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%I #21 Oct 15 2023 12:01:27
%S 1,2,3,3,4,4,5,5,6,6,7,7,7,7,7,7,8,8,9,9,9,9,10,10,10,10,10,10,11,11,
%T 11,11,12,12,13,13,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,15,
%U 15,15,15,15,15,15,15,15,15,15,15,16,16,17,17,17,17,17
%N Index of the least binary palindrome >=n; also the "upper inverse" of A006995.
%C The least m such that A006995(m)>=n;
%C n is palindromic iff a(n)=A206915(n);
%C a(n) is the number of the binary palindrome A206914(n);
%C if n is a binary palindrome, then A006995(a(n))=n, so a(n) is 'inverse' with respect to A006995
%F a(n)=min(m|A006995(m)>=n);
%F a(A006995(n))=n;
%F A006995(a(n))>=n, equality holds true iff n is a binary palindrome;
%F Let p=A206914(n), m=floor(log_2(p)) and p>2, then:
%F a(n)=(((5-(-1)^m)/2) + sum_{k=1..floor(m/2)} (floor(p/2^k) mod 2)/2^k))*2^floor(m/2);
%F a(n)=(1/2)*((6-(-1)^m)*2^floor(m/2)-1-sum_ {k=1..floor(m/2)} (-1)^floor(p/2^k)*2^(floor(m/2)-k)));
%F a(n)=(5-(-1)^m)*2^floor(m/2)/2-3*sum_{k=2..floor(m/2)} floor(p/2^k)*2^floor(m/2)/2^k)+(floor(p/2)*2^floor(m/2)/2-2*floor((p/2)*2^floor(m/2))*floor((m-1)/m+1/2).
%F Partial sums S(n) = sum_{k=0..n} a(k):
%F S(n) = 1+n*a(n)-A206920(a(n)-1), valid for n>0.
%F G.f.: g(x)=(x+x^2+x^3+sum_{j=1..infinity} x^(3*2^j)*(f_j(x)+f_j(1/x)))/(x(1-x)), where the f_j(x) are defined as follows:
%F f_1(x)=x, and for j>1,
%F f_j(x)=x^3*product_{k=1..floor((j-1)/2)} (1+x^b(j,k)), where b(j,k)=2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k>1, and b(j,1)=(2+(-1)^j)*2^(floor((j-1)/2)+1).
%e a(2)=3 since 3 is the index number of the least binary palindrome >= 2;
%e a(5)=4 since 4 is the index number of the least binary palindrome >= 5;
%e a(10)=7 since A006995(7)=15>=10, but A006995(6)=9<10, and so that, 7 is the index number of least binary palindrome >= 10;
%Y Cf. A006995, A206920.
%K nonn,base
%O 0,2
%A _Hieronymus Fischer_, Feb 17 2012
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