A257905 -id:A257905 - cp's OEIS frontend

A256283 Smallest m such that A257905(m) = n.

1, 2, 4, 3, 7, 5, 9, 11, 14, 8, 13, 6, 17, 10, 19, 12, 21, 15, 23, 25, 27, 29, 31, 33, 38, 18, 36, 43, 41, 20, 35, 45, 47, 22, 51, 16, 53, 24, 49, 26, 40, 28, 50, 30, 57, 32, 59, 34, 55, 39, 65, 61, 63, 37, 67, 44, 71, 42, 75, 69, 77, 73, 79, 46, 81, 48, 85

Offset: 0

Reinhard Zumkeller, Jun 03 2015

nonn

Conjectured inverse of A257905.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A257905.

import Data.List (elemIndex); import Data.Maybe (fromJust)
a256283 = (+ 1) . fromJust . (`elemIndex` a257905_list)

{a, f} = {{1}, {0}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]];
AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {1000}]; a;
Flatten[Table[Position[a, n], {n, 1, 150}]] (* Peter J. C. Moses, May 14 2015 *)

A257883 Sequence (a(n)) generated by Algorithm (in Comments) with a(1) = 0 and d(1) = 0.

Original entry on oeis.org

0, 1, 3, 2, 5, 9, 4, 10, 6, 11, 8, 15, 7, 16, 14, 22, 12, 23, 17, 27, 13, 25, 18, 31, 19, 33, 20, 35, 24, 40, 21, 38, 29, 47, 26, 45, 28, 48, 30, 51, 36, 58, 32, 55, 39, 63, 34, 59, 37, 64, 41, 67, 42, 70, 43, 72, 44, 74, 50, 81, 46, 78, 111, 49, 83, 52, 87

Offset: 1

Clark Kimberling, May 13 2015

Algorithm: For k >= 1, let A(k) = {a(1), ..., a(k)} and D(k) = {d(1), ..., d(k)}. Begin with k = 1 and nonnegative integers a(1) and d(1). Let h be the least integer > -a(k) such that h is not in D(k) and a(k) + h is not in A(k). Let a(k+1) = a(k) + h and d(k+1) = h. Replace k by k+1 and repeat inductively.
Conjecture: if a(1) is a nonnegative integer and d(1) is an integer, then (a(n)) is a permutation of the nonnegative integers (if a(1) = 0) or a permutation of the positive integers (if a(1) > 0). Moreover, (d(n)) is a permutation of the integers if d(1) = 0, or of the nonzero integers if d(1) > 0.
Guide to related sequences:
a(1)   d(1)    (a(n))       (d(n))
0      0      A257883      A175499 except for initial terms
0      1      A257884      A175499
0      2      A257885      A257902
0      3      A257903      A257904
1      0      A175498      A175499 except for first term
1      1      A257905      A175499
2      0      A257908      A257909
2      1      A257910      A257909 except for initial terms
2      2      A257911      A257912

			a(1) = 0, d(1) = 0;
a(2) = 1, d(2) = 1;
a(3) = 3, d(3) = 2;
a(4) = 2, d(4) = -1.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A081145, A175498, A257705.

a[1] = 0; d[1] = 0; k = 1; z = 10000; zz = 120;
A[k_] := Table[a[i], {i, 1, k}]; diff[k_] := Table[d[i], {i, 1, k}];
c[k_] := Complement[Range[-z, z], diff[k]];
T[k_] := -a[k] + Complement[Range[z], A[k]]
Table[{h = Min[Intersection[c[k], T[k]]], a[k + 1] = a[k] + h,
   d[k + 1] = h, k = k + 1}, {i, 1, zz}];
u = Table[a[k], {k, 1, zz}]  (* A257883, = -1 + A175498 *)
Table[d[k], {k, 1, zz}] (* A175499 except that here first term is 0 *)

a(k+1) - a(k) = d(k+1) for k >= 1.

Also, A257883(n) = -1 + A175498(n) for n >= 1.

