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%I A048702 #48 Aug 10 2024 03:37:57
%S A048702 0,1,3,5,11,15,17,21,43,51,55,63,65,73,77,85,171,187,195,211,215,231,
%T A048702 239,255,257,273,281,297,301,317,325,341,683,715,731,763,771,803,819,
%U A048702 851,855,887,903,935,943,975,991
%N A048702 Binary palindromes of even length divided by 3.
%C A048702 Let the length of A048701(n) in binary be 2k. Since it is a palindrome of even length, its digits come in pairs which are equal: one in the left half and the other in the right half. Thus, A048701(n) is a sum of numbers of the form d * 2^m * (2^(2k-2m-1) + 1). The number 2^(2k-2m-1) = 2 * 4^(k-m-1) is congruent to 2 (mod 3), so 2^(2k-2m-1) + 1 is divisible by 3. This means A048701(n) is divisible by 3, and therefore a(n) is an integer. - _Michael B. Porter_, Jun 18 2019
%H A048702 G. C. Greubel, <a href="/A048702/b048702.txt">Table of n, a(n) for n = 0..1000</a>
%F A048702 a(n) = A048701(n)/3.
%F A048702 Conjecture: a(n) = 2^floor(log_2(n)) * Sum_{i=1..n} 1/(2^v_2(i)), for n >= 1, where v_2(i) = A007814(i) is the exponent of the highest power of 2 dividing i.
%F A048702 Conjecture: a(n) = n*2^floor(log_2(n)) - Sum_{i=1..floor(log_2(n))} 2^(floor(log_2(n)) - i)*floor(n/(2^i)).
%p A048702 # Two unproved formulas which are not based upon first generating a palindrome and then dividing by 3, recursive and more direct:
%p A048702 # Here d is 2^(the distance between the most and least significant 1-bit of n):
%p A048702 bper3_rec := proc(n) option remember; local d; if(0 = n) then RETURN(0); fi; d := 2^([ log2(n) ]-A007814[ n ]);
%p A048702 if(1 = d) then RETURN((2*bper3_rec(n-1))+d); else RETURN(bper3_rec(n-1)+d); fi; end;
%p A048702 # or more directly (after K. Atanassov's formula for partial sums of A007814):
%p A048702 bper3_direct := proc(n) local l,j; l := [ log2(n) ]; RETURN((2/3*((2^(2*l))-1))+1+ sum('(2^(l-j)*floor((n-(2^l)+2^j)/(2^(j+1))))','j'=0..l)); end;
%p A048702 # Can anybody find an even simpler closed form? See A005187 for inspiration.
%t A048702 Join[{0}, Reap[For[k = 1, k < 3000, k += 2, bb = IntegerDigits[k, 2]; If[bb == Reverse[bb], If[EvenQ[Length[bb]], Sow[k/3]]]]][[2, 1]]] (* _Jean-François Alcover_, Mar 04 2016 *)
%Y A048702 Cf. A048701, A048704 (base 4 palindromes of even length divided by 5), A044051 (binary palindromes plus one divided by 2: (A006995(n)+1)/2), A000975.
%K A048702 nonn,base
%O A048702 0,3
%A A048702 _Antti Karttunen_, Mar 07 1999

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