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Search: id:A065450
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| A065450 |
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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. |
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| 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; listen)
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OFFSET
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0,2
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COMMENT
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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 (vim(AT)gmx.li), Apr 16 2008
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FORMULA
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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 (ralf(AT)ark.in-berlin.de), 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 (vim(AT)gmx.li), Apr 12 2008
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CROSSREFS
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Cf. A098498.
Adjacent sequences: A065447 A065448 A065449 this_sequence A065451 A065452 A065453
Sequence in context: A080881 A138298 A119153 this_sequence A090288 A032526 A096183
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KEYWORD
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nonn,easy
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AUTHOR
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Bodo Zinser (BodoZinser(AT)Compuserve.com), Nov 18 2001
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EXTENSIONS
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More terms from Don Reble (djr(AT)nk.ca), Nov 28 2001
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