A029479 Numbers k that divide the (left) concatenation of all numbers <= k written in base 10 (most significant digit on left).

1, 3, 9, 19, 27, 41, 103, 147, 189, 441, 567, 711, 6759, 15353, 24441, 59823, 209903, 1430217, 2848851, 2969973, 13358067, 146247471, 289542573, 1891846557, 2388085659, 4489093899, 5345125899, 5455876131, 9843149241

Offset: 1

Author

Olivier Gérard

Keywords

• nonn
• base
• more

Comments

No other terms below 10^10.

Examples

			19 is a term since 19181716151413121110987654321 is divisible by 19.
		

Links

• Index entries for related sequences

Crossrefs

Cf. A029447-A029470, A029471-A029494, A029495-A029518, A029519-A029542, A061931-A061954, A061955-A061978.

Programs

Mathematica

b = 10; c = {}; Select[Range[10^4], Divisible[FromDigits[c = Join[IntegerDigits[#, b], c], b], #] &] (* Robert Price, Mar 12 2020 *)
Select[Range[134*10^5],Divisible[FromDigits[Flatten[IntegerDigits/@Range[#,1,-1]]],#]&] (* Harvey P. Dale, Oct 09 2022 *)

Python

def concat_mod(base, k, mod):
  total, offset, digits, n1 = 0, 0, 1, 1
  while n1 <= k:
    n2, p = min(n1*base-1, k), n1*base
    # Compute ((p-1)*n2-1)*p**(n2-n1+1)-(n1-1)*p+n1 divided by (p-1)**2.
    # Since (a//b)%mod == (a%(b*mod))//b, compute the numerator mod (p-1)**2*mod.
    tmp = pow(p,n2-n1+1,(p-1)**2*mod)
    tmp = ((p-1)*n2-1)*tmp-(n1-1)*p+n1
    tmp = (tmp%((p-1)**2*mod))//(p-1)**2
    total += tmp*pow(base,offset,mod)
    offset, digits, n1 = offset+digits*(n2-n1+1), digits+1, p
  return total%mod
for k in range(1,10**10):
  if concat_mod(10, k, k) == 0: print(k) # Jason Yuen, Jan 14 2024

Extensions

6759 from Andrew Gacek (andrew(AT)dgi.net), Feb 20 2000

More terms from Larry Reeves (larryr(AT)acm.org), May 24 2001

Edited and updated by Larry Reeves (larryr(AT)acm.org), Apr 12 2002

a(18)-a(21) from Max Alekseyev, May 15 2011

a(22)-a(29) from Jason Yuen, Jan 14 2024