A005565 Number of walks on square lattice.
20, 75, 189, 392, 720, 1215, 1925, 2904, 4212, 5915, 8085, 10800, 14144, 18207, 23085, 28880, 35700, 43659, 52877, 63480, 75600, 89375, 104949, 122472, 142100, 163995, 188325, 215264, 244992, 277695, 313565, 352800, 395604, 442187, 492765, 547560, 606800
Offset: 0
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- R. K. Guy, Letter to N. J. A. Sloane, May 1990
- R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6.
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Programs
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Magma
[1/4*(n^4+14*n^3+69*n^2+136*n+80): n in [0..40]]; // Vincenzo Librandi, May 24 2012
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Maple
seq(add (k^3-n^2, k =0..n), n=4..28 ); # Zerinvary Lajos, Aug 26 2007 A005565:=(-20+25*z-14*z**2+3*z**3)/(z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation
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Mathematica
CoefficientList[Series[(20-25x+14x^2-3x^3)/(1-x)^5,{x,0,40}],x] (* Vincenzo Librandi, May 24 2012 *) LinearRecurrence[{5,-10,10,-5,1},{20,75,189,392,720},40] (* Harvey P. Dale, Dec 04 2020 *) Differences[Table[Sum[x^3 - y^2, {x, 0, g}, {y, x, g}], {g, 3, 30}]] (* Horst H. Manninger, Jun 19 2025 *) -
PARI
a(n)=(n^4+14*n^3+69*n^2+136*n)/4+20 \\ Charles R Greathouse IV, Nov 22 2011
Formula
From Ralf Stephan, Apr 23 2004: (Start)
a(n) = (1/4)*(n^4+14n^3+69n^2+136n+80).
G.f.: (20-25x+14x^2-3x^3)/(1-x)^5. (End)
a(n) = binomial(n+4,2)^2 - binomial(n+4,1)^2. - Gary Detlefs, Nov 22 2011
Using two consecutive triangular numbers t(n) and t(n+1), starting at n=3, compute the determinant of a 2 X 2 matrix with the first row t(n), t(n+1) and the second row t(n+1), 2*t(n+1). This gives (n+1)^2*(n-2)*(n+2)/4 = a(n-3). - J. M. Bergot, May 17 2012
E.g.f.: exp(x)*(80 + 220*x + 118*x^2 + 20*x^3 + x^4)/4. - Stefano Spezia, Jun 20 2025