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EXAMPLE
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For n = 1, there is only one permutation, p = (1), whence a(1) = 1.
For n = 2, the permutation p = (1,2) yields C(p) = {(1,2), (12)} with M(p) = min {1+2, 12} = 3, while p = (2,1) yields C(p) = {(21)} (no other increasing v can be constructed), with M(p) = 21, whence a(n) = concat(2,1) = 21.
For n = 3, the permutation p = (1,3,2) yields C(p) = {(1,32), (132)} with M(p) = 1+32 = 33, and no other permutation yields a larger minimum (p = (2,3,1) yields the same as 2+31, but comes lexicographically later). Therefore a(3) = 33.
a(4) = M((4,3,2,1)) = 4+321 = 325.
a(5) = M((4,3,5,2,1)) = 43+521 = 564.
a(6) = M((1,5,4,6,3,2)) = 1+54+632 = 687.
a(7) = M((6,5,1,7,4,3,2)) = 6+51+7432 = 7489.
a(8) = M((7,6,5,1,8,4,3,2)) = 7+651+8432 = 9090.
a(9) = M((7,6,8,5,1,9,4,3,2)) = 76+851+9432 = 10359.
a(10) = M((9,8,4,7,1,10,65,3,2)) = 98+471+106532 = 107101.
a(11) = M((1,6,5,11,4,2,9,10,8,7,3)) = 1+65+1142+910873 = 912081.
a(12) = M((4,1,11,2,9,10,12,8,7,6,5,3)) = 41+112+910+1287653 = 1288716.
a(13) = M((9,7,8,6,4,1,2,5,3,13,12,11,10)) = 97+864+1253+13121110 = 13123324.
a(14) = M((4,2,1,11,13,10,9,14,12,8,7,6,5,3)) = 42+111+13109+141287653 = 141300915.
As an example for how to construct this, consider the 7 numbers (14,12,8,7,6,5,3) with concatenation 141287653. If the preceding term is larger than 1412, then this cannot be split up. This is the case for concat(13,10,9) = 13109. The whole cannot be split up as 1310+91412+87653 (last term too small), but to prevent splitting up into 13+109+... the preceding term must be larger than 13. This is true for concat(1,11) (or 11,1), and so on. Restricting the initial choice to a permutation of 7 among 14 narrows the search space compared to considering all permutations of [1..14] and possible increasing concatenations. - M. F. Hasler, Oct 29 2019
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