A257909 Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 2.

Original entry on oeis.org

2, 1, 3, -2, 4, -3, 5, -1, 8, -10, 6, 12, -14, 10, -9, 11, -8, 14, -15, 13, -11, 15, -13, 17, -16, 18, -17, 19, -7, -6, -4, 21, -19, 23, -5, -12, 29, -31, 27, -22, 32, -29, 35, -37, 33, -27, 39, -41, 37, -34, 40, -36, 44, -45, 43, -39, 47, -49, 45, -42, 48

Offset: 1

Clark Kimberling, May 16 2015

Rule 3 follows.  For k >= 1, let  A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}.  Begin with k = 1 and nonnegative integers a(1) and d(1).
Step 1:   If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.
Step 2:  Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k).  Let a(k+1) = a(k) + h and d(k+1) = h.  Replace k by k+1 and do Step 1.
See A257905 for a guide to related sequences and conjectures.

			a(1) = 0, d(1) = 2;
a(2) = 1, d(2) = 1;
a(3) = 4, d(3) = 3;
a(4) = 2, d(4) = -2.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A257905, A257908.

{a, f} = {{0}, {2}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)

A258047 Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 1 and d(1) = 0.

Original entry on oeis.org

0, 1, 2, -1, 3, 6, -7, 5, -3, 7, -6, 8, -5, -2, 9, 18, -23, 13, -11, 15, -13, 17, -15, 19, -18, 20, -19, 21, -20, 22, -21, 23, -22, 24, -17, -4, 27, -29, 25, -9, -12, 29, -30, 28, -24, 32, -31, 33, -27, 4, -8, 35, -33, 37, -25, 49, -53, 45, -43, 47, -42, 52

Offset: 1

Clark Kimberling, Jun 05 2015

Rule 3 follows.  For k >= 1, let  A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}.  Begin with k = 1 and nonnegative integers a(1) and d(1).
Step 1:   If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.
Step 2:  Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k).  Let a(k+1) = a(k) + h and d(k+1) = h.  Replace k by k+1 and do Step 1.
See A257905 for a guide to related sequences and conjectures.

			a(1) = 1, d(1) = 0;
a(2) = 2, d(2) = 1;
a(3) = 4, d(3) = 2;
a(4) = 3, d(4) = -1.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A257905, A258046, A258049, A258050.

{a, f} = {{1}, {0}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)

A257906 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 1.

Original entry on oeis.org

0, 2, 5, 3, 7, 4, 9, 8, 15, 6, 14, 10, 19, 11, 21, 16, 31, 12, 23, 13, 25, 18, 35, 17, 33, 20, 39, 22, 43, 27, 53, 24, 47, 26, 51, 28, 55, 29, 57, 30, 59, 34, 67, 32, 63, 41, 81, 36, 71, 37, 73, 40, 79, 38, 75, 44, 87, 45, 89, 42, 83, 46, 91, 48, 95, 49, 97

Offset: 1

Clark Kimberling, May 16 2015

Rule 3 follows.  For k >= 1, let  A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}.  Begin with k = 1 and nonnegative integers a(1) and d(1).
Step 1:  If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.
Step 2:  Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k).  Let a(k+1) = a(k) + h and d(k+1) = h.  Replace k by k+1 and do Step 1.
See A257905 for a guide to related sequences and conjectures.

			a(1) = 0, d(1) = 1;
a(2) = 2, d(2) = 2;
a(3) = 5, d(3) = 3;
a(4) = 3, d(4) = -2.

Clark Kimberling (first 1000 terms) & Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A257907 (the associated d-sequence, 1 followed by the first differences of this sequence).
Cf. A258105 (inverse to a conjectured permutation of natural numbers which is obtained from this sequence by substituting 1 for the initial value a(1) instead of 0).
Cf. also A257905.

import Data.List ((\\))
a257906 n = a257906_list !! (n-1)
a257906_list = 0 : f [0] [1] where
   f xs@(x:_) ds = g [2 - x .. -1] where
     g [] = y : f (y:xs) (h:ds) where
                  y = x + h
                  (h:_) = [z | z <- [1..] \\ ds, x - z `notElem` xs]
     g (h:hs) | h `notElem` ds && y `notElem` xs = y : f (y:xs) (h:ds)
              | otherwise = g hs
              where y = x + h
-- Reinhard Zumkeller, Jun 03 2015

{a, f} = {{0}, {1}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)

A257907 After 1, the first differences of A257906 (its d-sequence).

