A344822 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, 505, 50505, 246802, 482604, 628406, 864208, 5050505, 505050505, 12345678901, 24680246802, 36925814703, 48260482604, 50505050505, 62840628406, 74185296307, 86420864208, 98765432109, 5050505050505, 505050505050505, 2468024680246802

Offset: 1

Author

Rémy Sigrist, May 29 2021

Keywords

• nonn
• base

Comments

This sequence is infinite as it contains 5 * A094028(k) for any k > 0.

Also contains terms with patterns 2(46802)^k, 4(82604)^k, 6(28406)^k, 8(64208)^k, 1(2345678901)^k, 3(6925814703)^k, 7(4185296307)^k, 9(8765432109)^k for k >= 0, where ^ denotes repeated concatenation; all terms have first and last digits the same. - Michael S. Branicky, May 29 2021

Examples

			- 4 * 1 = 4 mod 10,
- 4 * 2 = 8 mod 10,
- 4 * 3 = 2 mod 10,
- 4 * 4 = 6 mod 10,
- 4 * 5 = 0 mod 10,
- 4 * 6 = 4 mod 10,
so 482604 is a term.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..426
• Rémy Sigrist, PARI program for A344822

Crossrefs

Cf. A094028, A344748, A344823.

Programs

PARI

is(n) = { my (d=digits(n)); for (k=1, #d, if (d[k] != (n*k)%10, 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(16)) # Michael S. Branicky, May 29 2021