%I #8 Oct 19 2017 03:13:57
%S 1,2,10,22,37,54,76,100,129,160,196,234,277,322,372,424,481,540,604,
%T 670,741,814,892,972,1057,1144,1236,1330,1429,1530,1636,1744,1857,
%U 1972,2092,2214,2341,2470,2604,2740,2881,3024,3172,3322,3477,3634,3796,3960
%N Make an infinite chessboard from the squares in the first quadrant; sequence gives number of squares a knight can reach in n moves starting at the origin.
%C The first conjecture is true: Partial sums of A047356 = b(n) = (14*(n*(n+1)) + 2*n + 5 + 3*(-1)^n )/8, since A047356(n)=(14*n+1+3*(-1)^n)/4. And b(n) has g.f. (4*x^2 + 2*x + 1)/(-x^4 + 2*x^3 - 2*x + 1). The difference a(n) - b(n) = 0,2,2,0,0,0,0,0,0..., which has g.f. 2*x^2 + 2*x. Then (4*x^2 + 2*x + 1)/(-x^4 + 2*x^3 - 2*x + 1) + 2*x^2 + 2*x = (-2*x^6 + 2*x^5 + 4*x^4 - 4*x^3 + 2*x^2 + 4*x + 1)/(-x^4 + 2*x^3 - 2*x + 1). - _Vim Wenders_, Apr 16 2008
%F Conjectures: G.f.: [1+6x^2+4x^3-4x^4-2x^5+2x^6]/[(1+x)*(1-x)^3]. For n>3, partial sums of A047356. - _Ralf Stephan_, Mar 06 2004
%F The second conjecture "For n>3, partial sums of A047356" is also true. From the last possible move, we can either move back to the second last possible move or to b(n)=A047883(n) new squares. So a(n) = a(n-2)+b(n). For n>6, b(n)=7(n-1)+4=A017029(n-1). But a number of the form 7n+4 is naturally the sum of two consecutive terms in A047356 (4=1+3,11=3+8,18=8+10, ...). The conjecture follows. - _Vim Wenders_, Apr 12 2008
%Y Cf. A098498.
%K nonn,easy
%O 0,2
%A _Bodo Zinser_, Nov 18 2001
%E More terms from _Don Reble_, Nov 28 2001