A244232 Sum of "digit values" in Semigreedy Catalan Representation of n, A244159.

0, 1, 1, 2, 3, 1, 2, 2, 3, 4, 4, 5, 6, 4, 1, 2, 2, 3, 4, 2, 3, 3, 4, 5, 5, 6, 7, 5, 6, 6, 7, 8, 8, 9, 10, 8, 5, 6, 6, 7, 8, 6, 1, 2, 2, 3, 4, 2, 3, 3, 4, 5, 5, 6, 7, 5, 2, 3, 3, 4, 5, 3, 4, 4, 5, 6, 6, 7, 8, 6, 7, 7, 8, 9, 9, 10, 11, 9, 6, 7, 7, 8, 9, 7, 8, 8, 9, 10, 10, 11, 12, 10, 11, 11, 12, 13, 13, 14, 15, 13, 10, 11, 11, 12, 13, 11, 6, 7, 7, 8, 9, 7, 8, 8, 9, 10, 10, 11, 12, 10, 7, 8, 8, 9, 10, 8, 9, 9, 10, 11, 11, 12, 1

Offset: 0

Author

Antti Karttunen, Jun 25 2014

Keywords

• nonn

Comments

Note that a(33604) = A000217(10) = 55 because the sum is computed from the underlying list (vector) of numbers, and thus is not subject to any corruption by decimal representation as A244159 itself is.

Equivalent description: partition n "greedily" as terms of A197433, i.e. n = A197433(i) + A197433(j) + ... + A197433(k), always using the largest term of A197433 that still "fits in" (i.e. is <= n remaining). Then a(n) = A000120(i) + A000120(j) + ... + A000120(k).

Examples

			For n=18, using the alternative description, we see that it is partitioned  into the terms of A197433 as a greedy sum A197433(11) + A197433(1) = 17 + 1. Thus a(18) = A000120(11) + A000120(1) = 3+1 = 4.
For n=128, we see that is likewise represented as A197433(31) + A197433(31) = 64 + 64. Thus a(128) = 2*A000120(31) = 10.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..16796

Crossrefs

Cf. A000108, A000120, A176137, A197433, A244230, A244159, A244231, A244233, A244234, A014420, A236855.

Formula

If A176137(n) = 1, a(n) = A000120(A244230(n)), otherwise a(n) = A000120(A244230(n)-1) + a(n-A197433(A244230(n)-1)).

For all n, a(A000108(n)) = 1. [And moreover, Catalan numbers, A000108, give all such k that a(k) = 1].

For all n, a(A014138(n)) = n and a(A014143(n)) = A000217(n+1).