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

Author

Georg Fischer, Aug 06 2019

Keywords

• nonn
• easy

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.

Examples

			  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
		

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).

Crossrefs

Cf. A308709.

Programs

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]

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

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)