OFFSET
0,2
COMMENTS
Define an infinite array by m(n,k) = 2^n-n+k for n>=k>=0 (in the lower left triangle) and by m(n,k) = 2^k+k-n for k>=n>=0 (in the upper right triangle). The antidiagonal sums of this array are a(n) = sum_{k=0..n} m(n-k,k). - J. M. Bergot, Aug 16 2013
LINKS
FORMULA
a(2n) = 2^n(2^(n+2)-3), a(2n+1) = 2^n(2^(n+3)-4).
G.f.: (1+2*x)/[(1-2*x)*(1-2*x^2)].
a(0)=1, a(1)=4, a(2)=10, a(n)=2*a(n-1)+2*a(n-2)-4*a(n-3). - Harvey P. Dale, Nov 10 2013
a(n) = 2^(n+2) - (2 + mod(n+1, 2)) * 2^floor((n+1)/2). - Michael Somos, Jun 24 2018
a(n) = - (2^(n+2)) * A052955(-n-3) for all n in Z. - Michael Somos, Jun 24 2018
EXAMPLE
G.f. = 1 + 4*x + 10*x^2 + 24*x^3 + 52*x^4 + 112*x^5 + 232*x^6 + 480*x^7 + ... - Michael Somos, Jun 24 2018
MATHEMATICA
Table[Sum[2^Max[k, n-k], {k, 0, n}], {n, 0, 30}] (* or *) LinearRecurrence[ {2, 2, -4}, {1, 4, 10}, 30] (* Harvey P. Dale, Nov 10 2013 *)
a[ n_] := 2^(n + 2) - (2 + Mod[n + 1, 2]) 2^Quotient[n + 1, 2]; (* Michael Somos, Jun 24 2018 *)
PROG
(PARI) {a(n) = 2^(n+2) - (2 + (n+1)%2) * 2^((n+1)\2)}; /* Michael Somos, Jun 24 2018 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved