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A309523
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Start with a(1) = 1 and apply certain patterns of operations on a(n-1) to obtain a(n) as described in comments.
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1
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (1,-1,1,0,0,0,0,4,-4,4,-4).
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FORMULA
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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)
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EXAMPLE
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| 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
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MATHEMATICA
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LinearRecurrence[{1, -1, 1, 0, 0, 0, 0, 4, -4, 4, -4}, {1, 7, 8, 2, 16, 4, 5, 17, 10, 34, 35}, 50]
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PROG
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(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]
(Python 3)
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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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