A276331 -id:A276331 - cp's OEIS frontend

A276335 Discard the most significant digit when n is expressed in greedy A001563-base (A276326), then convert back to decimal: a(n) = n - A276334(n).

0, 0, 0, 0, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 0, 1, 2, 3, 4, 5, 0

Offset: 0

Antti Karttunen, Aug 30 2016

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

Cf. A001563, A276326, A276328, A276334, A276331, A276337.

Cf. also A257687.

{0}~Join~Table[n - # Floor[n/#] &@(# #!) &@ NestWhile[# + 1 &, 0, # #! <= n &[# + 1] &], {n, 96}] (* Michael De Vlieger, Aug 31 2016 *)

Scheme
```
(define (A276335 n) (- n (A276334 n)))
```

a(n) = n - A276334(n).

A276330 a(n) = largest term of A001563 that divides n, a(0) = 0.

Original entry on oeis.org

0, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 18, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 18, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 18, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 18, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 18, 1, 4, 1, 1, 1, 96

Offset: 0

Antti Karttunen, Aug 30 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..4319

Cf. A001563, A258199, A276328, A276329, A276331.

f[n_] := Block[{a = {{0, n}}}, Do[AppendTo[a, {First@ #, Last@ #} &@ QuotientRemainder[a[[-1, -1]], (# #!) &[# - i]]], {i, 0, # - 1}] &@ NestWhile[# + 1 &, 0, (# #!) &[# + 1] <= n &]; Rest[a][[All, 1]]]; {0}~Join~Table[# #! &[Length@ TakeWhile[Reverse@ f@ n, # == 0 &] + 1], {n, 120}] (* Michael De Vlieger, Aug 31 2016 *)

(define (A276330 n) (if (zero? n) n (A001563 (A276329 n))))

a(0) = 0; for n >= 1, a(n) = A001563(A276329(n)).
Other identities. For all n >= 0:
A276331(n) = n - a(n).

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.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Aug 30 2016

nonn

			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.

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

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

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.

cp's OEIS Frontend

A276335 Discard the most significant digit when n is expressed in greedy A001563-base (A276326), then convert back to decimal: a(n) = n - A276334(n).

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A276330 a(n) = largest term of A001563 that divides n, a(0) = 0.

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Mathematica

Scheme

Formula

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.

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