A338789 Lexicographically earliest sequence of positive integers such that the sum of any number of consecutive terms can only repeat as sum of other consecutive terms after two or more terms between them.

1, 2, 4, 1, 9, 2, 1, 21, 2, 3, 1, 8, 2, 23, 18, 2, 8, 7, 9, 2, 1, 20, 16, 1, 21, 36, 3, 32, 2, 3, 4, 11, 13, 8, 2, 13, 27, 4, 18, 3, 7, 5, 8, 133, 3, 22, 31, 46, 19, 8, 47, 14, 3, 2, 14, 10, 44, 3, 5, 1, 10, 3, 9, 6, 19, 73, 39, 22, 36, 6, 1, 60, 17, 32, 227, 2, 134, 9, 45, 11, 33, 3, 37, 1, 8, 12, 14, 8, 1, 67

Offset

1,2

Links

S. Brunner, Table of n, a(n) for n = 1..10000

Example

The solution for a(5):
We look through the numbers step by step and if groups with the same sum are less than 2 terms apart they are put in brackets:
   1,2,4,[1],[1?] - not possible
   [1,2],4,[1,2?] - not possible
   1,2,[4],[1,3?] - not possible
   1,2,[4],1,[4?] - not possible
   1,2,[4,1],[5?] - not possible
   1,[2,4],1,[6?] - not possible
   1,[2,4,1],[7?] - not possible
   [1,2,4,1],[8?] - not possible
   1,2,4,1,9?
There are no 2 sums which contradict the definition of this sequence with a(5) = 9, so this is the next term. In this case we knew it must be the solution because the upper bound of a(n) is always the sum of all previous terms + 1.
Another example for a(8) = 21:
  1,2,4,1,9,2,[1],[1?]   ;  1,2,4,1,9,[2],1,[2?]   ;  1,2,4,1,9,[2,1],[3?]
  1,[2,4,1],9,[2,1,4?]   ;  [1,2,4,1],9,[2,1,5?]   ;  1,2,4,1,[9],[2,1,6?]
  1,2,4,[1,9],[2,1,7?]   ;  1,2,4,1,[9],2,[1,8?]   ;  1,2,4,[1,9],2,[1,9?]
  1,2,4,1,[9,2],[1,10?]  ;  1,2,4,1,[9,2],1,[11?]  ;  1,2,4,1,[9,2,1],[12?]
  1,2,4,[1,9,2,1],[13?]  ;  [1,2,4,1,9],[2,1,14?]  ;  1,2,[4,1,9,2],[1,15?]
  1,2,[4,1,9,2],1,[16?]  ;  1,2,[4,1,9,2,1],[17?]  ;  1,[2,4,1,9,2],1,[18?]
  1,[2,4,1,9,2,1],[19?]  ;  [1,2,4,1,9,2,1],[20?]  ;  1,2,4,1,9,2,1,21?
  -> a(8) = 21.

Prog

(Python)
def A(lastn):
    n, a, chk, nchk=1, [], [], []
    while n<=lastn:
        i=1
        while i in chk: i+=1
        for x, v in enumerate(chk): chk[x]=v-i
        chk.extend(nchk)
        for x, v in enumerate(nchk): nchk[x]=v+i
        nchk.append(i)
        chk.extend(nchk)
        chk=[x for x in chk if x>0]
        chk=list(set(chk))
        a.append(i)
        print(i)
        n += 1
    return a

Crossrefs

Cf. A002048.

Sequence in context: A077387 A057551 A019823 * A174135 A182903 A169840

Adjacent sequences: A338786 A338787 A338788 * A338790 A338791 A338792

Keyword

nonn,look

Author

S. Brunner, Nov 09 2020

Status

approved