Displaying 1-10 of 28 results found.
a(n) = Sum_{k=0..n-1} binomial(n-1,k)*binomial(n+k,k). Also a(n) = T(n,n), array T as in A049600.
+20
21
0, 1, 4, 19, 96, 501, 2668, 14407, 78592, 432073, 2390004, 13286043, 74160672, 415382397, 2333445468, 13141557519, 74174404608, 419472490257, 2376287945572, 13482186743203, 76598310928096, 435730007006341, 2481447593848524, 14146164790774359
COMMENTS
Also main diagonal of array: m(i,1)=1, m(1,j)=j, m(i,j)=m(i,j-1)+m(i-1,j-1)+m(i-1,j): 1 2 3 4 ... / 1 4 9 16 ... / 1 6 19 44 ... / 1 8 33 96 ... /. - Benoit Cloitre, Aug 05 2002 This array is now listed as A142978, where some conjectural congruences for the present sequence are given. - Peter Bala, Nov 13 2008 Define a finite triangle T(r,c) with T(r,0) = binomial(n,r) for 0 <= r <= n and the other terms recursively with T(r,c) = T(r-1,c-1) + 2*T(r,c-1). The sum of the last terms in the rows is Sum_{r=0..n} T(r,r) = a(n+1). Example: For n=4 the triangle has the rows 1; 4 9; 6 16 41; 4 14 44 129; 1 6 26 96 321 having sum of last terms 1 + 9 + 41 + 129 + 321 = 501 = a(5). - J. M. Bergot, Feb 15 2013 a(n-1) for n > 1 is the number of assembly trees with the connected gluing rule for cycle graphs with n vertices. - Nick Mayers, Aug 16 2018
FORMULA
D-finite with recurrence n*(2*n-3)*a(n) - (12*n^2-24*n+8)*a(n-1) + (2*n-1)*(n-2)*a(n-2) = 0. - Vladeta Jovovic, Aug 29 2004 a(n+1) = Sum_{k=0..n} binomial(n, k)*binomial(n+1, k+1)*2^k. - Paul Barry, Sep 20 2004 a(n) = Sum_{k=0..n} T(n, k), array T as in A008288. If shifted one place left, the third binomial transform of A098660. - Paul Barry, Sep 20 2004 E.g.f. for sequence shifted left: Sum_{n>=0} a(n+1)*x^n/n! = exp(3*x)*(BesselI(0, 2*sqrt(2)*x)+BesselI(1, 2*sqrt(2)*x)/sqrt(2)). - Paul Barry, Sep 20 2004 a(n) = Sum_{k=0..n-1} C(n,k)*C(n-1,k)*2^(n-k-1); a(n+1) = 2^n*Hypergeometric2F1(-n,-n-1;1;1/2). - Paul Barry, Feb 08 2011 Recurrence (an alternative): n*a(n) = (6-n)*a(n-6) + 2*(5*n-27)*a(n-5) + (84-15*n)*a(n-4) + 52*(3-n)*a(n-3) + 3*(2-5*n)*a(n-2) + 2*(5*n-3)*a(n-1), n >= 7. - Fung Lam, Feb 05 2014 a(n) = Hyper2F1([-n, n], [1], -1)/2 for n > 0. - Peter Luschny, Aug 02 2014 n^2*a(n) = Sum_{k=0..n-1} (2*k^2+2*k+1)*binomial(n-1,k)*binomial(n+k,k). By the Zeilberger algorithm, both sides of the equality satisfy the same recurrence. - Zhi-Wei Sun, Aug 30 2014 a(n) = [x^n] (1/2) * ((1+x)/(1-x))^n for n > 0. - Seiichi Manyama, Jun 07 2018
MAPLE
a := proc(n) local k; add(binomial(n-1, k)*binomial(n+k, k), k=0..n-1); end;
MATHEMATICA
Table[SeriesCoefficient[x*((1+x)-Sqrt[1-6*x+x^2])/(4*x*Sqrt[1-6*x+x^2]), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 08 2012 *) a[n_] := Hypergeometric2F1[1-n, n+1, 1, -1]; a[0] = 0; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Feb 26 2013 *) a[n_] := Sum[ Binomial[n - 1, k] Binomial[n + k, k], {k, 0, n - 1}]; Array[a, 25] (* Robert G. Wilson v, Aug 08 2018 *)
PROG
(Maxima) makelist(if n=0 then 0 else sum(binomial(n-1, k)*binomial(n+k, k), k, 0, n-1), n, 0, 22); \\ Bruno Berselli, May 19 2011 (Magma) [n eq 0 select 0 else &+[Binomial(n-1, k)*Binomial(n+k, k): k in [0..n-1]]: n in [0..22]]; // Bruno Berselli, May 19 2011 (Haskell)
(Python)
from sympy import binomial
def a(n):
return sum(binomial(n - 1, k) * binomial(n + k, k) for k in range(n))
(Python)
from math import comb
def A047781(n): return sum(comb(n, k)**2*k<<k-1 for k in range(1, n+1))//n if n else 0 # Chai Wah Wu, Mar 22 2023
a(n) = T(n,2), array T as in A049600.
