A055249 Triangle of partial row sums (prs) of triangle A055248 (prs of Pascal's triangle A007318).
1, 3, 1, 8, 4, 1, 20, 12, 5, 1, 48, 32, 17, 6, 1, 112, 80, 49, 23, 7, 1, 256, 192, 129, 72, 30, 8, 1, 576, 448, 321, 201, 102, 38, 9, 1, 1280, 1024, 769, 522, 303, 140, 47, 10, 1, 2816, 2304, 1793, 1291, 825, 443, 187, 57, 11, 1, 6144, 5120, 4097, 3084, 2116, 1268, 630
Offset: 0
Examples
1; 3,1; 8,4,1; 20,12,5,1; ... Fourth row polynomial (n=3): p(3,x)= 20+12*x+5*x^2+x^3
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1274
Programs
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Mathematica
a[n_, m_] := Binomial[n, m]*Hypergeometric2F1[2, m-n, m+1, -1]; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Mar 11 2014 *)
Formula
a(n, m) = Sum_{k=m,..,n} ( A055248(n, k) ), n >= m >= 0, a(n, m) := 0 if n
Column m recursion: a(n, m) = Sum_{j=m,..,(n-1)} ( a(j, m) ) + A055248(n, m), n >= m >= 0, a(n, m) := 0 if n
G.f. for column m: ((1-x)/(1-2*x)^2)*(x/(1-x))^m, m >= 0.
a(n, m) = binomial(n, m) * 2F1(2, m-n; m+1; -1) where 2F1 is the hypergeometric function. Jean-François Alcover, Mar 11 2014
A034009 Convolution of A000295(n+2) (n>=0) with itself.
1, 8, 38, 140, 443, 1268, 3384, 8584, 20965, 49744, 115402, 262996, 590831, 1311900, 2884956, 6293040, 13633305, 29362200, 62916910, 134220380, 285215651, 603983108, 1275072128, 2684358680, 5637149133, 11811165088
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (8,-26,44,-41,20,-4).
Programs
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Magma
[(16*(n-3)*2^n+(n+7)*(n^2+11*n+42) div 6): n in [0..30]]; // Vincenzo Librandi, Sep 20 2014
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Maple
seq(16*(n-3)*2^n+(n+7)*(n^2+11*n+42)/6, n=0..100); # Robert Israel, Sep 19 2014
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Mathematica
Table[Sum[ k Binomial[n + 5, k + 4], {k, 0, n+1}], {n, 0, 26}] (* Zerinvary Lajos, Jul 08 2009 *) Table[(16 (n-3) 2^n + (n + 7) (n^2 + 11 n + 42) / 6), {n, 0, 40}] (* Vincenzo Librandi, Sep 20 2014 *)
Formula
(2^(n+2)-n-3) '*' (2^(n+2)-n-3) where '*' denotes the convolution product.
G.f.: 1/((1-2*x)*(1-x)^2)^2.
Partial sums of A045889.
a(n) = (n-3)*2^(n+4)+binomial(n+3,3)+4*(binomial(n+1,2)+4*n+12)
= 2^(n+4)*(n-3)+(n+7)*(n*(n+11)+42)/6.
a(n) = binomial(n+3,3)*hypergeom([2,-n],[-n-3],2). - Peter Luschny, Sep 19 2014
a(n) = Sum_{k=0..n+4} Sum_{i=0..n+4} (i-k) * C(n-k+4,i+2). - Wesley Ivan Hurt, Sep 19 2017
Extensions
Edited by Peter Luschny, Sep 20 2014
A188553 T(n,k) = Number of n X k binary arrays without the pattern 0 1 diagonally, vertically, antidiagonally or horizontally.
2, 3, 3, 4, 5, 4, 5, 8, 7, 5, 6, 12, 12, 9, 6, 7, 17, 20, 16, 11, 7, 8, 23, 32, 28, 20, 13, 8, 9, 30, 49, 48, 36, 24, 15, 9, 10, 38, 72, 80, 64, 44, 28, 17, 10, 11, 47, 102, 129, 112, 80, 52, 32, 19, 11, 12, 57, 140, 201, 192, 144, 96, 60, 36, 21, 12, 13, 68, 187, 303, 321, 256, 176
Offset: 1
Comments
From Miquel A. Fiol, Feb 06 2024: (Start)
Also, T(n,k) is the number of words of length k, x(1)x(2)...x(k), on the alphabet {0,1,...,n}, such that, for i=2,...,k, x(i)=either x(i-1) or x(i)=x(i-1)-1.
