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 1-3 of 3 results.

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

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Author

Clark Kimberling, May 16 2015

Keywords

Comments

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).

Examples

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

Crossrefs

After initial 1, the first differences of A257906 (the associated a-sequence for this rule).
Cf. A257905.

Programs

  • Haskell
    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
  • Mathematica
    {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 *)

A258105 Inverse to sequence defined as {b(1) = 1, and for n>1: b(n) = A257906(n)}, which is conjectured to be a permutation of natural numbers.

Original entry on oeis.org

1, 2, 4, 6, 3, 10, 5, 8, 7, 12, 14, 18, 20, 11, 9, 16, 24, 22, 13, 26, 15, 28, 19, 32, 21, 34, 30, 36, 38, 40, 17, 44, 25, 42, 23, 48, 50, 54, 27, 52, 46, 60, 29, 56, 58, 62, 33, 64, 66, 71, 35, 69, 31, 75, 37, 77, 39, 68, 41, 81, 73, 79, 45, 83, 85, 89, 43, 87, 93, 97, 49, 91, 51, 99, 55, 101, 95, 103, 53, 105, 47, 107, 61
Offset: 1

Views

Author

Antti Karttunen, May 20 2015

Keywords

Comments

In case the sequence b is not a permutation of natural numbers, then neither this will be and we set a(n) = -1 for those n > 1 that do not occur in A257906.

Crossrefs

A257905 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 0.

Original entry on oeis.org

0, 1, 3, 2, 5, 11, 4, 9, 6, 13, 7, 15, 10, 8, 17, 35, 12, 25, 14, 29, 16, 33, 18, 37, 19, 39, 20, 41, 21, 43, 22, 45, 23, 47, 30, 26, 53, 24, 49, 40, 28, 57, 27, 55, 31, 63, 32, 65, 38, 42, 34, 69, 36, 73, 48, 97, 44, 89, 46, 93, 51, 103, 52, 105, 50, 101
Offset: 1

Views

Author

Clark Kimberling, May 16 2015

Keywords

Comments

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.
Conjecture: suppose that a(1) is an nonnegative integer and d(1) is an integer.
If a(1) = 0 and d(1) != 1, then (a(n)) is a permutation of the nonnegative integers;
if a(1) = 0 and d(1) = 1, then (a(n)) is a permutation of the nonnegative integers excluding 1;
if a(1) = 1, then (a(n)) is a permutation of the positive integers;
if a(1) > 1, then (a(n)) is a permutation of the integers >1;
if d(1) = 0, then (d(n)) is a permutation of the integers;
if d(1) !=0, then (d(n)) is a permutation of the nonzero integers.
Guide to related sequences:
a(1) d(1) (a(n)) (d(n))

Examples

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

Crossrefs

Cf. A256283 (putative inverse).

Programs

  • Haskell
    import Data.List ((\\))
    a257905 n = a257905_list !! (n-1)
    a257905_list = 0 : f [0] [0] 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
  • Mathematica
    {a, f} = {{0}, {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 *)

Formula

a(n) = A258046(n) - 1 for n >= 1.
Showing 1-3 of 3 results.