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A051382
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Numbers k whose base 3 expansion matches (0|1)*(02)?(0|1)* (no more than one "02" allowed in midst of 0's and 1's).
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13
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0, 1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 13, 18, 19, 21, 22, 27, 28, 29, 30, 31, 33, 34, 36, 37, 38, 39, 40, 54, 55, 57, 58, 63, 64, 66, 67, 81, 82, 83, 84, 85, 87, 88, 90, 91, 92, 93, 94, 99, 100, 102, 103, 108, 109, 110, 111, 112, 114, 115, 117, 118, 119, 120, 121, 162, 163
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OFFSET
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1,3
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COMMENTS
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Representation of 2n in base 3 consists entirely of 0's and 2's, except possibly for a single pair of adjacent 1's among them.
9 divides neither C(2s-1,s) [= A001700(s)] nor C(2s,s) [= A000984(s)] if and only if s = a(n). [Cf. also A249721].
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LINKS
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EXAMPLE
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In base 3 the terms look like 0, 1, 2, 10, 11, 20, 21, 100, 101, 102, 110, 111, 200, 201, 210, 211, 1000, 1001, 1002, 1010, 1011, 1020, 1021, 1100, 1101, 1102, 1110, 1111, 2000, 2001, 2010, 2011, 2100, 2101, 2 110, 2111, 10000
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MAPLE
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q:= n-> (l-> (h-> h=0 or h=1 and l[1+ListTools[Search](2, l)]
=0 )(numboccur(l, 2)))([convert(n, base, 3)[], 0]):
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PROG
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(Perl) sub conv_x_base_n { my($x, $b) = @_; my ($r, $z) = (0, ''); do { $r = $x % $b; $x = ($x - $r)/$b; $z = "$r" . $z; } while(0 != $x); return($z); }
(Perl) for($i=1; $i <= 201; $i++) { if(("0" . conv_x_base_n($i, 3)) =~ /^(0|1)*(02)?(0|1)*$/) { print $i, ", "; } }
(define (in_A051382? n) (let loop ((n n) (seen02yet? #f)) (cond ((zero? n) #t) ((= 1 n) #t) ((modulo n 3) => (lambda (r) (cond ((= r 2) (if (or seen02yet? (not (zero? (modulo (/ (- n r) 3) 3)))) #f (loop (/ (- n r) 3) #t))) (else (loop (/ (- n r) 3) seen02yet?))))))))
(Python)
import re
from sympy.ntheory.digits import digits
def b3(n): return "".join(map(str, digits(n, 3)[1:]))
def ok(n): return re.fullmatch('2(0|1)*|(0|1)*(02)?(0|1)*', b3(n)) != None
(PARI) is(n)=my(v=digits(n, 3)); for(i=1, #v, if(v[i]==2, if(i>1 && v[i-1], return(0)); for(j=i+1, #v, if(v[j]==2, return(0))); return(1))); 1 \\ Charles R Greathouse IV, Feb 23 2024
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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