Original entry on oeis.org

1, 2, 3, -2, 4, -3, 5, -1, 7, -9, 8, -4, 9, -8, 10, -5, 15, -19, 11, -10, 12, -7, 17, -18, 16, -13, 19, -17, 21, -16, 26, -29, 23, -21, 25, -23, 27, -26, 28, -27, 29, -25, 33, -35, 31, -22, 40, -45, 35, -34, 36, -33, 39, -41, 37, -31, 43, -42, 44, -47, 41

Offset: 1

Clark Kimberling, May 16 2015

This is sequence (d(n)) generated by the following generic rule, "Rule 3" (see below) with parameters a(1) = 0 and d(1) = 1, while A257906 gives the corresponding sequence a(n).
Rule 3 follows.  For k >= 1, let  A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}.  Begin with k = 1 and nonnegative integers a(1) and d(1).
Step 1:  If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.
Step 2:  Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k).  Let a(k+1) = a(k) + h and d(k+1) = h.  Replace k by k+1 and do Step 1.
See A257905 for a guide to related sequences and conjectures.
An informal version of Rule 3 follows:  the sequences "d" and "a" are jointly generated, the "driving idea" idea being that each new term of "d" is obtained by a greedy algorithm.  See A131388 for a similar procedure (Rule 1).

			a(1) = 0, d(1) = 1;
a(2) = 2, d(2) = 2;
a(3) = 5, d(3) = 3;
a(4) = 3, d(4) = -2.

Clark Kimberling (first 1000 terms) & Antti Karttunen, Table of n, a(n) for n = 1..10000

After initial 1, the first differences of A257906 (the associated a-sequence for this rule).

Cf. A257905.

import Data.List ((\\))
a257907 n = a257907_list !! (n-1)
a257907_list = 1 : f [0] [1] where
   f xs@(x:_) ds = g [2 - x .. -1] where
     g [] = h : f ((x + h) : xs) (h : ds) where
                  (h:_) = [z | z <- [1..] \\ ds, x - z `notElem` xs]
     g (h:hs) | h `notElem` ds && y `notElem` xs = h : f (y:xs) (h:ds)
              | otherwise = g hs
              where y = x + h
-- Reinhard Zumkeller, Jun 03 2015

{a, f} = {{0}, {1}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)

A257908 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 2.

Original entry on oeis.org

0, 1, 4, 2, 6, 3, 8, 7, 15, 5, 11, 23, 9, 19, 10, 21, 13, 27, 12, 25, 14, 29, 16, 33, 17, 35, 18, 37, 30, 24, 20, 41, 22, 45, 40, 28, 57, 26, 53, 31, 63, 34, 69, 32, 65, 38, 77, 36, 73, 39, 79, 43, 87, 42, 85, 46, 93, 44, 89, 47, 95, 48, 97, 49, 99, 55, 111

Offset: 1

Clark Kimberling, May 16 2015

Rule 3 follows.  For k >= 1, let  A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}.  Begin with k = 1 and nonnegative integers a(1) and d(1).
Step 1:   If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.
Step 2:  Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k).  Let a(k+1) = a(k) + h and d(k+1) = h.  Replace k by k+1 and do Step 1.
See A257905 for a guide to related sequences and conjectures.

			a(1) = 0, d(1) = 2;
a(2) = 1, d(2) = 1;
a(3) = 4, d(3) = 3;
a(4) = 2, d(4) = -2.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A257905, A257909.

{a, f} = {{0}, {2}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)

A257910 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 3.