+20
20
0, 1, 4, 13, 38, 104, 272, 688, 1696, 4096, 9728, 22784, 52736, 120832, 274432, 618496, 1384448, 3080192, 6815744, 15007744, 32899072, 71827456, 156237824, 338690048, 731906048, 1577058304, 3388997632, 7264534528, 15535702016
COMMENTS
We may ask for the number of one-step transitions (NOOST) between all unlabeled hierarchies of n elements with the restriction that no subhierarchies are allowed. As an example, consider n = 4 and the hierarchy H1 = [[2,2]] with two elements on level 1 and two on level 2. Starting from H1 the hierarchies [[1, 3]], [[2, 1, 1]], [[1, 2, 1]] can be reached by moving one element only, but [[1, 1, 2]] cannot be reached in a one-step transitition. The solution is n = 1, NOOST = 0; n = 2, NOOST = 1; n = 3, NOOST = 4; n = 4, NOOST = 13; n = 5, NOOST = 38; n = 6, NOOST = 104; n = 7, NOOST = 272; n = 8, NOOST = 688; n = 9, NOOST = 1696. This is sequence A049611. - Thomas Wieder, Feb 28 2004 If X_1,X_2,...,X_n are 2-blocks of a (2n+2)-set X then, for n>=1, a(n+1) is the number of (n+2)-subsets of X intersecting each X_i, (i=1,2,...,n). - Milan Janjic, Nov 18 2007 In each composition (ordered partition) of the integer n, circle the first summand once, circle the second summand twice, etc. a(n) is the total number of circles in all compositions of n (that is, add k*(k+1)/2 for each composition into k parts). Note the O.g.f. is B(A(x)) where A(x)= x/(1-x) and B(x)= x/(1-x)^3.
This is the Riordan transform with the Riordan matrix A097805 (of the associated type) of the triangular number sequence A000217. See a Feb 17 2017 comment on A097805. - Wolfdieter Lang, Feb 17 2017
FORMULA
G.f.: x*(1-x)^2/(1-2*x)^3.
Binomial transform of quarter squares A002620(n+1): a(n) = Sum_{k=0..n} binomial(n, k)*floor((k+1)^2/4). - Paul Barry, May 27 2003 a(n) = 2^(n-4)*(n^2+5*n+2) - 0^n/8. - Paul Barry, Jun 09 2003 a(n) = Hyper2F1([-n+1, 3], [1], -1) for n>0. - Peter Luschny, Aug 02 2014 a(n) = Sum_{k=0..n-1} Sum_{j=0..n-1} Sum_{i=0..n-1} binomial(n-1, i+j+k). - Yalcin Aktar, Aug 27 2023
MATHEMATICA
CoefficientList[Series[x (1-x)^2/(1-2x)^3, {x, 0, 40}], x] (* Harvey P. Dale, Sep 24 2013 *)
a(n) = T(n,3), array T as in A049600.
+20
11
0, 1, 5, 19, 63, 192, 552, 1520, 4048, 10496, 26624, 66304, 162560, 393216, 940032, 2224128, 5214208, 12124160, 27983872, 64159744, 146210816, 331350016, 747110400, 1676673024, 3746562048, 8338276352, 18488492032
COMMENTS
If X_1, X_2, ..., X_n are 2-blocks of a (2n+3)-set X then, for n >= 1, a(n+1) is the number of (n+3)-subsets of X intersecting each X_i, (i=1,2,...,n). - Milan Janjic, Nov 18 2007
REFERENCES
Robert Cori, Gabor Hetyei, Genus one partitions, in 26th International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2014), 2014, Chicago, United States. Discrete Mathematics and Theoretical Computer Science, DMTCS Proceedings vol. AT, pp. 333-344, <hal-01207612>
LINKS
R. Cori and G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344. S. Kitaev and J. Remmel, p-Ascent Sequences, arXiv preprint arXiv:1503.00914 [math.CO], 2015.
FORMULA
G.f.: x*(1-x)^3 /(1-2*x)^4.
a(n) = Sum_{k=0..floor((n+3)/2)} C(n+3, 2k)*C(k+1, 3). - Paul Barry, May 15 2003 a(n+1) = 2^n*n^3/48 + 5*2^n*n^2/16 + 7*2^n*n/6 + 2^n, n>=1. - Milan Janjic, Nov 18 2007 Binomial transform of the tetrahedral numbers A000292 when omitting the initial 0 in both sequences. - Carl Najafi, Sep 08 2011
MATHEMATICA
CoefficientList[Series[x (1-x)^3/(1-2x)^4, {x, 0, 30}], x] (* or *) Join[ {0}, LinearRecurrence[{8, -24, 32, -16}, {1, 5, 19, 63}, 30]] (* Harvey P. Dale, Jan 07 2014 *)
Triangle with columns built from row sums of the partial row sums triangles obtained from Pascal's triangle A007318. Essentially A049600 formatted differently.
