A344823 Numbers m with decimal expansion (d_1, ..., d_k) such that d_i = m ^ i mod 10 for i = 1..k.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 55, 66, 111, 464, 555, 666, 919, 1111, 5555, 6666, 11111, 24862, 39713, 46464, 55555, 66666, 79317, 84268, 91919, 111111, 555555, 666666, 1111111, 4646464, 5555555, 6666666, 9191919, 11111111, 55555555, 66666666, 111111111

Offset

1,3

Comments

This sequence is infinite as it contains d * A002275(k) for any d in {1, 5, 6} and k > 0.

Also contains terms with patterns 4(64)^k, 9(19)^k, 2(4862)^k, 3(9713)^k, 7(9317)^k, 8(4268)^k for k >= 0, where ^ denotes repeated concatenation; all terms have first and last digits the same. - Michael S. Branicky, May 29 2021

Links

Rémy Sigrist, Table of n, a(n) for n = 1..1253

Rémy Sigrist, PARI program for A344823

Example

- 4^1 = 4 mod 10,
- 4^2 = 6 mod 10,
- 4^3 = 4 mod 10,
- so 464 belongs to the sequence.

Prog

(PARI) is(n) = { my (d=digits(n), m=Mod(n, 10)); for (k=1, #d, if (d[k] != m^k, return (0))); return (1) }
(PARI) See Links section.
(Python)
def ok(m):
  d = str(m)
  return all(d[i-1] == str((m**i)%10) for i in range(1, len(d)+1))
print(list(filter(ok, range(10**6)))) # Michael S. Branicky, May 29 2021
(Python)
def auptod(maxdigits):
  alst = [0]
  for k in range(1, maxdigits+1):
    for d1 in range(1, 10):
      d = [(d1**i)%10 for i in range(1, k+1)]
      if d1 == d[-1]: alst.append(int("".join(map(str, d))))
  return alst
print(auptod(9)) # Michael S. Branicky, May 29 2021

Crossrefs

Cf. A002275, A344555, A344749, A344822.

Sequence in context: A303002 A167152 A246008 * A064704 A083136 A369127

Adjacent sequences: A344820 A344821 A344822 * A344824 A344825 A344826

Keyword

nonn,base

Author

Rémy Sigrist, May 29 2021

Status

approved