A309523 Start with a(1) = 1 and apply certain patterns of operations on a(n-1) to obtain a(n) as described in comments.

1, 7, 8, 2, 16, 4, 5, 17, 10, 34, 35, 11, 70, 22, 23, 71, 46, 142, 143, 47, 286, 94, 95, 287, 190, 574, 575, 191, 1150, 382, 383, 1151, 766, 2302, 2303, 767, 4606, 1534, 1535, 4607, 3070, 9214, 9215, 3071, 18430, 6142, 6143, 18431, 12286, 36862

Offset

1,2

Comments

a(2) = 7 is obtained from a(1) = 1 by (((1) +1) *3) +1. We abbreviate this to the operation pattern "+1 *3 +1". The 8 patterns for a(3..10), a(11..18) etc. are:

+1

+1 /3 -1

+1 *3 *2 -2

-1 /3 -1

+1

+1 *3 -1

+1 /3 *2 -2

+1 *3 +1

A308709 uses similar, but simpler patterns in blocks of 4 (cf. the example, below). A308709 contains the set {2^k | k>=0} union {3*2^k | k>=0}, so all terms are different. This sequence contains the terms {6*A308709 - 2} union {6*A308709 - 1}, therefore all terms are also different.

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,-1,1,0,0,0,0,4,-4,4,-4).

Formula

From Colin Barker, Aug 06 2019: (Start)

G.f.: x*(1 + 6*x + 2*x^2 + 15*x^4 - 18*x^5 + 15*x^6 - 10*x^8 + 12*x^9 - 14*x^10) / ((1 - x)*(1 + x^2)*(1 - 2*x^4)*(1 + 2*x^4)).

a(n) = a(n-1) - a(n-2) + a(n-3) + 4*a(n-8) - 4*a(n-9) + 4*a(n-10) - 4*a(n-11) for n>11.

(End)

Example

  A308709 | this sequence
          |   1
          |   7    +1 *3 +1
          |   8    +1
          |   2    +1 /3 -1
   3      |  16    +1 *3 *2 -2
   1   /3 |   4    -1 /3 -1
          |   5    +1
          |  17    +1 *3 -1
   2   *2 |  10    +1 /3 *2 -2
   6   *3 |  34    +1 *3 +1
          |  35    +1
          |  11    +1 /3 -1
  12   *2 |  70    +1 *3 *2 -2
   4   /3 |  22    -1 /3 -1
          |  23    +1
          |  71    +1 *3 -1
   8   *2 |  46    +1 /3 *2 -2
  24   *3 | 142    +1 *3 +1
          | 143    +1

Mathematica

LinearRecurrence[{1, -1, 1, 0, 0, 0, 0, 4, -4, 4, -4}, {1, 7, 8, 2, 16, 4, 5, 17, 10, 34, 35}, 50]

Prog

(PARI) Vec(x*(1 + 6*x + 2*x^2 + 15*x^4 - 18*x^5 + 15*x^6 - 10*x^8 + 12*x^9 - 14*x^10) / ((1 - x)*(1 + x^2)*(1 - 2*x^4)*(1 + 2*x^4)) + O(x^40)) \\ Colin Barker, Aug 06 2019
(Perl) use integer;
  my @a; my $n = 1; $a[$n ++] = 1;
  $a[$n ++] =   (($a[$n-1] +1) *3) +1;    #  7
  while ($n < 50) {
    $a[$n ++] = (($a[$n-1] +1)   );       #  8
    $a[$n ++] = (($a[$n-1] +1) /3) -1;    #  2
    $a[$n ++] = (($a[$n-1] +1) *3) *2 -2; # 16
    $a[$n ++] = (($a[$n-1] -1) /3) -1;    #  4
    $a[$n ++] = (($a[$n-1] +1)   );       #  5
    $a[$n ++] = (($a[$n-1] +1) *3) -1;    # 17
    $a[$n ++] = (($a[$n-1] +1) /3) *2 -2; # 10
    $a[$n ++] = (($a[$n-1] +1) *3) +1;    # 34
  } # while
  shift(@a); # remove $a[0]
  print join(", ", @a) . "\n"; # Georg Fischer, Aug 07 2019
(Python)
def A309523():
    k, j, a = 0, 0, 1
    def b(a): return a + 1
    def c(a): return a + 2
    def d(a): return a - 1
    def e(a): return a - 2
    def f(a): return a << 1
    def g(a): return a * 3
    def h(a): return a // 3
    O = [c, g, e, b, b, h, d, b, g, f, e, c, h, e, b, b, g, d, b, h, f, e]
    L = [3, 1, 3, 4]
    while True:
        yield(a)
        for _ in range(L[j]):
            a = O[k](a)
            k += 1; k %= 22
        j += 1; j %= 4
a = A309523()
print([next(a) for _ in range(50)]) # Peter Luschny, Aug 06 2019

Crossrefs

Cf. A308709.

Sequence in context: A342486 A245758 A266566 * A153622 A257576 A378129

Adjacent sequences: A309520 A309521 A309522 * A309524 A309525 A309526

Keyword

nonn,easy

Author

Georg Fischer, Aug 06 2019

Status

approved