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PROJECT EULER · #0506

Clock Sequence

Statement only · SolvedOriginal problem ↗

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

Let S(n) be v1+v2++vn. For example, S(11)=36120, and S(1000)mod123454321=18232686.

Find S(1014)mod123454321.

Write-up coming later

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