A320667 First differences of A066194.

1, 2, -1, 5, -1, -2, 1, 10, -1, -2, 1, -5, 1, 2, -1, 21, -1, -2, 1, -5, 1, 2, -1, -10, 1, 2, -1, 5, -1, -2, 1, 42, -1, -2, 1, -5, 1, 2, -1, -10, 1, 2, -1, 5, -1, -2, 1, -21, 1, 2, -1, 5, -1, -2, 1, 10, -1, -2, 1, -5, 1, 2, -1, 85, -1, -2, 1, -5, 1, 2, -1, -10

Offset: 1

Author

John Erickson, Oct 18 2018

Keywords

• sign

Examples

			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}.
		

Crossrefs

Cf. A066194.

Programs

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 *)

Formula

a(n) = A066194(n+1) - A066194(n).

a(n) = (1/6)*(-3 + (-1)^A007814(n) + 2^(A007814(n)+3))*(-1)^(A000120(n)+1).