+20
9
1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 8, 4, 1, 1, 16, 20, 13, 5, 1, 1, 32, 48, 38, 19, 6, 1, 1, 64, 112, 104, 63, 26, 7, 1, 1, 128, 256, 272, 192, 96, 34, 8, 1, 1, 256, 576, 688, 552, 321, 138, 43, 9, 1, 1, 512, 1280, 1696, 1520, 1002, 501, 190, 53, 10, 1, 1, 1024, 2816, 4096
COMMENTS
In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is 1/((1-z)*(1-x*z*(1-z)/(1-2*z))). Column m (without leading zeros) is obtained from convolution of A000012 (powers of 1) with m-fold convoluted A011782.
FORMULA
a(n, m)= Am(n, 0) if n >= m >= 0 and a(n, m) := 0 if n<m; i.e. first column of the lower triangular matrix Am := prs^(m)( A007318) with the lower triangular matrix A007318 (Pascal triangle) and prs^(m) is the partial row sums (prs) mapping for triangular matrices applied m times. See e.g. A055584 for m=4. G.f. for column m: (1/(1-x))*(x*(1-x)/(1-2*x))^m, m >= 0.
T(n, k) = sum_{j=0..n-k} C(n-k, j)*C(k+j-1, k-1). - Paul D. Hanna, Jan 14 2004
EXAMPLE
{1}; {1, 1}; {1, 2, 1}; {1, 4, 3, 1}; {1, 8, 8, 4, 1}; ...
Fourth row polynomial (n=3): p(3,x)= 1+4*x+3*x^2+x^3
MATHEMATICA
t[n_, k_] := Hypergeometric2F1[k, k-n, 1, -1]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2014, after Paul D. Hanna *)
PROG
(PARI) {T(n, k) = if( n<0 || k<0, 0, polcoeff( polcoeff( 1 / ((1 - z) * (1 - x*z * (1 - z) / (1 - 2*z) + z * O(z^n) + x * O(x^k))), k), n))}; /* Michael Somos, Sep 30 2003 */ (PARI) {T(n, k)=if(k>n||n<0||k<0, 0, if(k==0||k==n, 1, sum(j=0, n-k, binomial(n-k, j)*binomial(k+j-1, k-1)); ); )} (Hanna)
Number of loopless rooted planar maps with 3 faces and n vertices and no isthmuses. Also a(n)=T(4,n-3), array T as in A049600.
(Formerly M4490)
+20
4
1, 8, 20, 38, 63, 96, 138, 190, 253, 328, 416, 518, 635, 768, 918, 1086, 1273, 1480, 1708, 1958, 2231, 2528, 2850, 3198, 3573, 3976, 4408, 4870, 5363, 5888, 6446, 7038, 7665, 8328, 9028, 9766, 10543, 11360, 12218, 13118, 14061, 15048
COMMENTS
If Y_i (i=1,2,3) are 2-blocks of an n-set X then, for n>=6, a(n-3) is the number of (n-3)-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Nov 09 2007 a(n) is also the number of triangle subgraphs in a complete graph on n+3 vertices, minus 3 non-incident edges, for n > 2. - Robert H Cowen, Jun 23 2018
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
G.f.: x^2*(1+4*x-6*x^2+2*x^3)/(1-x)^4.
a(n-3) = (1/6)*n^3-(1/2)*n^2-(8/3)*n+6, n=6,7,... - Milan Janjic, Nov 09 2007 a(2)=1, a(3)=8, a(4)=20, a(5)=38, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Aug 25 2013 a(n+2) = Hyper2F1([-3, n], [1], -1). - Peter Luschny, Aug 02 2014
MAPLE
a := n -> hypergeom([-3, n-2], [1], -1);
seq(round(evalf(a(n), 32)), n=2..41); # Peter Luschny, Aug 02 2014
MATHEMATICA
CoefficientList[Series[(1+4x-6x^2+2x^3)/(1-x)^4, {x, 0, 50}], x] (* or *) LinearRecurrence[{4, -6, 4, -1}, {1, 8, 20, 38}, 50] (* Harvey P. Dale, Aug 25 2013 *) f[n_]:= Binomial[n, 3] - 3(n-2); Table[{n, f[n]}, {n, 5, 100}]//TableForm (* Robert H Cowen, Jun 23 2018 *)
PROG
(PARI) Vec((1+4*x-6*x^2+2*x^3)/(1-x)^4 + O(x^40)) \\ Andrew Howroyd, Jul 15 2018
a(n)=Sum{T(2i,n-2i): i=0,1,...,[ n/2 ]}, array T as in A049600.
+20
1
0, 0, 2, 3, 12, 25, 76, 182, 504, 1275, 3410, 8811, 23256, 60580, 159094, 415715, 1089648, 2850645, 7466468, 19541994, 51170460, 133951675, 350713222, 918141623, 2403786672, 6293097000, 16475700746, 43133687427, 112925875764
FORMULA
G.f.: x^2*(-2+x) / ( (x^2-x-1)*(x^2-3*x+1) ).
a(n) = (Fibonacci(2*n)+(-1)^n*Fibonacci(n))/2. - Vladeta Jovovic, Aug 30 2004
a(n)=T(n,n+2), array T as in A049600.