For the bijection between arrays and sequences, notice that the i-th column consists of 1's and then 0's, and there are x(i)=0 to n of 1's.
Such a bijection implies that all the empirical/conjectured formulas in A188554, A188555, A188556, A188557, A188558, and A188559 become correct.
(End)
Examples
Table starts ..2..3..4..5...6...7...8...9...10...11...12....13....14....15....16.....17 ..3..5..8.12..17..23..30..38...47...57...68....80....93...107...122....138 ..4..7.12.20..32..49..72.102..140..187..244...312...392...485...592....714 ..5..9.16.28..48..80.129.201..303..443..630...874..1186..1578..2063...2655 ..6.11.20.36..64.112.192.321..522..825.1268..1898..2772..3958..5536...7599 ..7.13.24.44..80.144.256.448..769.1291.2116..3384..5282..8054.12012..17548 ..8.15.28.52..96.176.320.576.1024.1793.3084..5200..8584.13866.21920..33932 ..9.17.32.60.112.208.384.704.1280.2304.4097..7181.12381.20965.34831..56751 .10.19.36.68.128.240.448.832.1536.2816.5120..9217.16398.28779.49744..84575 .11.21.40.76.144.272.512.960.1792.3328.6144.11264.20481.36879.65658.115402 Some solutions for 5 X 3: 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 Some solutions for T(5,3): By taking the sums of the columns in the above arrays we get 555, 100, 000, 543, 322, 432, 554. - _Miquel A. Fiol_, Feb 04 2024
Links
- R. H. Hardin, Table of n, a(n) for n = 1..9934
Crossrefs
Diagonal is A045623.
Column 4 is A086570.
Upper diagonals T(n,n+i) for i=1..8 give: A001792, A001787(n+1), A000337(n+1), A045618, A045889, A034009, A055250, A055251.
Lower diagonals T(n+i,n) for i=1..7 give: A045891(n+1), A034007(n+2), A111297(n+1), A159694(n-1), A159695(n-1), A159696(n-1), A159697(n-1).
Antidiagonal sums give A065220(n+5).
Programs
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Maple
T:= (n,k)-> `if`(k<=n+1, (2*n+3-k)*2^(k-2), (n+1-k)*binomial(k-1, n) * add(binomial(n, j-1)/(k-j)*T(n, j)*(-1)^(n-j), j=1..n+1)): seq(seq(T(n, 1+d-n), n=1..d), d=1..15); #Alois P. Heinz in the Sequence Fans Mailing List, Apr 04 2011 [We do not permit programs based on conjectures, but this program is now justified by Fiol's comment. - N. J. A. Sloane, Mar 09 2024]
Formula
Empirical: T(n,k) = (n+1)*2^(k-1) + (1-k)*2^(k-2) for k < n+3, and then the entire row n is a polynomial of degree n in k.
From Miquel A. Fiol, Feb 06 2024: (Start)
The above empirical formula is correct.
It can be proved that T(n,k) satisfies the recurrence
T(n,k) = Sum_{r=1..n+1} (-1)^(r+1)*binomial(n+1,r)*T(n,k-r)
with initial values
T(n,k) = Sum_{r=0..k-1} (n+1-r)*binomial(k-1,r) for k = 1..n+1. (End)
A055581 Fifth column of triangle A055252.
1, 8, 39, 150, 501, 1524, 4339, 11762, 30705, 77808, 192495, 466926, 1114093, 2621420, 6094827, 14024682, 31981545, 72351720, 162529255, 362807270, 805306341, 1778384868, 3909091299, 8556380130, 18656264161, 40533753824
Offset: 0
Comments
a(n) = number of directed column-convex polyominoes of area n+5 having along the lower contour exactly two reentrant corners. - Emeric Deutsch, May 21 2003
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..3297
- Robert Davis, Greg Simay, Further Combinatorics and Applications of Two-Toned Tilings, arXiv:2001.11089 [math.CO], 2020.
