+9+4+1=22 --> a(552)=2+2=4.
A134941(552)=134941 --> a(552) = MAX{1,3,4,9} = 9.
The A-number of this sequence (A134951) is a mountain prime because 134951 is a mountain number and it is also a prime number. . . . 9 . . . . . . . . . . . . . . . . . . . . . . . . 5 . . . 4 . . . . 3 . . . . . . . . . . 1 . . . . 1
Illustration using the final term of this sequence: . . . . . . . . 9 . . . . . . . . . . . . . . . 8 . 8 . . . . . . . . . . . . . 7 . . . 7 . . . . . . . . . . . 6 . . . . . 6 . . . . . . . . . 5 . . . . . . . 5 . . . . . . . 4 . . . . . . . . . 4 . . . . . 3 . . . . . . . . . . . 3 . . . 2 . . . . . . . . . . . . . 2 . 1 . . . . . . . . . . . . . . . 1
ups = Flatten[Table[Range[i, j - 1], {i, 1, 9}, {j, i + 1, 10}], 1];afull = Sort[ Map[ToExpression@StringJoin@Map[ToString, #[[;; -2]] ~Join~ Reverse[#]] &, ups]];afull (* James C. McMahon, Apr 11 2025 *)
ups = [tuple(range(i, j)) for i in range(1, 10) for j in range(i+1, 11)]
afull = sorted(int("".join(map(str, u[:-1] + u[::-1]))) for u in ups)
print(afull) # Michael S. Branicky, Aug 02 2022
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
The number of this sequence (A134853) is a generalized mountain number. . . . . . . . . . 8 . . . . . . . . . . . . . . . . . . 5 . . . 4 . . . . 3 . . . 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[-1] == R[-1])
afull = list(range(1, 10))
afull += sorted(int("".join(map(str, t))) for t in s if t[0] != 0)
print(afull[:60]) # 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
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