OFFSET
0,3
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
FORMULA
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 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Mar 07 1999
STATUS
approved