|
Search: id:A048702
|
|
|
| A048702 |
|
Binary palindromes of even length divided by 3. a(n) = A048701[ n ]/3 |
|
+0
6
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
0,3
|
|
|
COMMENT
|
Two unproved formulae 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.
|
|
CROSSREFS
|
A048704 base 4 palindromes of even length divided by 5, A044051 binary palindromes plus one divided by 2 (A006995[ n ]+1)/2. Cf. A000975.
Adjacent sequences: A048699 A048700 A048701 this_sequence A048703 A048704 A048705
Sequence in context: A086284 A136500 A024897 this_sequence A003546 A082421 A105772
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Antti Karttunen, Mar 07 1999
|
|
|
Search completed in 0.002 seconds
|
