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%I #15 Nov 25 2016 05:29:05
%S 1,1,2,3,4,8,20,33,64,64,192,256,768,1024,2304,4353,8192,16384,32768,
%T 65536,131072,262144,524288,1048576,2097152,4194034,8388608,16777216,
%U 33554432,67108864,134217728,268435457,536870912,1073741824,2147483648
%N 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.
%C 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.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PascalsTriangle.html">Pascal's Triangle</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AND.html">Bitwise AND</a>
%e Consider the n = 7 row of Pascal's Triangle:
%e in decimal notation: 1-7-21-35-35-21-7-1;
%e in binary notation: 1-111-10101-100011-...
%e Note: only distinct digits are of any importance.
%e Now add 1's to the left of the most significant digit and "AND" all terms:
%e 111111 AND 111111 AND 110101 AND 100011 = 100001
%e which is 33 in decimal, thus a(7)=33.
%o (PARI) nexttwo(n)=local(r=1);while(r<=n,r*=2);r
%o a(n)={local(v=vector(ceil(n/2),i,binomial(n,i)),r);
%o r=nexttwo(v[#v])-1;
%o for(i=1,#v,r=bitand(r,nexttwo(v[#v])-nexttwo(v[i])+v[i]));
%o r} \\ _Franklin T. Adams-Watters_, Mar 29 2014
%Y Cf. A001405.
%K nonn,base
%O 0,3
%A Andrew G. West (WestA(AT)wlu.edu), Apr 04 2005
%E Edited by _Franklin T. Adams-Watters_, Mar 29 2014
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