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A002002
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a(n) = Sum_{k=0..n-1} binomial(n,k+1) * binomial(n+k,k).
(Formerly M3938 N1621)
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29
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0, 1, 5, 25, 129, 681, 3653, 19825, 108545, 598417, 3317445, 18474633, 103274625, 579168825, 3256957317, 18359266785, 103706427393, 586889743905, 3326741166725, 18885056428537, 107347191941249, 610916200215241
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OFFSET
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0,3
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COMMENTS
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Array interpretation (first row and column are the natural numbers):
1 2 3 ..j ... if b(i,j) = b(i-1,j) + b(i-1,j-1) + b(i,j-1) then a(n+1) = b(n,n)
2 5 .........
.............
i........... b(i,j)
(End)
Number of ordered trees with 2n edges, having root of even degree, nonroot nodes of outdegree at most 2 and branches of odd length. - Emeric Deutsch, Aug 02 2002
Coefficient of x^n in ((1-x)/(1-2x))^n, n>0. - Michael Somos, Sep 24 2003
Number of peaks in all Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0). Example: a(2)=5 because HH, HU*D, U*DH, UHD, U*DU*D, UU*DD contain 5 peaks (indicated by *). - Emeric Deutsch, Dec 06 2003
a(n) is the total number of HHs in all Schroeder (n+1)-paths. Example: a(2)=5 because UH*HD, H*H*H, UDH*H, H*HUD contain 5 HHs (indicated by *) and the other 18 Schroeder 3-paths contain no HHs. - David Callan, Jul 03 2006
a(n) is the total number of Hs in all Schroeder n-paths. Example: a(2)=5 as the Schroeder 2-paths are HH, DUH, DHU, HDU, DUDU and DDUU, and there are 5 H's. In general, a(n) is the total number of H..Hs (m+1 H's) in all Schroeder (n+m)-paths. - FUNG Cheok Yin, Jun 19 2021
a(n) is the number of points in Z^(n+1) that are L1 (Manhattan) distance <= n from the origin, or the number of points in Z^n that are L1 distance <= n+1 from the origin. These terms occur in the crystal ball sequences: a(n) here is the n-th term in the sequence for the (n+1)-dimensional cubic lattice as well as the (n+1)-st term in the sequence for the n-dimensional cubic lattice. See A008288 for a list of crystal ball sequences (rows or columns of A008288). - Shel Kaphan, Dec 25 2022 [Edited by Peter Munn, Jan 05 2023]
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REFERENCES
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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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E.g.f.: exp(3*x)*(BesselI(0, 2*sqrt(2)*x)+sqrt(2)*BesselI(1, 2*sqrt(2)*x)). - Vladeta Jovovic, Mar 28 2004
a(n) = Sum_{k=0..n-1} binomial(n-1, k)*binomial(n+k, k+1). - Paul Barry, Sep 20 2004
a(n) = n * hypergeom([n + 1, -n + 1], [2], -1) = (n+1)*LegendreP(n+1,3) - (5*n+3)*LegendreP(n,3))/(2*n) for n > 0. - Mark van Hoeij, Jul 12 2010
G.f.: -1 + 1 / ( 1 - x / (1 - 4*x / (1 - x^2 / (1 - 4*x / (1 - x^2 / (1 - 4*x / ...)))))). - Michael Somos, Jan 03 2013
D-finite with recurrence: 2*(6*n^2-12*n+5)*a(n-1)-(n-2)*(2*n-1)*a(n-2)-n*(2*n-3)*a(n)=0. - Vaclav Kotesovec, Oct 04 2012
D-finite (an alternative): n*a(n) = (6-n)*a(n-6) + (14*n-72)*a(n-5) + (264-63*n)*a(n-4) + 100*(n-3)*a(n-3) + (114-63*n)*a(n-2) + 2*(7*n-6)*a(n-1), n >= 7. - Fung Lam, Feb 05 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} (-2)^k*binomial(n-1,k)*binomial(n+k,k) and n^3*a(n) = Sum_{k=0..n-1} (4*k^3+4*k^2+4*k+1)*binomial(n-1,k)*binomial(n+k,k). For each of the two equalities, both sides satisfy the same recurrence -- this follows from the Zeilberger algorithm. - Zhi-Wei Sun, Aug 30 2014
a(n) = hypergeom([1-n, -n], [1], 2) for n >= 1. - Peter Luschny, Nov 19 2014
L.g.f.: L(x) = Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1) * (1-x)^(-n) / n! = Sum_{n>=1} a(n)*x^n/n where exp(L(x)) = g.f. of A001003. - Paul D. Hanna, May 17 2015
a(n+1) = (1/2^(n+1)) * Sum_{k >= 0} (1/2^k) * binomial(n + k, n)*binomial(n + k, n + 1). - Peter Bala, Mar 02 2017
a(n) = (LegendreP(n,3) - LegendreP(n-1,3))/2. - Mark van Hoeij, Jul 14 2022
D-finite with recurrence n*a(n) +(-7*n+5)*a(n-1) +(7*n-16)*a(n-2) +(-n+3)*a(n-3)=0. - R. J. Mathar, Aug 01 2022
a(n) = (-1)^(n+1)*hypergeom( [n+1, -n+1], [1], 2) for n >= 1.
The Gauss congruences hold: a(n*p^r) == a(n^p^(r-1)) (mod p^r) for all primes p and all positive integers n and r. (End)
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EXAMPLE
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G.f. = x + 5*x^2 + 25*x^3 + 129*x^4 + 681*x^5 + 3653*x^6 + 19825*x^7 + 108545*x^8 + ...
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MAPLE
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A064861 := proc(n, k) option remember; if n = 1 then 1; elif k = 0 then 0; else A064861(n, k-1)+(3/2-1/2*(-1)^(n+k))*A064861(n-1, k); fi; end; seq(A064861(i, i+1), i=1..40);
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MATHEMATICA
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CoefficientList[Series[((1-x)/Sqrt[1-6x+x^2]-1)/2, {x, 0, 30}], x] (* Harvey P. Dale, Mar 17 2011 *)
a[ n_] := n Hypergeometric2F1[ n + 1, -n + 1, 2, -1] (* Michael Somos, Aug 09 2011 *)
a[ n_] := With[{m = Abs@n}, Sign[n] Sum[ Binomial[ m, k] Binomial[ m + k - 1, m], {k, m}]]; (* Michael Somos, Aug 09 2011 *)
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PROG
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(PARI) {a(n) = my(m = abs(n)); sign( n) * sum( k=0, m-1, binomial( m, k+1) * binomial( m+k, k))}; /* Michael Somos, Aug 09 2011 */
(Maxima) makelist(sum(binomial(n, k+1)*binomial(n+k, k), k, 0, n), n, 0, 21); \\ Bruno Berselli, May 19 2011
(Magma) [&+[Binomial(n, k+1)*Binomial(n+k, k): k in [0..n]]: n in [0..21]]; // Bruno Berselli, May 19 2011
(Sage)
a = lambda n: hypergeometric([1-n, -n], [1], 2) if n>0 else 0
(PARI) /* L.g.f.: Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1)*(1-x)^(-n)/n! */
{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=1); A=(sum(m=1, n+1, Dx(m-1, x^(2*m-1)/(1-x)^m/m!)+x*O(x^n))); n*polcoeff(A, n)}
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CROSSREFS
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a(n)=T(n, n+1), array T as in A050143.
a(n)=T(n, n+1), array T as in A064861.
Half the first differences of central Delannoy numbers (A001850).
a(n)=T(n, n+1), array T as in A008288.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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