cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A276328 Digit sum when n is expressed in greedy A001563-base (A276326).

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, 3, 4, 2, 3, 4, 5, 3, 4, 5, 6, 4, 5, 6, 7, 5, 6, 2, 3, 4, 5, 3, 4, 5, 6, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 3, 4, 5, 6, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 10, 8, 9, 5, 6, 7, 8, 6, 7, 1
Offset: 0

Views

Author

Antti Karttunen, Aug 30 2016

Keywords

Comments

a(n) is the number of terms of A001563 needed to sum to n using the greedy algorithm.
This seems to give also the minimal number of terms of A001563 that sum to n (checked empirically up to n=3265920), but it would be nice to know for sure whether this holds for all n.

Examples

			For n=1, the largest term of A001563 <= 1 is A001563(1) = 1, thus a(1) = 1.
For n=2, the largest term of A001563 <= 2 is A001563(1) = 1, thus a(2) = 1 + a(2-1) = 2.
For n=18, the largest term of A001563 <= 18 is A001563(3) = 18, thus a(18) = 1.
For n=20, the largest term of A001563 <= 20 is A001563(3) = 18, thus a(20) = 1 + a(20-18) = 3.
For n=36, the largest term of A001563 <= 36 is A001563(3) = 18, thus a(36) = 1 + a(18) = 2.

Links

Crossrefs

Cf. A001563, A258199.
Cf. A000120, A276329, A276330, A276332, A276333, A276335.
Cf. A276091 (gives all n for which a(n) = A276337(n)).
Cf. also A007895, A034968, A265744, A265745 for similar sequences.

Programs

  • Mathematica
    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[Total@ f@ n, {n, 120}] (* Michael De Vlieger, Aug 31 2016 *)

Formula

a(0) = 0; for n >= 1, a(n) = 1 + a(n-A258199(n)).
a(0) = 0; for n >= 1, a(n) = A276333(n) + a(A276335(n)).
Other identities and observations. For all n >= 0:
a(A276091(n)) = A000120(n).
a(n) >= A276337(n).
It also seems that a(n) <= A276332(n) for all n.

A276329 a(n) = largest k such that A001563(k) divides n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2
Offset: 1

Views

Author

Antti Karttunen, Aug 30 2016

Keywords

Links

Crossrefs

Cf. A001563, A258198, A276330.

Programs

  • Mathematica
    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]]]; Table[Length@ TakeWhile[Reverse@ f@ n, # == 0 &] + 1, {n, 120}] (* Michael De Vlieger, Aug 31 2016 *)
  • Scheme
    (define (A276329 n) (let loop ((i (A258198 n))) (cond ((zero? i) 1) ((zero? (modulo n (A001563 i))) i) (else (loop (- i 1))))))

A276331 a(0) = 0 and for n >= 1, a(n) = n minus the largest term of A001563 that divides n.

Original entry on oeis.org

0, 0, 1, 2, 0, 4, 5, 6, 4, 8, 9, 10, 8, 12, 13, 14, 12, 16, 0, 18, 16, 20, 21, 22, 20, 24, 25, 26, 24, 28, 29, 30, 28, 32, 33, 34, 18, 36, 37, 38, 36, 40, 41, 42, 40, 44, 45, 46, 44, 48, 49, 50, 48, 52, 36, 54, 52, 56, 57, 58, 56, 60, 61, 62, 60, 64, 65, 66, 64, 68, 69, 70, 54, 72, 73, 74, 72, 76, 77, 78, 76, 80, 81, 82, 80
Offset: 0

Views

Author

Antti Karttunen, Aug 30 2016

Keywords

Links

Crossrefs

Cf. A001563, A276330, A276332.

Programs

  • Mathematica
    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[n - # #! &[Length@ TakeWhile[Reverse@ f@ n, # == 0 &] + 1], {n, 84}] (* Michael De Vlieger, Aug 31 2016 *)
  • Scheme
    (define (A276331 n) (- n (A276330 n)))

Formula

a(n) = n - A276330(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

Views

Author

Antti Karttunen, Aug 30 2016

Keywords

Examples

			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.

Links

Crossrefs

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

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.
Showing 1-4 of 4 results.

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