+20
1
0, 1, 6, 34, 190, 1059, 5908, 33028, 185076, 1039525, 5851626, 33006438, 186519138, 1055789511, 5985405000, 33979107336, 193143097288
FORMULA
Conjecture: (n-1)*(2*n-1)*(n+2)*a(n) -12*n*(n-1)*(n+1)*a(n-1) +(n-2)*(2*n+1)*(n+1)*a(n-2)=0. - R. J. Mathar, Oct 26 2015
a(n)=T(n,n+3), array T as in A049600.
+20
0
0, 1, 7, 43, 253, 1462, 8378, 47818, 272422, 1550927, 8829033, 50276013, 286430763, 1632808572, 9313861092, 53163187748, 303653552188
FORMULA
Conjecture: -(n+3)*(n-1)*a(n) +(7*n^2+8*n-9)*a(n-1) +(-7*n^2+8*n+9)*a(n-2) +(n+1)*(n-3)*a(n-3)=0. - R. J. Mathar, Oct 26 2015 Conjecture: +n*(n+3)*(n-1)*a(n) -(2*n+1)*(3*n^2+3*n-4)*a(n-1) +(n-2)*(n+2)*(n+1)*a(n-2)=0. - R. J. Mathar, Oct 26 2015
a(n) = (n+2)*2^(n-1).
(Formerly M2739 N1100)
+10
211
1, 3, 8, 20, 48, 112, 256, 576, 1280, 2816, 6144, 13312, 28672, 61440, 131072, 278528, 589824, 1245184, 2621440, 5505024, 11534336, 24117248, 50331648, 104857600, 218103808, 452984832, 939524096, 1946157056, 4026531840, 8321499136, 17179869184, 35433480192
COMMENTS
Number of parts in all compositions (ordered partitions) of n + 1. For example, a(2) = 8 because in 3 = 2 + 1 = 1 + 2 = 1 + 1 + 1 we have 8 parts. Also number of compositions (ordered partitions) of 2n + 1 with exactly 1 odd part. For example, a(2) = 8 because the only compositions of 5 with exactly 1 odd part are 5 = 1 + 4 = 2 + 3 = 3 + 2 = 4 + 1 = 1 + 2 + 2 = 2 + 1 + 2 = 2 + 2 + 1. - Emeric Deutsch, May 10 2001 Binomial transform of natural numbers [1, 2, 3, 4, ...].
For n >= 1 a(n) is also the determinant of the n X n matrix with 3's on the diagonal and 1's elsewhere. - Ahmed Fares (ahmedfares(AT)my-deja.com), May 06 2001
The arithmetic mean of first n terms of the sequence is 2^(n-1). - Amarnath Murthy, Dec 25 2001, corrected by M. F. Hasler, Dec 17 2016 Also the number of "winning paths" of length n across an n X n Hex board. Satisfies the recursion a(n) = 2a(n-1) + 2^(n-2). - David Molnar (molnar(AT)stolaf.edu), Apr 10 2002
Let M_n be the n X n matrix m_(i, j) = 1 + abs(i-j) then det(M_n) = (-1)^(n-1)*a(n-1). - Benoit Cloitre, May 28 2002 Absolute value of determinant of n X n matrix of form: [1 2 3 4 5 / 2 1 2 3 4 / 3 2 1 2 3 / 4 3 2 1 2 / 5 4 3 2 1]. - Benoit Cloitre, Jun 20 2002 This sequence also emerges as a floretion force transform of powers of 2 (see program code). Define a(-1) = 0 (as the sequence is returned by FAMP). Then a(n-1) + A098156(n+1) = 2*a(n) (conjecture). - Creighton Dement, Mar 14 2005 This sequence gives the absolute value of the determinant of the Toeplitz matrix with first row containing the first n integers. - Paul Max Payton, May 23 2006 Equals sums of rows right of left edge of A102363 divided by three, + 2^K. - David G. Williams (davidwilliams(AT)paxway.com), Oct 08 2007 If X_1, X_2, ..., X_n are 2-blocks of a (2n+1)-set X then, for n >= 1, a(n) is the number of (n+1)-subsets of X intersecting each X_i, (i = 1, 2, ..., n). - Milan Janjic, Nov 18 2007 Also, a(n-1) is the determinant of the n X n matrix with A[i, j] = n - |i-j|. - M. F. Hasler, Dec 17 2008 1/2 the number of permutations of 1 .. (n+2) arranged in a circle with exactly one local maximum. - R. H. Hardin, Apr 19 2009 The first corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n) is the number of runs of consecutive 1's in all binary sequences of length (n+1). - Geoffrey Critzer, Jul 02 2009 Let X_j (0 < j <= 2^n) all the subsets of N_n; m(i, j) := if {i} in X_j then 1 else 0. Let A = transpose(M).M; Then a(i, j) = (number of elements)|X_i intersect X_j|. Determinant(X*I-A) = (X-(n+1)*2^(n-2))*(X-2^(n-2))^(n-1)*X^(2^n-n).
Eigenvector for (n+1)*2^(n-2) is V_i=|X_i|.
