A371144 The smallest number such that the concatenation of n, a(n), n+1 is divisible by the concatenation of n and n+1.

3, 5, 7, 0, 37, 76, 48, 98, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 560, 571, 582, 593, 604, 615, 626, 637, 648, 361, 670, 681, 692, 703, 714, 725, 736, 747, 758, 769, 780

Offset: 1

Author

Scott R. Shannon, Mar 12 2024

Keywords

• nonn
• base

Comments

For n > 10, when n starts with the digits 1, 2, 3, or 4, then a(n) = 2*n + 1. When n starts with the digits 5, 6, 7, or 8, then a(n) = 11*n + 10 for the vast majority of terms, although some outliers exist e.g., a(749) = 2251. When n starts with the digit 9, the values are somewhat more varied.

The maximum possible value for any term is the concatenation of n+1 and n, see the example for a(6) below. However except for a(6) and a(8), for the terms studied this only occurs four times for every order of magnitude increase in n, namely the four numbers consisting of all 9's except for the final digit of 0, 2, 6, or 8.

The first duplicate term is a(5) = a(18) = 37. There are 234 duplicates in the first 10000 terms.

Examples

			a(1) != 1 as "1"+"1"+"2" = 112 is not divisible by "1"+"2" = 12.
a(1) != 2 as "1"+"1"+"2" = 122 is not divisible by "1"+"2" = 12.
a(1) = 3 as "1"+"3"+"2" = 132 is divisible by "1"+"2" = 12.
a(5) = 37 as "5"+"37"+"6" = 5376 is divisible by "5"+"6" = 56.
a(6) = 76 as "6"+"76"+"7" = 6767 is divisible by "6"+"7" = 67. This is the first time the maximum possible value is required.
		

Links

• Scott R. Shannon, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001704, A215027, A248378.