- A. F. Y. Zhao, Pattern Popularity in Multiply Restricted Permutations, Journal of Integer Sequences, 17 (2014), #14.10.3.
- Index entries for linear recurrences with constant coefficients, signature (8, -25, 38, -28, 8).
Programs
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Mathematica
Table[(n^2-n+4)2^(n+1)-7-n,{n,0,30}] (* or *) LinearRecurrence[ {8,-25,38,-28,8},{1,8,39,150,501},30] (* Harvey P. Dale, Nov 07 2011 *)
Formula
G.f.: 1/(((1-2*x)^3)*(1-x)^2).
a(n) = (n^2-n+4)2^(n+1)-7-n - Emeric Deutsch, May 21 2003
a(0)=1, a(1)=8, a(2)=39, a(3)=150, a(4)=501, a(n) = 8*a(n-1)- 25*a(n-2)+ 38*a(n-3)-28*a(n-4)+8*a(n-5). [Harvey P. Dale, Nov 07 2011]
A055251 Eighth column of triangle A055249.
1, 10, 57, 244, 874, 2772, 8054, 21920, 56751, 141326, 341303, 804276, 1858080, 4223784, 9474444, 21018144, 46195149, 100734354, 218190469, 469866964, 1006759110, 2147634364, 4563581746, 9663887808, 20401343003, 42949963286, 90194651043, 188978952404
Offset: 0
Comments
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (10,-43,104,-155,146,-85,28,-4).
Programs
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Maple
a:= n-> (Matrix(8, (i,j)-> if (i=j-1) then 1 elif j=1 then [10,-43,104,-155, 146,-85,28,-4][i] else 0 fi)^(n))[1,1]: seq(a(n), n=0..25); # Alois P. Heinz, Aug 05 2008
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Mathematica
Table[Sum[(-1)^(n - k) k (-1)^(n - k) Binomial[n + 6, k + 6], {k, 0, n}], {n, 1, 26}] (* Zerinvary Lajos, Jul 08 2009 *) -
PARI
Vec(1 / ((1 - x)^6*(1 - 2*x)^2) + O(x^30)) \\ Colin Barker, Sep 20 2017
Formula
G.f.: 1 / (((1-2*x)^2)*(1-x)^6).
a(n) = A055249(n+7, 7).
For n >= 1, a(n) = A035039(n+7) + Sum_{j=0..n-1} a(j).
a(n) = Sum_{k=0..n+6} Sum_{i=0..n+6} (i-k) * C(n-k+6,i+4). - Wesley Ivan Hurt, Sep 19 2017
a(n) = (1/120)*(38520 - 75*2^(9+n) + 2*(9637 + 15*2^(8+n))*n + 4285*n^2 + 525*n^3 + 35*n^4 + n^5). - Colin Barker, Sep 20 2017
A058394 A square array based on natural numbers (A000027) with each term being the sum of 2 consecutive terms in the previous row.
1, 0, 1, 2, 1, 1, 0, 2, 2, 1, 3, 2, 3, 3, 1, 0, 3, 4, 5, 4, 1, 4, 3, 5, 7, 8, 5, 1, 0, 4, 6, 9, 12, 12, 6, 1, 5, 4, 7, 11, 16, 20, 17, 7, 1, 0, 5, 8, 13, 20, 28, 32, 23, 8, 1, 6, 5, 9, 15, 24, 36, 48, 49, 30, 9, 1, 0, 6, 10, 17, 28, 44, 64, 80, 72, 38, 10, 1, 7, 6, 11, 19, 32, 52, 80, 112, 129
Offset: 0
Comments
Examples
Rows are (1,0,2,0,3,0,4,...), (1,1,2,2,3,3,...), (1,2,3,4,5,6,...), (1,3,5,7,9,11,...), etc.
Crossrefs
Formula
T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(2n, 0)=T(n, 2) and T(2n+1, 0)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2)^2.
A106194 Triangle read by rows, generated from binomial transforms of odd numbers.