Sum_{k=1..2^n} |X_i intersect X_k|*|X_k| = (n+1)*2^(n-2)*|X_i|.
Eigenvectors for 2^(n-2) are {line(M)[i] - line(M)[j], 1 <= i, j <= n}. - CLARISSE Philippe (clarissephilippe(AT)yahoo.fr), Mar 24 2010
The sequence b(n) = 2* A001792(n), for n >= 1 with b(0) = 1, is an elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 187, 190, 250 and 442, lead to the b(n) sequence. For the corner squares these vectors lead to the companion sequence A134401. - Johannes W. Meijer, Aug 15 2010 Equals partial sums of A045623: (1, 2, 5, 12, 28, ...); where A045623 = the convolution square of (1, 1, 2, 4, 8, 16, 32, ...). - Gary W. Adamson, Oct 26 2010 Equals (1, 2, 4, 8, 16, ...) convolved with (1, 1, 2, 4, 8, 16, ...); e.g., a(3) = 20 = (1, 1, 2, 4) dot (8, 4, 2, 1) = (8 + 4 + 4 + 4). - Gary W. Adamson, Oct 26 2010 This sequence seems to give the first x+1 nonzero terms in the sequence derived by subtracting the m-th term in the x_binacci sequence (where the first term is one and the y-th term is the sum of x terms immediately preceding it) from 2^(m-2). - Dylan Hamilton, Nov 03 2010 Recursive formulas for a(n) are in many cases derivable from its property wherein delta^k(a(n)) - a(n) = k*2^n where delta^k(a(n)) represents the k-th forward difference of a(n). Provable with a difference table and a little induction. - Ethan Beihl, May 02 2011 Let f(n,k) be the sum of numbers in the subsets of size k of {1, 2, ..., n}. Then a(n-1) is the average of the numbers f(n, 0), ... f(n, n). Example: (f(3, 1), f(3, 2), f(3, 3)) = (6, 12, 6), with average (6+12+6)/3. - Clark Kimberling, Feb 24 2012 a(n) is the number of length-2n binary sequences that contain a subsequence of ones with length n or more. To derive this result, note that there are 2^n sequences where the initial one of the subsequence occurs at entry one. If the initial one of the subsequence occurs at entry 2, 3, ..., or n + 1, there are 2^(n-1) sequences since a zero must precede the initial one. Hence a(n) = 2^n + n*2^(n-1)=(n+2)2^(n-1). An example is given in the example section below. - Dennis P. Walsh, Oct 25 2012 As the total number of parts in all compositions of n+1 (see the first line in Comments) the equivalent sequence for partitions is A006128. On the other hand, as the first differences of A001787 (see the first line in Crossrefs) the equivalent sequence for partitions is A138879. - Omar E. Pol, Aug 28 2013 a(n) is the number of spanning trees of the complete tripartite graph K_{n,1,1}. - James Mahoney, Oct 24 2013 a(n-1) = denominator of the mean (2n/(n+1), after reduction), of the compositions of n; numerator is given by A022998(n). - Clark Kimberling, Mar 11 2014 The shifted array belongs to an interpolated family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the interpolating o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x(1-x)/(1+(t-1)x(1-x)). See A091867 for more info on this family. Here the interpolation is t=-3 (mod signs in the results). Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t). Shifted o.g.f: G(x) = x*(1-x)/(1 - 4x*(1-x)) = P[Cinv(x),-4].
Inverse o.g.f: Ginv(x) = [1 - sqrt(1 - 4*x/(1+4x))]/2 = C[P(x, 4)] (signed shifted A001700). Cf. A030528. (End) For n > 0, element a(n) of the sequence is equal to the gradients of the (n-1)-th row of Pascal triangle multiplied with the square of the integers from n+1,...,1. I.e., row 3 of Pascal's triangle 1,3,3,1 has gradients 1,2,0,-2,-1, so a(4) = 1*(5^2) + 2*(4^2) + 0*(3^2) - 2*(2^2) - 1*(1^2) = 48. - Jens Martin Carlsson, May 18 2017 Number of self-avoiding paths connecting all the vertices of a convex (n+2)-gon. - Ivaylo Kortezov, Jan 19 2020 a(n-1) is the total number of elements of subsets of {1,2,..,n} that contain n. For example, for n = 3, a(2) = 8, and the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, with a total of 8 elements. - Enrique Navarrete, Aug 01 2020
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
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).
A. M. Stepin and A. T. Tagi-Zade, Words with restrictions, pp. 67-74 of Kvant Selecta: Combinatorics I, Amer. Math. Soc., 2001 (G_n on p. 70).
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. Milan Janjić and Boris Petković, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. C. W. Jones, J. C. P. Miller, J. F. C. Conn and R. C. Pankhurst, Tables of Chebyshev polynomials, Proc. Roy. Soc. Edinburgh. Sect. A., Vol. 62, No. 2 (1946), pp. 187-203. Sergey Kitaev and Jeffrey Remmel, p-Ascent Sequences, arXiv preprint arXiv:1503.00914 [math.CO], 2015.
FORMULA
a(n) = (n+2)*2^(n-1).
