|
%I #23 Jun 04 2021 10:51:49
%S 1,7,8,2,16,4,5,17,10,34,35,11,70,22,23,71,46,142,143,47,286,94,95,
%T 287,190,574,575,191,1150,382,383,1151,766,2302,2303,767,4606,1534,
%U 1535,4607,3070,9214,9215,3071,18430,6142,6143,18431,12286,36862
%N Start with a(1) = 1 and apply certain patterns of operations on a(n-1) to obtain a(n) as described in comments.
%C 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:
%C +1
%C +1 /3 -1
%C +1 *3 *2 -2
%C -1 /3 -1
%C +1
%C +1 *3 -1
%C +1 /3 *2 -2
%C +1 *3 +1
%C 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.
%H Colin Barker, <a href="/A309523/b309523.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (1,-1,1,0,0,0,0,4,-4,4,-4).
%F From _Colin Barker_, Aug 06 2019: (Start)
%F 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)).
%F 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.
%F (End)
%e A308709 | this sequence
%e | 1
%e | 7 +1 *3 +1
%e | 8 +1
%e | 2 +1 /3 -1
%e 3 | 16 +1 *3 *2 -2
%e 1 /3 | 4 -1 /3 -1
%e | 5 +1
%e | 17 +1 *3 -1
%e 2 *2 | 10 +1 /3 *2 -2
%e 6 *3 | 34 +1 *3 +1
%e | 35 +1
%e | 11 +1 /3 -1
%e 12 *2 | 70 +1 *3 *2 -2
%e 4 /3 | 22 -1 /3 -1
%e | 23 +1
%e | 71 +1 *3 -1
%e 8 *2 | 46 +1 /3 *2 -2
%e 24 *3 | 142 +1 *3 +1
%e | 143 +1
%t LinearRecurrence[{1, -1, 1, 0, 0, 0, 0, 4, -4, 4, -4},{1, 7, 8, 2, 16, 4, 5, 17, 10, 34, 35}, 50]
%o (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
%o (Perl) use integer;
%o my @a; my $n = 1; $a[$n ++] = 1;
%o $a[$n ++] = (($a[$n-1] +1) *3) +1; # 7
%o while ($n < 50) {
%o $a[$n ++] = (($a[$n-1] +1) ); # 8
%o $a[$n ++] = (($a[$n-1] +1) /3) -1; # 2
%o $a[$n ++] = (($a[$n-1] +1) *3) *2 -2; # 16
%o $a[$n ++] = (($a[$n-1] -1) /3) -1; # 4
%o $a[$n ++] = (($a[$n-1] +1) ); # 5
%o $a[$n ++] = (($a[$n-1] +1) *3) -1; # 17
%o $a[$n ++] = (($a[$n-1] +1) /3) *2 -2; # 10
%o $a[$n ++] = (($a[$n-1] +1) *3) +1; # 34
%o } # while
%o shift(@a); # remove $a[0]
%o print join(", ", @a) . "\n"; # _Georg Fischer_, Aug 07 2019
%o (Python 3)
%o def A309523():
%o k, j, a = 0, 0, 1
%o def b(a): return a + 1
%o def c(a): return a + 2
%o def d(a): return a - 1
%o def e(a): return a - 2
%o def f(a): return a << 1
%o def g(a): return a * 3
%o def h(a): return a // 3
%o O = [c,g,e,b,b,h,d,b,g,f,e,c,h,e,b,b,g,d,b,h,f,e]
%o L = [3,1,3,4]
%o while True:
%o yield(a)
%o for _ in range(L[j]):
%o a = O[k](a)
%o k += 1; k %= 22
%o j += 1; j %= 4
%o a = A309523()
%o print([next(a) for _ in range(50)]) # _Peter Luschny_, Aug 06 2019
%Y Cf. A308709.
%K nonn,easy
%O 1,2
%A _Georg Fischer_, Aug 06 2019
|
