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A048702
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Binary palindromes of even length divided by 3.
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7
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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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)).
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MAPLE
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# 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.
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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