cp's OEIS Frontend

A217100 Greatest number such that the number of contiguous palindromic bit patterns in its binary representation is n, or 0, if there is no such number.

%I A217100 #17 Mar 15 2016 08:01:30
%S A217100 1,2,3,6,0,13,14,26,29,52,58,105,116,211,233,422,467,845,934,1690,
%T A217100 1869,3380,3738,6761,7476,13523,14953,27046,29907,54093,59814,108186,
%U A217100 119629,216372,239258,432745,478516,865491,957033,1730982,1914067,3461965,3828134
%N A217100 Greatest number such that the number of contiguous palindromic bit patterns in its binary representation is n, or 0, if there is no such number.
%C A217100 The set of numbers which have n contiguous palindromic bit patterns (in their binary representation) is not empty and finite, provided n<>5. Proof: compare A217101 for the proof of existence. The finiteness follows from A206925, since each number p > 2^floor((n+1)/2) has at least A206925(p) >= 2*floor(log_2(p)) > 2*floor((n+1)/2) = n + n mod 2 palindromic substrings. Thus, there is a boundary b <= 2^floor((n+1)/2) such that all numbers > b have more than n palindromic substrings. It follows, that the set of numbers with n palindromic substrings is finite.
%C A217100 a(5)=0, and this is the only zero term. Proof: cf. A217101.
%C A217100 The binary expansion of a(n) has 1 + floor(n/2) digits (n<>5).
%C A217100 a(2n) is the maximal number with n+1 binary digits such that the number of contiguous palindromic bit patterns in the binary representation is minimal (cf. A206926).
%H A217100 Hieronymus Fischer, <a href="/A217100/b217100.txt">Table of n, a(n) for n = 1..1000</a>
%F A217100 a(n) = max(k | A206925(k) = n), for n<>5.
%F A217100 A206925(a(n)) = n, n<>5.
%F A217100 a(n) => A217101(n), equality holds for n = 1, 2, 3 and 5, only.
%F A217100 Iteration formula:
%F A217100 a(n+2) = 2*a(n) + floor(52*2^floor((n+4-(-1)^n)/2)/63) mod 2, n>5.
%F A217100 a(n+2) = 2*a(n) + floor(52*2^((2*n+7-3*(-1)^n)/4)/63) mod 2, n>5.
%F A217100 Direct calculation formula:
%F A217100 a(n) = floor(52*2^floor((n+1+(-1)^n)/2)/63) mod 2^floor((n+1+(-1)^n)/2)) + (1-(-1)^n)*2^floor((n-2)/2)), for n>5.
%F A217100 a(n) = floor(52*2^((2*n + 1 + 3*(-1)^n)/4)/63) + (1-(-1)^n)*2^((2*n - 5 + (-1)^n)/4), for n>5.
%F A217100 a(2k) = A206926(6k-9), k>2.
%F A217100 G.f.: x*(1 +2*x +x^2 +2*x^3 -6*x^4 +x^5 +14*x^6 +x^8 +x^11 -x^12 -x^13 -2*x^15 +7*x^16 -14*x^18)/((1-2*x^2)*(1-x^3)*(1+x^3)*(1+x^6)); also:
%F A217100 x*(1 -x +3*x^2 +x*(3+14*x^5)/(1-2*x^2) +x^5*(1+x^3)*(1+x^6+x^8)/((1-2*x^2)*(1-x^12))).
%e A217100 a(3)=3, since 3=11_2 has 3 contiguous palindromic bit patterns, and this is the greatest such number.
%e A217100 a(6)=13. Since 13=1101_2 has 6 contiguous palindromic bit patterns, and this is the greatest such number.
%e A217100 a(8)=26. Since 26=11010_2 has 8 contiguous palindromic bit patterns (1, 1, 0, 1, 0, 11, 101 and 010), and this is the greatest such number.
%e A217100 a(9)=27. Since 17=11011_2 has 9 contiguous palindromic bit patterns (1, 1, 0, 1, 1, 11, 11, 101 and 11011), and this is the greatest such number.
%Y A217100 Cf. A006995, A206923, A206924, A206925, A206926, A070939, A217101.
%K A217100 base,nonn
%O A217100 1,2
%A A217100 _Hieronymus Fischer_, Jan 23 2013