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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
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OFFSET
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0,3
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COMMENTS
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For many n, a(n) = 2^k, where k = floor(log_2(A001405(n))). For n from 9 through 40, k = n-3, but it is smaller thereafter; no formula of the form k = n-c will work for sufficiently large n. There are infinitely many exceptions to this formula, since a(2^m-1) is always odd, hence not equal to 2^k for m > 1. - Franklin T. Adams-Watters, Mar 29 2014, replacing assertions by the author.
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LINKS
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EXAMPLE
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Consider the n = 7 row of Pascal's Triangle:
in decimal notation: 1-7-21-35-35-21-7-1;
in binary notation: 1-111-10101-100011-...
Note: only distinct digits are of any importance.
Now add 1's to the left of the most 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
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(PARI) nexttwo(n)=local(r=1); while(r<=n, r*=2); r
a(n)={local(v=vector(ceil(n/2), i, binomial(n, i)), r);
r=nexttwo(v[#v])-1;
for(i=1, #v, r=bitand(r, nexttwo(v[#v])-nexttwo(v[i])+v[i]));
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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Andrew G. West (WestA(AT)wlu.edu), Apr 04 2005
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EXTENSIONS
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STATUS
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approved
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