G.f.: (1 - x)/(1 - 2*x)^2 = 2F1(1,3;2;2x).
a(n) = 4*a(n-1) - 4*a(n-2).
a(n) = Sum_{k=0..n+2} binomial(n+2, 2k)*k. - Paul Barry, Mar 06 2003 With a leading 0, this is ((n+1)2^n - 0^n)/4 = Sum_{m=0..n} binomial(n - 1, m - 1)*m, the binomial transform of A004526(n+1). - Paul Barry, Jun 05 2003 a(n) = 1 + Sum_{k=1..n} (n - k + 4)2^(n - k - 1). This follows from the result that the number of parts equal to k in all compositions of n is (n - k + 3)2^(n - k - 2) for 0 < k < n. - Geoffrey Critzer, Sep 21 2008 a(n) = 2^(n-1) + 2 a(n-1) ; a(n-1) = det(n - |i - j|)_{i, j = 1..n}. - M. F. Hasler, Dec 17 2008 a(n) = Sum_{i=1..2^n} gcd(i, 2^n) = A018804(2^n). So we have: 2^0 * phi(2^n) + ... + 2^n * phi(2^0) = (n + 2)*2^(n-1), where phi is the Euler totient function. - Jeffrey R. Goodwin, Nov 11 2011 a(n) = Sum_{j=0..n} Sum_{i=0..n} binomial(n, i + j). - Yalcin Aktar, Jan 17 2012 Eigensequence of an infinite lower triangular matrix with 2^n as the left border and the rest 1's. - Gary W. Adamson, Jan 30 2012 G.f.: 1 + 2*x*U(0) where U(k) = 1 + (k + 1)/(2 - 8*x/(4*x + (k + 1)/U(k + 1))); (continued fraction, 3 - step). - Sergei N. Gladkovskii, Oct 19 2012 a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n,j). - Peter Luschny, Dec 03 2013 G.f.: 1 / (1 - 3*x / (1 + x / (3 - 4*x))). - Michael Somos, Aug 26 2015 a(n) = - A053120(2+n, n), n >= 0, the negative of the third (sub)diagonal of the triangle of Chebyshev's T polynomials. - Wolfdieter Lang, Nov 26 2019 Sum_{n>=0} 1/a(n) = 8*log(2) - 4.
Sum_{n>=0} (-1)^n/a(n) = 4 - 8*log(3/2). (End)
EXAMPLE
a(0) = 1, a(1) = 2*1 + 1 = 3, a(2) = 2*3 + 2 = 8, a(3) = 2*8 + 4 = 20, a(4) = 2*20 + 8 = 48, a(5) = 2*48 + 16 = 112, a(6) = 2*112 + 32 = 256, ... - Philippe Deléham, Apr 19 2009 a(2) = 8 since there are 8 length-4 binary sequences with a subsequence of ones of length 2 or more, namely, 1111, 1110, 1101, 1011, 0111, 1100, 0110, and 0011. - Dennis P. Walsh, Oct 25 2012 G.f. = 1 + 3*x + 8*x^2 + 20*x^3 + 48*x^4 + 112*x^5 + 256*x^6 + 576*x^7 + ...
MAPLE
spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, B, B)}, labeled]: seq(combstruct[count](spec, size=n)/4, n=2..30); # Zerinvary Lajos, Oct 09 2006 A001792:=-(-3+4*z)/(2*z-1)^2; # Simon Plouffe in his 1992 dissertation, which gives the sequence without the initial 1
G(x):=1/exp(2*x)*(1-x): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n-1], x) od: x:=0: seq(abs(f[n]), n=0..28 ); # Zerinvary Lajos, Apr 17 2009 a := n -> hypergeom([-n, 2], [1], -1);
seq(round(evalf(a(n), 32)), n=0..31); # Peter Luschny, Aug 02 2014
MATHEMATICA
matrix[n_Integer /; n >= 1] := Table[Abs[p - q] + 1, {q, n}, {p, n}]; a[n_Integer /; n >= 1] := Abs[Det[matrix[n]]] (* Josh Locker (joshlocker(AT)macfora.com), Apr 29 2004 *)
g[n_, m_, r_] := Binomial[n - 1, r - 1] Binomial[m + 1, r] r; Table[1 + Sum[g[n, k - n, r], {r, 1, k}, {n, 1, k - 1}], {k, 1, 29}] (* Geoffrey Critzer, Jul 02 2009 *) LinearRecurrence[{4, -4}, {1, 3}, 40] (* Harvey P. Dale, Aug 29 2011 *) CoefficientList[Series[(1 - x) / (1 - 2 x)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *) b[i_]:=i; a[n_]:=Abs[Det[ToeplitzMatrix[Array[b, n], Array[b, n]]]]; Array[a, 40] (* Stefano Spezia, Sep 25 2018 *)
PROG
(Haskell)
a001792 n = a001792_list !! n
a001792_list = scanl1 (+) a045623_list
(Python) for n in range(0, 40): print(int((n+2)*2**(n-1)), end=' ') # Stefano Spezia, Oct 16 2018
Central Delannoy numbers: a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k).