1, 4, 1, 12, 5, 1, 32, 17, 6, 1, 80, 49, 23, 7, 1, 192, 129, 72, 30, 8, 1, 448, 321, 201, 102, 38, 9, 1, 1024, 769, 522, 303, 140, 47, 10, 1, 2304, 1793, 1291, 825, 443, 187, 57, 11, 1, 5120, 4097, 3084, 2116, 1268, 630, 244, 68, 12, 1
Offset: 0
Comments
Appending the binomial transform of the natural numbers, (A001792: 1, 3, 8, 20, 48...) to A106194 as a leftmost column creates triangle A055249.
Placing zeros into the offset spaces, column 1: 0, 1, 5, 17, 49...; is the binomial transform of 0, 1, 3, 5...; and alternatively the binomial transform of 0, 0, 1, 2, 3...
n-th column is the binomial transform of 1, 3, 5...prefaced by n zeros. n-th column is alternatively the binomial transform of 1, 2, 3...prefaced by (n+1) zeros. The triangle of A106194 is identical to the binomial transform (of natural numbers, prefaced with zeros) triangle: A055249, deleting the leftmost column.
Examples
First few rows of the triangle are: 1; 4, 1; 12, 5, 1; 32, 17, 6, 1; 80, 49, 23, 7, 1; 192, 129, 72, 30, 8, 1; 448, 321, 201, 102, 38, 9, 1; ...
A331969 T(n, k) = [x^(n-k)] 1/(((1 - 2*x)^k)*(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.
1, 1, 1, 1, 4, 1, 1, 11, 7, 1, 1, 26, 30, 10, 1, 1, 57, 102, 58, 13, 1, 1, 120, 303, 256, 95, 16, 1, 1, 247, 825, 955, 515, 141, 19, 1, 1, 502, 2116, 3178, 2310, 906, 196, 22, 1, 1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1
Offset: 0
Comments
The triangle is the matrix inverse of the Riordan square (see A321620) generated by (1 + x - sqrt(1 - 6*x + x^2))/(4*x) (see A172094), where we take the absolute value of the terms.
T(n,k) is the number of evil-avoiding (2413, 3214, 4132, and 4213 avoiding) permutations of length (n+2) that start with 1 and whose inverse has k descents. - Donghyun Kim, Aug 16 2021
Examples
Triangle starts: [0] [1] [1] [1, 1] [2] [1, 4, 1] [3] [1, 11, 7, 1] [4] [1, 26, 30, 10, 1] [5] [1, 57, 102, 58, 13, 1] [6] [1, 120, 303, 256, 95, 16, 1] [7] [1, 247, 825, 955, 515, 141, 19, 1] [8] [1, 502, 2116, 3178, 2310, 906, 196, 22, 1] [9] [1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1] ... Seen as a square array (the triangle is formed by descending antidiagonals): 1, 1, 1, 1, 1, 1, 1, 1, 1, ... [A000012] 1, 4, 11, 26, 57, 120, 247, 502, 1013, ... [A000295] 1, 7, 30, 102, 303, 825, 2116, 5200, 12381, ... [A045889] 1, 10, 58, 256, 955, 3178, 9740, 28064, 77093, ... [A055583] 1, 13, 95, 515, 2310, 9078, 32354, 106970, 333295, ... 1, 16, 141, 906, 4746, 21504, 87374, 326084, 1136799, ... 1, 19, 196, 1456, 8722, 44758, 204204, 849180, 3275931, ...
Links
- Donghyun Kim and Lauren Williams, Schubert polynomials and the inhomogeneous TASEP on a ring, arXiv:2102.00560 [math.CO], 2021.
Crossrefs
Programs
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Maple
gf := k -> 1/(((1-2*x)^k)*(1-x)^(k+1)): ser := k -> series(gf(k), x, 32): # Prints the triangle: seq(lprint(seq(coeff(ser(k), x, n-k), k=0..n)), n=0..6); # Prints the square array: seq(lprint(seq(coeff(ser(k), x, n), n=0..8)), k=0..6);
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Mathematica
(* The function RiordanSquare is defined in A321620; returns the triangle as a lower triangular matrix. *) M := RiordanSquare[(1 + x - Sqrt[1 - 6 x + x^2])/(4 x), 9]; Abs[#] & /@ Inverse[PadRight[M]]
Comments