A283512
Last block of length-n to appear for the first time in the Kolakoski sequence K (A000002).
Original entry on oeis.org
2, 11, 212, 2121, 21211, 121121, 2212212, 22122121, 221221211, 2212212112, 22122121122, 112212211211, 1122122112112, 11221221121121, 112212211211212, 1122122112112122, 11211212212211211, 112212211211212211, 1211212212112212212, 12112122121122122121, 121121221211221221211
Offset: 1
For n = 4 the last block of length 4 to appear is 2121.
A379017
a(n) is the number of distinct sums s(m) + s(m+1) + ... + s(m+n-1), where s = A000002, and m >= 1.
Original entry on oeis.org
2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 4, 3, 4, 5, 4, 3, 4, 3, 4, 5, 4, 5, 6, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 6, 5, 6, 5, 4, 5, 6, 5, 6, 7, 6, 5, 6, 5, 6, 7, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 7, 6, 5, 6, 5, 6, 7, 6, 7, 6, 5, 6, 7, 6, 7, 8, 7, 8, 7, 6, 7
Offset: 1
Starting with s = (1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, ...) we form a shifted partial sum array:
(row 1) = (1,2,2,1,1,2,1,2,2,...)
(row 2) = (s(1)+s(2), s(2)+s(3), s(3)+s(4), ...) = (3,4,3,2,3,3,3,4,...) = A333229
(row 3) = (s(1)+s(2)+s(3), s(2)+s(3)+s(4), s(3)+s(4)+s(5), ...) = (5,5,4,4,4,5,5,5,5,5,5,4,...)
The number of distinct numbers in (row 3) is 2, so a(3) = 2.
The first 12 rows of the shifted partial sum array: (1, 2), (2, 3, 4), (4, 5), (5, 6, 7), (6, 7, 8, 9), (8, 9, 10), (9, 10, 11, 12), (11, 12, 13), (13, 14), (14, 15, 16), (15, 16, 17, 18), (17, 18, 19). These rows illustrate that fact that the integers in each row are consecutive.
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s = Prepend[Nest[Flatten[Partition[#, 2] /. {{2, 2} -> {2, 2, 1, 1}, {2, 1} -> {2, 2, 1}, {1, 2} -> {2, 1, 1}, {1, 1} -> {2, 1}}] &, {2, 2}, 24], 1]; (* A000002 *)
Length[s]
r[1] = s;
r[n_] := r[n] = Rest[r[n - 1]];
c[n_] := c[n] = Take[r[n], 1000];
sum[n_] := Sum[c[k], {k, 1, n}];
t = Table[Union[sum[n]], {n, 1, 100}]
Map[Length, t]
Showing 1-2 of 2 results.