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A322510 a(1) = 0, and for any n > 0, a(2*n) = a(n) + k(n) and a(2*n+1) = a(n) - k(n) where k(n) is the least positive integer not leading to a duplicate term in sequence a. 5
0, 1, -1, 4, -2, 2, -4, 5, 3, 6, -10, 7, -3, 8, -16, 15, -5, 12, -6, 19, -7, -9, -11, 22, -8, 9, -15, 28, -12, -14, -18, 16, 14, 10, -20, 13, 11, 17, -29, 20, 18, 21, -35, 23, -41, 24, -46, 57, -13, 26, -42, 35, -17, 25, -55, 29, 27, 30, -54, 31, -59, 32, -68 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

The point is that the same k(n) must be used for both a(2*n) and a(2*n+1). - N. J. A. Sloane, Dec 17 2019

Apparently every signed integer appears in the sequence.

LINKS

Rémy Sigrist, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = (a(2*n) + a(2*n+1))/2.

EXAMPLE

The first terms, alongside k(n) and associate children, are:

  n   a(n)  k(n)  a(2*n)  a(2*n+1)

  --  ----  ----  ------  --------

   1     0     1       1        -1

   2     1     3       4        -2

   3    -1     3       2        -4

   4     4     1       5         3

   5    -2     8       6       -10

   6     2     5       7        -3

   7    -4    12       8       -16

   8     5    10      15        -5

   9     3     9      12        -6

  10     6    13      19        -7

PROG

(PARI) lista(nn) = my (a=[0], s=Set(0)); for (n=1, ceil(nn/2), for (k=1, oo, if (!setsearch(s, a[n]+k) && !setsearch(s, a[n]-k), a=concat(a, [a[n]+k, a[n]-k]); s=setunion(s, Set([a[n]+k, a[n]-k])); break))); a[1..nn]

CROSSREFS

For k(n) see A330337, A330338.

See A305410, A304971 and A322574-A322575 for similar sequences.

Sequence in context: A100854 A194688 A317389 * A021707 A126560 A289762

Adjacent sequences:  A322507 A322508 A322509 * A322511 A322512 A322513

KEYWORD

sign,look

AUTHOR

Rémy Sigrist, Dec 13 2018

STATUS

approved

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Last modified May 29 16:43 EDT 2020. Contains 334704 sequences. (Running on oeis4.)