Problem 506

Clock sequence

Consider the infinite repeating sequence of digits:

1234321234321234321...

Amazingly, you can break this sequence of digits into a sequence of integers such that the sum of the digits in the n’th value is n.

The sequence goes as follows:

1, 2, 3, 4, 32, 123, 43, 2123, 432, 1234, 32123, ...

Let v_n be the n’th value in this sequence. For example, v₂ = 2, v₅ = 32 and v₁₁ = 32123.

Let S(n) be v₁ + v₂ + … + v_n. For example, S(11) = 36120, and S(1000) mod 123454321 = 18232686.

Find S(10¹⁴) mod 123454321.

钟摆序列
考虑下面这个无限重复的数字序列：

1234321234321234321...

神奇的是，你可以将这个序列分解成一个整数的序列，使得第n个整数的各位数字之和恰好是n。

这个整数序列如下所示：

1, 2, 3, 4, 32, 123, 43, 2123, 432, 1234, 32123, ...

记v_n是这个整数序列的第n个整数，例如，v₂ = 2，v₅ = 32，以及v₁₁ = 32123。

记S(n)为v₁ + v₂ + … + v_n。例如，S(11) = 36120，以及S(1000) mod 123454321 = 18232686。

求S(10¹⁴) mod 123454321。