(Formerly M2942 N1184)
+10
190
1, 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823, 252055236609, 1425834724419, 8079317057869, 45849429914943, 260543813797441, 1482376214227923, 8443414161166173, 48141245001931263
COMMENTS
Number of paths from (0,0) to (n,n) in an n X n grid using only steps north, northeast and east (i.e., steps (1,0), (1,1), and (0,1)).
Also the number of ways of aligning two sequences (e.g., of nucleotides or amino acids) of length n, with at most 2*n gaps (-) inserted, so that while unnecessary gappings: - -a a- - are forbidden, both b- and -b are allowed. (If only other of the latter is allowed, then the sequence A000984 gives the number of alignments.) There is an easy bijection from grid walks given by Dickau to such set of alignments (e.g., the straight diagonal corresponds to the perfect alignment with no gaps). - Antti Karttunen, Oct 10 2001 Also main diagonal of array A008288 defined by m(i,1) = m(1,j) = 1, m(i,j) = m(i-1,j-1) + m(i-1,j) + m(i,j-1). - Benoit Cloitre, May 03 2002 So, as a special case of Dmitry Zaitsev's Dec 10 2015 comment on A008288, a(n) is the number of points in Z^n that are L1 (Manhattan) distance <= n from any given point. These terms occur in the crystal ball sequences: a(n) here is the n-th 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 26 2022 a(n) is the number of n-matchings of a comb-like graph with 2*n teeth. Example: a(2) = 13 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has 13 2-matchings: any of the six possible pairs of teeth and {Aa, BC}, {Aa, CD}, {Bb, CD}, {Cc, AB}, {Dd, AB}, {Dd, BC}, {AB, CD}. - Emeric Deutsch, Jul 02 2002 Number of ordered trees with 2*n+1 edges, having root of odd degree, nonroot nodes of outdegree at most 2 and branches of odd length. - Emeric Deutsch, Aug 02 2002 The sum of the first n coefficients of ((1 - x) / (1 - 2*x))^n is a(n-1). - Michael Somos, Sep 28 2003 Also number of paths from (0,0) to (n,0) using only steps U = (1,1), H = (1,0) and D =(1,-1), U can have 2 colors and H can have 3 colors. - N-E. Fahssi, Jan 27 2008 Number of overpartitions in the n X n box (treat a walk of the type in the first comment as an overpartition, by interpreting a NE step as N, E with the part thus created being overlined). - William J. Keith, May 19 2017 Diagonal of rational functions 1/(1 - x - y - x*y), 1/(1 - x - y*z - x*y*z). - Gheorghe Coserea, Jul 03 2018 Dimensions of endomorphism algebras End(R^{(n)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023
REFERENCES
Frits Beukers, Arithmetic properties of Picard-Fuchs equations, Séminaire de Théorie des nombres de Paris, 1982-83, Birkhäuser Boston, Inc.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 593.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
L. Moser and W. Zayachkowski, Lattice paths with diagonal steps, Scripta Math., 26 (1961), 223-229.
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).
R. P. Stanley, Enumerative Combinatorics, Wadsworth, Vol. 2, 1999; see Example 6.3.8 and Problem 6.49.
D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 28.
LINKS
Cyril Banderier and Sylviane Schwer, Why Delannoy numbers?, Journal of Statistical Planning and Inference, 135(1) (2005), 40-54.
FORMULA
a(n) = P_n(3), where P_n is n-th Legendre polynomial.
G.f.: 1 / sqrt(1 - 6*x + x^2).
