A029455 Numbers k that divide the (right) concatenation of all numbers <= k written in base 10 (most significant digit on left).

1, 2, 3, 5, 6, 9, 10, 12, 15, 18, 20, 25, 27, 30, 36, 45, 50, 54, 60, 69, 75, 90, 100, 108, 120, 125, 135, 150, 162, 180, 200, 216, 225, 248, 250, 270, 300, 324, 360, 375, 405, 450, 470, 500, 540, 558, 600, 648, 675, 710, 750, 810, 900, 1000, 1053, 1080, 1116

Offset: 1

Author

Olivier Gérard

Keywords

• nonn
• base

Comments

Numbers k such that k divides A007908(k).

Examples

			k = 13 is not a term since 12345678910111213 is not divisible by 13.
		

Links

• Jason Yuen, Table of n, a(n) for n = 1..1195 (terms 1..236 from M. F. Hasler, terms 237..637 from Chai Wah Wu)

Crossrefs

Cf. A007908.

Cf. A029447-A029470, A029471-A029494, A029495-A029518, A029519-A029542, A061931-A061954, A061955-A061978, A332580.

See A171785 for numbers that divide the concatenation of a(1) through a(n).

Programs

Mathematica

b = 10; c = {}; Select[Range[10^5], Divisible[FromDigits[c = Join[c, IntegerDigits[#, b]], b], #] &] (* Robert Price, Mar 11 2020 *)
Select[Range[1200],Divisible[FromDigits[Flatten[IntegerDigits/@Range[#]]],#]&] (* Harvey P. Dale, Dec 31 2020 *)
nxt[{rc_,n_}]:={rc*10^IntegerLength[n+1]+n+1,n+1}; Select[NestList[nxt,{1,1},1200],Mod[#[[1]],#[[2]]]==0&][[;;,2]] (* Harvey P. Dale, Sep 26 2023 *)

PARI

c=0;for(d=1,1e9,for(n=d,-1+d*=10,(c=c*d+n)%n || print1(n","));d--) \\ M. F. Hasler, Sep 11 2011

Python

A029455_list, r = [], 0
for n in range(1,10**4+1):
    r = r*10**len(str(n))+n
    if not (r % n):
        A029455_list.append(n) # Chai Wah Wu, Nov 05 2014

Python

def concat_mod(base, k, mod):
  total, digits, n1 = 0, 1, 1
  while n1 <= k:
    n2, p = min(n1*base-1, k), n1*base
    # Compute ((p-1)*n1+1)*p**(n2-n1+1)-(n2+1)*p+n2 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)*n1+1)*tmp-(n2+1)*p+n2
    tmp = (tmp%((p-1)**2*mod))//(p-1)**2
    total = (total*pow(p, n2-n1+1, mod)+tmp)%mod
    digits, n1 = digits+1, p
  return total
for k in range(1, 10**10+1):
  if concat_mod(10, k, k) == 0: print(k) # Jason Yuen, Jan 27 2024