A316935 a(n)=1; for n > 1, a(n) is the smallest number > n such that the concatenation of all terms from a(1) through a(n) is divisible by the concatenation of the integers 1 through n.

1, 20, 54, 946, 8180, 93504, 878732, 6841732, 102829509, 19305995230, 1822646098871, 35208071275344, 8691465582891615, 2131922062844429082, 190058192685217102545, 9285111636083665154512, 565278857209893562444229, 49237824030642874847017458, 15301141018410914663693576388

Offset: 1

Author

Christopher Hohl, Dec 12 2018

Keywords

• nonn
• easy
• base
• nice

Examples

			Let I(n) be the concatenation of the integers 1 through n, and let T(n) be the concatenation of all terms from a(1) through a(n). Then a(n) is the least number which, when appended to T(n-1) -- with leading zeros disallowed -- forms a number T(n) that is strictly greater than I(n) and is a multiple of I(n).
a(1)=1, so T(1)=1. I(2)=12, so a(2)=20 because a(2) is the least number which, when appended to T(1), creates a number T(2) that is greater than I(2)=12 and evenly divisible by it, and the least number greater than 12 that begins with a 1 and is divisible by 12 is 120.
a(2) is not 08 because leading zeros are not allowed.
a(3)=54 because at a(3), I(3)=123, and T(2)=120, which means that T(3) must be 12054, since the least number greater than 123 that is divisible by it but has leading digits 120 is 12054.
From _David A. Corneth_, Dec 18 2018: (Start)
Let concat(v) be the concatenation of the elements of vector v. To find a(n) for n >= 3, we find the concatenation of the first n-1 terms. For n = 3, that's 120. We also find the concatenation of the first n positive integers, in this case, 123.
Concatenating 1 gives 1201. 1201 mod 123 = 94 < 123 - 8 so concatenating a 1-digit number without leading zeros doesn't work.
We carry on and compute 12010 mod 123 = 79 >= 123 - 89 so concatenating a 2-digit number without leading zeros works and we stop to find a(3) = 10 + 123 - 79 = 54. (End)
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..369

Crossrefs

Cf. A007908 (concatenation of the numbers from 1 to n).

Programs

Mathematica

getRes[m_,k_,e_] := Module[{}, r = k-Mod[m*10^e,k]; If[r < 10^e && m*10^e+r != k, r += k*(1 + Floor[(10^(e - 1) - r)/k]); If[r>10^e, r=-1], r=-1]; r]; f[m_,k_] := Module[{e=1}, While[(r=getRes[m,k,e])<0, e++]; r]; t[n_] := FromDigits[ Flatten[IntegerDigits[ Range[n]]]]; a[1]=1; a[n_] := a[n] = f[FromDigits[ Flatten[IntegerDigits[Array[a, n-1]]]], t[n]]; Array[a, 20] (* Amiram Eldar, Dec 13 2018 *)

PARI

first(n) = {my(res = [1,20]); for(i = 3, n, res = concat(res, [nxt(i, res)])); res}
nxt(n, v) = {my(div = concatnums([1..n]), start = concatnums(v)); start = 10*start+1; f = 1; subt = 8; c = start % div; while(c < div - subt, start *= 10; f*=10; subt = 10*subt + 9; c = start % div); f + div - c}
concatnums(v) = {my(res = v[1]); for(i = 2, #v, res *= 10^#Str(v[i]); res += v[i]); res} \\ David A. Corneth, Dec 18 2018

Extensions

a(8)-a(18) from Jon E. Schoenfield, Dec 13 2018

a(19) from David A. Corneth, Dec 18 2018