Dominant term in asymptotic expansion is binomial(2*n, n)/2^(1/4)*((sqrt(2) + 1)/2)^(2*n + 1)*(1 + c_1/n + c_2/n^2 + ...). - Michael David Hirschhorna(n) = Sum_{i=0..n} ( A000079(i)* A008459(n, i)) = Sum_{i=0..n} (2^i * C(n, i)^2). - Antti Karttunen, Oct 10 2001 a(n) = Sum_{k=0..n} C(n+k, n-k)*C(2*k, k). - Benoit Cloitre, Feb 13 2003 a(n) = Sum_{k=0..n} C(n, k)^2 * 2^k. - Michael Somos, Oct 08 2003 G.f. of a(n-1) = 1 / (1 - x / (1 - 2*x / (1 - 2*x / (1 - x / (1 - 2*x / (1 - x / ...)))))). - Michael Somos, May 11 2012 a(n) = Sum_{k=0..n} C(2*n-k, n)*C(n, k). - Paul Barry, Apr 23 2005 D-finite with recurrence: a(-1) = a(0) = 1; n*a(n) = 3*(2*n-1)*a(n-1) - (n-1)*a(n-2). Eq (4) in T. D. Noe's article in JIS 9 (2006) #06.2.7. Define general Delannoy numbers by (i,j > 0): d(i,0) = d(0,j) = 1 =: d(0,0) and d(i,j) = d(i-1,j-1) + d(i-2,j-1) + d(i-1,j). Then a(k) = Sum_{j >= 0} d(k,j)^2 + d(k-1,j)^2 = A026933(n)+ A026933(n-1). This is a special case of the following formula for general Delannoy numbers: d(k,j) = Sum_{i >= 0, p=0..n} d(p, i) * d(n-p, j-i) + d(p-1, i) * d(n-p-1, j-i-1). - Peter E John, Oct 19 2006 Coefficient of x^n in (1 + 3*x + 2*x^2)^n. - N-E. Fahssi, Jan 11 2008 G.f.: 1/(1 - x - 2*x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ... (continued fraction). - Paul Barry, May 28 2009 G.f.: 1/(2*Q(0) + x - 1) where Q(k) = 1 + k*(1-x) - x - x*(k + 1)*(k + 2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013 a(n) = Sum_{k=0..n} C(n,k) * C(n+k,k). - Joerg Arndt, May 11 2013 G.f.: G(0), where G(k) = 1 + x*(6 - x)*(4*k + 1)/(4*k + 2 - 2*x*(6-x)*(2*k + 1)*(4*k + 3)/(x*(6 - x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013 G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k - 1)/(x*(6 - x)*(2*k - 1) + 2*(k + 1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013 G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k + 1)/(x*(6 - x)*(2*k + 1) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013 a(n) = Sum_{k=0..n/2} C(n-k,k) * 3^(n-2*k) * 2^k * C(n,k). - Vladimir Kruchinin, Jun 29 2015 a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)*C(n,j)*C(n+k,k)*C(n+k+j,k+j). Cf. A126086 and A274668. - Peter Bala, Jan 15 2020 a(n) ~ c * (3 + 2*sqrt(2))^n / sqrt(n), where c = 1/sqrt(4*Pi*(3*sqrt(2)-4)) = 0.572681... (Banderier and Schwer, 2005). - Amiram Eldar, Jun 07 2020 a(n+1) = 3*a(n) + 2*Sum_{l=1..n} A006318(l)*a(n-l). [Eq. (1.16) in Qi-Shi-Guo (2016)] a(n) ~ (1 + sqrt(2))^(2*n+1) / (2^(5/4) * sqrt(Pi*n)). - Vaclav Kotesovec, Jan 09 2023
EXAMPLE
G.f. = 1 + 3*x + 13*x^2 + 63*x^3 + 321*x^4 + 1683*x^5 + 8989*x^6 + ...
MAPLE
seq(add(multinomial(n+k, n-k, k, k), k=0..n), n=0..20); # Zerinvary Lajos, Oct 18 2006
MATHEMATICA
f[n_] := Sum[ Binomial[n, k] Binomial[n + k, k], {k, 0, n}]; Array[f, 21, 0] (* Or *)
a[0] = 1; a[1] = 3; a[n_] := a[n] = (3(2 n - 1)a[n - 1] - (n - 1)a[n - 2])/n; Array[a, 21, 0] (* Or *)
a[n_] := Hypergeometric2F1[-n, n+1, 1, -1]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Feb 26 2013 *) a[ n_] := With[ {m = If[n < 0, -1 - n, n]}, SeriesCoefficient[ (1 - 6 x + x^2)^(-1/2), {x, 0, m}]]; (* Michael Somos, Jun 10 2015 *)
PROG
(PARI) {a(n) = if( n<0, n = -1 - n); polcoeff( 1 / sqrt(1 - 6*x + x^2 + x * O(x^n)), n)}; /* Michael Somos, Sep 23 2006 */ (PARI) {a(n) = if( n<0, n = -1 - n); subst( pollegendre(n), x, 3)}; /* Michael Somos, Sep 23 2006 */ (PARI) {a(n) = if( n<0, n = -1 - n); n++; subst( Pol(((1 - x) / (1 - 2*x) + O(x^n))^n), x, 1); } /* Michael Somos, Sep 23 2006 */ (PARI) a(n)=if(n<0, 0, polcoeff((1+3*x+2*x^2)^n, n)) \\ Paul Barry, Aug 22 2007 (PARI) /* same as in A092566 but use */ steps=[[1, 0], [0, 1], [1, 1]]; /* Joerg Arndt, Jun 30 2011 */ (PARI) a(n)=sum(k=0, n, binomial(n, k)*binomial(n+k, k)); \\ Joerg Arndt, May 11 2013 (PARI) x='x+O('x^100); Vec(1/sqrt(1 - 6*x + x^2)) \\ Altug Alkan, Oct 17 2015 (Python) # from Nick Hobson.
def f(a, b):
if a == 0 or b == 0:
return 1
return f(a, b - 1) + f(a - 1, b) + f(a - 1, b - 1)
[f(n, n) for n in range(7)]
(Python)
from gmpy2 import divexact
for n in range(2, 10**3):
(Maxima) a(n):=coeff(expand((1+3*x+2*x^2)^n), x, n);
(Sage)
a = lambda n: hypergeometric([-n, -n], [1], 2)
CROSSREFS
Cf. A008288, bisection of A026003, A027618, A047665, A052141, A084773, A152250, A109980, A000129, A078057, A241023, A243949.
EXTENSIONS
New name and reference Sep 15 1995
Formula and more references from Don Knuth, May 15 1996
Search completed in 0.044 seconds
|