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A064686
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a(n) = number of n-digit base-3 biquams.
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3
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0, 2, 7, 23, 73, 227, 697, 2123, 6433, 19427, 58537, 176123, 529393, 1590227, 4774777, 14332523, 43013953, 129074627, 387289417, 1161999323, 3486260113, 10459304627, 31378962457, 94138984523, 282421147873, 847271832227
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| A biquam or biquanimous number (A064544) is a number whose digits can be split into two groups with equal sum.
This is the same as A083313 (apart from the initial term). Proof. Let sum(w) denote the sum of the digits of w. There are 2*3^(n-1) n-digit base-3 numbers: w = (w_1,w_2,...,w_n) with w_i in {0,1,2} for all i and w_1 != 0. Partition them into 4 classes: (i) sum(w) is odd, (ii) sum(w) is even, w contains no 1's and has an odd number of 2s, (iii) sum(w) is even, w contains no 1's and has an even number of 2s and (iv) sum(w) is even and w contains some 1's. Clearly, no biquams occur in cases (i) and (ii), case (iii) consists entirely of biquams and, we claim, so does case (iv). For case (iv) forces an even number, say 2k, of 1's. An even number of 2s clearly gives a biquam and an odd number 2m+1 of 2s does too because {m 2s, (k+1) 1's} and {(m+1) 2s, (k-1) 1's} is a biquam split. There are 3^(n-1) w's in case (i) and 2^(n-2) w's in case (ii) and hence 2*3^(n-1) - (3^(n-1) + 2^(n-2)) = 3^(n-1) - 2^(n-2) (A083313) biquams among n-digit base-3 numbers. - David Callan (callan(AT)stat.wisc.edu), Sep 15 2004
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CROSSREFS
| Essentially the same as A083313.
Sequence in context: A027139 A192906 * A083313 A077832 A030282 A042575
Adjacent sequences: A064683 A064684 A064685 * A064687 A064688 A064689
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KEYWORD
| base,nonn
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AUTHOR
| David W. Wilson (davidwwilson(AT)comcast.net), Oct 10 2001
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