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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

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Author

Clark Kimberling, Dec 16 2024

Keywords

Examples

			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.
		

Crossrefs

Cf. A000002, A007782 (subword complexity), A283511, A333229, A376677.

Programs

  • Mathematica
    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]

Formula

|a(n+1)-a(n)| = 1 for every n.

Extensions

More terms from Jinyuan Wang, Jan 22 2025