OFFSET
1,2
FORMULA
EXAMPLE
To obtain the first 2^n-1 entries if you have the first 2^(n-1)-1 entries, adjoin 1/6 (-3 + (-1)^(1 + n) + 2^(2 + n)) to the right end of the list, negate the signs of the first 2^(n-1)-1 entries, and then adjoin that list to the right. For example for n=3 {1,2,-1} becomes {1,2,-1,5,-1,-2,1}.
MATHEMATICA
Fold[Join[#1, {#2}, -#1] &, {1},
Table[1/6 (-3 + (-1)^(1 + n) + 2^(2 + n)), {n, 2, 6}]]
t[n_/; IntegerQ[Log2[n]]]:=1/6 (-3 + (-1)^IntegerExponent[n, 2] + 8*n);
t[n_/; Not[IntegerQ[Log2[n]]]]:=-t[n-2^Floor[Log2[n]]];
Table[t[j], {j, 1, 15}](* recursive formulation *)
Table[1/6 (-3+(-1)^IntegerExponent[j, 2]+2^(IntegerExponent[j, 2]+3))(-1)^(Total[IntegerDigits[j, 2]]+1), {j, 1, 15}] (* closed form *)
CROSSREFS
KEYWORD
sign
AUTHOR
John Erickson, Oct 18 2018
STATUS
approved