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A130515 In triangular peg solitaire, number of distinct feasible pairs starting with one peg missing and finishing with one peg. 2
1, 4, 3, 17, 29, 27, 80, 125, 108, 260, 356, 300, 637, 832, 675, 1341, 1665, 1323, 2500, 3025, 2352, 4304, 5072, 3888, 6929, 8036, 6075, 10625, 12125, 9075, 15616, 17629, 13068, 22212, 24804, 18252, 30685, 34000, 24843, 41405, 45521 (list; graph; refs; listen; history; text; internal format)
OFFSET

2,2

COMMENTS

Coincides with A130516 for n >= 6.

LINKS

George I. Bell, Table of n, a(n) for n = 2..52

George I. Bell, Solving Triangular Peg Solitaire [arXiv:math/0703865v4]

G. I. Bell, Solving Triangular Peg Solitaire, JIS 11 (2008) 08.4.8

Index entries for linear recurrences with constant coefficients, signature (0, 0, 3, 0, 0, -1, 0, 0, -5, 0, 0, 5, 0, 0, 1, 0, 0, -3, 0, 0, 1).

FORMULA

Reference gives an explicit formula for a(n).

G.f.: -x^2*(x^2+1) *(x^14 +4*x^13 +2*x^12 +10*x^11 +15*x^10 +8*x^9 +15*x^8 +34*x^7 +15*x^6 +8*x^5 +15*x^4 +10*x^3 +2*x^2 +4*x +1) / ( (1+x)^2 *(x^2-x+1)^2 *(x-1)^5 *(1+x+x^2)^5 ). - R. J. Mathar, Sep 07 2015

a(n) = 3*a(n-3) -a(n-6) -5*a(n-9) +5*a(n-12) +a(n-15) -3*a(n-18) +a(n-21). - R. J. Mathar, Sep 07 2015

MAPLE

A130515 := proc(n)

    t := n*(n+1)/2 ;

    if modp(n, 3) = 1 then

        (t-1)^2/27 ;

    elif type(n, 'even') then

        (4*t^2+9*n^2)/72 ;

    else

        (4*t^2+9*(n+1)^2)/72 ;

    fi;

end proc: # R. J. Mathar, Sep 07 2015

PROG

(PARI) a(n) = {my(T = n*(n+1)/2); if (n % 3 == 1, (T-1)^2/27, if ( n % 2 == 0, (4*T^2 + 9*n^2)/72, (4*T^2 + 9*(n+1)^2)/72; ); ); }  \\ Michel Marcus, Apr 21 2013

CROSSREFS

Cf. A130516.

Sequence in context: A060509 A113203 A034486 * A276083 A161893 A192773

Adjacent sequences:  A130512 A130513 A130514 * A130516 A130517 A130518

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane, Aug 09 2007

STATUS

approved

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Last modified December 2 23:29 EST 2016. Contains 278694 sequences.