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(PARI) A329623(n)=abs(n-A053393A053392(n)) \\ M. F. Hasler, Dec 02 2019
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a(9) = 9 as A053392(09) = 0 and | 9 - 0 | = 9.
(PARI) A329623(n)=abs(n-A053393(n)) \\ M. F. Hasler, Dec 02 2019
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As A040115 forms the basis of an iterative sequence leading to A329200 and A329201, this sequence forms the basis of a similar sequence A329624. As the concatenation of the digit sum can lead to a value larger than the original term we must take the absolute value of the difference to ensure subsequent terms are always positive. The largest value in the first 10000 terms is a(9991) = 171819.
Scott R. Shannon, <a href="/A329623/b329623.txt">Table of n, a(n) for n = 0..10000</a>
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