Original entry on oeis.org

0, 1, 3, 2, 6, 4, 9, 5, 11, 8, 17, 7, 15, 10, 21, 12, 25, 13, 27, 14, 29, 18, 37, 16, 33, 19, 39, 20, 41, 23, 47, 22, 45, 28, 57, 24, 49, 26, 53, 31, 63, 32, 65, 30, 61, 34, 69, 35, 71, 42, 36, 73, 43, 87, 38, 77, 40, 81, 55, 48, 97, 44, 89, 46, 93, 51, 103

Offset: 1

Clark Kimberling, May 16 2015

Rule 3 follows.  For k >= 1, let  A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}.  Begin with k = 1 and nonnegative integers a(1) and d(1).
Step 1:   If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.
Step 2:  Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k).  Let a(k+1) = a(k) + h and d(k+1) = h.  Replace k by k+1 and do Step 1.
See A257905 for a guide to related sequences and conjectures.

			a(1) = 0, d(1) = 3;
a(2) = 1, d(2) = 1;
a(3) = 3, d(3) = 2;
a(4) = 2, d(4) = -1.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A257905, A257980.

{a, f} = {{0}, {3}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)

A257982 Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 1 and d(1) = 1.

Original entry on oeis.org

1, 2, -1, 3, 5, -6, 4, -2, 6, -5, 7, -3, 11, -13, 9, 18, -23, 13, -11, 15, -14, 16, -15, 17, -10, -4, 20, -21, 19, -17, 21, -19, 23, -18, 28, -31, 25, -9, -12, 29, -27, 31, -35, 27, -7, -8, 39, -45, 33, -29, 37, -39, 35, -26, 44, -43, 45, -47, 43, -38, 48

Offset: 1

Clark Kimberling, May 19 2015

Rule 3 follows.  For k >= 1, let  A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}.  Begin with k = 1 and nonnegative integers a(1) and d(1).
Step 1:   If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.
Step 2:  Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k).  Let a(k+1) = a(k) + h and d(k+1) = h.  Replace k by k+1 and do Step 1.
See A257905 for a guide to related sequences and conjectures.

			a(1) = 1, d(1) = 1;
a(2) = 3, d(2) = 2;
a(3) = 2, d(3) = -1;
a(4) = 5, d(4) = 3.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A257905, A257981.

{a, f} = {{1}, {1}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)

A257980 Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 3.

Original entry on oeis.org

3, 1, 2, -1, 4, -2, 5, -4, 6, -3, 9, -10, 8, -5, 11, -9, 13, -12, 14, -13, 15, -11, 19, -21, 17, -14, 20, -19, 21, -18, 24, -25, 23, -17, 29, -33, 25, -23, 27, -22, 32, -31, 33, -35, 31, -27, 35, -34, 36, -29, -6, 37, -30, 44, -49, 39, -37, 41, -26, -7, 49

Offset: 1

Clark Kimberling, May 19 2015

Rule 3 follows.  For k >= 1, let  A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}.  Begin with k = 1 and nonnegative integers a(1) and d(1).
Step 1:   If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.
Step 2:  Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k).  Let a(k+1) = a(k) + h and d(k+1) = h.  Replace k by k+1 and do Step 1.
See A257905 for a guide to related sequences and conjectures.

			a(1) = 0, d(1) = 3;
a(2) = 1, d(2) = 1;
a(3) = 3, d(3) = 2;
a(4) = 2, d(4) = -1.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A257905, A257910.

{a, f} = {{0}, {3}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)

cp's OEIS Frontend

A256283 Smallest m such that A257905(m) = n.

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A257883 Sequence (a(n)) generated by Algorithm (in Comments) with a(1) = 0 and d(1) = 0.

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A257909 Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 2.

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A258047 Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 1 and d(1) = 0.

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Mathematica

A257906 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 1.

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Haskell

Mathematica

A257907 After 1, the first differences of A257906 (its d-sequence).

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Haskell

Mathematica

A257908 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 2.

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Mathematica

A257910 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 3.

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