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A105055
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a(n) = bitwise AND operation applied between every term of the n-th row of Pascal's triangle, with the stipulation that all bits left of the last significant bit in each element are turned "on" until all elements of a row contain the same number of bits (see example). Results represented in decimal notation.
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0
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1, 1, 2, 3, 4, 8, 20, 33, 64, 64, 192, 256, 768, 1024, 2304, 4353, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194034, 8388608, 16777216, 33554432, 67108864, 134217728, 268435457, 536870912, 1073741824, 2147483648
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| There are probably an infinite amount of anomalies that do not fit the "best-fit" formula above. Some of the early examples include a(3), a(6), a(10), a(12), a(14) and a(15). However, after this point the sequence dramatically, although not completely, stabilizes. Also interesting is the fact that a(8) = a(9).
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LINKS
| Eric Weisstein's World of Mathematics, Pascal's Triangle
Eric Weisstein's World of Mathematics, Bitwise AND
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FORMULA
| The sequence can be approximated with the function a(n)=2^(n-3), but there plenty of anomalies (see "Comments" below).
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EXAMPLE
| Consider the a=(7) row of Pascal's Triangle:
in decimal notation: 1-7-21-35-35-21-7-1
in binary notation: 000001-000111-010101-100011
Note: only distinct digits are of any importance
Now add 1's to the left of last significant digit and "AND" all terms:
111111 AND 111111 AND 110101 AND 100011 = 100001
which is 33 in decimal, thus a(7)=33.
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PROG
| Java
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CROSSREFS
| Sequence in context: A140974 A118841 A102276 * A108506 A129284 A100997
Adjacent sequences: A105052 A105053 A105054 * A105056 A105057 A105058
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KEYWORD
| nonn
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AUTHOR
| Andrew G. West (WestA(AT)wlu.edu), Apr 04 2005
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