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A276332 a(n) = number of terms of A001563 needed to sum to n when always choosing the largest term that divides the remaining n, a(0) = 0. 4
0, 1, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 5, 6, 4, 5, 1, 2, 5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 10, 8, 9, 10, 11, 2, 3, 4, 5, 3, 4, 5, 6, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 3, 4, 7, 8, 9, 10, 8, 9, 10, 11, 9, 10, 11, 12, 10, 11, 12, 13, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 10, 8, 9, 5, 6, 9, 10, 11, 12, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

LINKS

Antti Karttunen, Table of n, a(n) for n = 0..4320

FORMULA

a(0) = 0; for n >= 1, a(n) = 1 + a(n-A276330(n)) = 1 + a(A276331(n)).

Other identities and observations:

It seems that a(n) >= A276328(n) for all n.

EXAMPLE

For n=20, the largest term of A001563 that divides 20 is 4 and moreover, 4 is also the largest term of A001563 that divides 16, 12, 8 and 4, thus as 20 = 5*4, a(20) = 5.

For n=21, the largest term of A001563 that divides 21 is A001563(1) = 1, and case for 21 - 1 = 20 is shown above, thus a(21) = 1 + 5 = 6.

PROG

(Scheme, with memoization-macro definec)

(definec (A276332 n) (if (zero? n) n (+ 1 (A276332 (A276331 n)))))

CROSSREFS

Cf. A001563, A276328, A276329, A276330, A276331.

Sequence in context: A053737 A033924 A276328 * A274641 A003315 A194107

Adjacent sequences:  A276329 A276330 A276331 * A276333 A276334 A276335

KEYWORD

nonn

AUTHOR

Antti Karttunen, Aug 30 2016

STATUS

approved

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Last modified October 19 04:40 EDT 2019. Contains 328211 sequences. (Running on oeis4.)