A048702 Binary palindromes of even length divided by 3.

0, 1, 3, 5, 11, 15, 17, 21, 43, 51, 55, 63, 65, 73, 77, 85, 171, 187, 195, 211, 215, 231, 239, 255, 257, 273, 281, 297, 301, 317, 325, 341, 683, 715, 731, 763, 771, 803, 819, 851, 855, 887, 903, 935, 943, 975, 991

Offset: 0

Author

Antti Karttunen, Mar 07 1999

Keywords

• nonn
• base

Comments

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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

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.

Programs

Maple

# Two unproved formulas which are not based upon first generating a palindrome and then dividing by 3, recursive and more direct:
# Here d is 2^(the distance between the most and least significant 1-bit of n):
bper3_rec := proc(n) option remember; local d; if(0 = n) then RETURN(0); fi; d := 2^([ log2(n) ]-A007814[ n ]);
if(1 = d) then RETURN((2*bper3_rec(n-1))+d); else RETURN(bper3_rec(n-1)+d); fi; end;
# or more directly (after K. Atanassov's formula for partial sums of A007814):
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;
# Can anybody find an even simpler closed form? See A005187 for inspiration.

Mathematica

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 *)

Formula

a(n) = A048701(n)/3.

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.

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)).