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 A065450 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. 1
 1, 2, 10, 22, 37, 54, 76, 100, 129, 160, 196, 234, 277, 322, 372, 424, 481, 540, 604, 670, 741, 814, 892, 972, 1057, 1144, 1236, 1330, 1429, 1530, 1636, 1744, 1857, 1972, 2092, 2214, 2341, 2470, 2604, 2740, 2881, 3024, 3172, 3322, 3477, 3634, 3796, 3960 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 LINKS FORMULA 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 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 CROSSREFS Cf. A098498. Sequence in context: A339486 A273993 A225290 * A090288 A331132 A032526 Adjacent sequences:  A065447 A065448 A065449 * A065451 A065452 A065453 KEYWORD nonn,easy AUTHOR Bodo Zinser, Nov 18 2001 EXTENSIONS More terms from Don Reble, Nov 28 2001 STATUS approved

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Last modified September 22 02:48 EDT 2021. Contains 347605 sequences. (Running on oeis4.)