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.
Illustration using 654356: 6 . . . . 6 . 5 . . 5 . . . 4 . . . . . . 3 . .
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
Illustration of 751367 as a Canyon prime: . . . . . . . . . . . . 7 . . . . 7 . . . . 6 . . 5 . . . . . . . . . . . . . 3 . . . . . . . . . . 1 . . . . . . . . .
S = {}; c = 1;
For[n = 1, n <= 9, n++,
L = 2 n - 1;
d = Join[Reverse[Range[1, n - 1]], Range[0, n - 1]];
If[Mod[n, 2] != 0 && n != 5,
For[j = 1, j < 2^L, j++,
Dig = d[[Map[#[[1]] &, Position[IntegerDigits[j, 2, L], 1]]]];
min = Min[Dig];
If[Length[Position[Dig, min]] == 1,
p = FromDigits[Join[{n}, Dig, {n}]];
If[PrimeQ[p], S = Append[S, p]];
];
];
];
]; (* Kellen Myers, Jan 18 2011 *)
from sympy import isprime
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(filter(isprime, (int("".join(map(str, t))) for t in s if t[0] == t[-1])))
print(afull[:44]) # Michael S. Branicky, Jan 16 2023
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 using 6543333346: 6 . . . . . . . . 6 . 5 . . . . . . . . . . 4 . . . . . 4 . . . . 3 3 3 3 3 . .
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
Illustration using a(32)=7654567: 7 . . . . . 7 . 6 . . . 6 . . . 5 . 5 . . . . . 4 . . .
Flatten[Table[FromDigits/@(Join[Reverse[Rest[#]],#]&/@Partition[ Range[ 0,9],n,1]),{n,10}]] (* Harvey P. Dale, Dec 27 2018 *)
ups = [tuple(range(i, j)) for i in range(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
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