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
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)
A055582 Sixth column of triangle A055252.
1, 9, 48, 198, 699, 2223, 6562, 18324, 49029, 126837, 319332, 786258, 1900351, 4521771, 10616598, 24641280, 56622825, 128974545, 291503800, 654311070, 1459617411, 3238002279, 7147093578, 15703473708, 34359737869, 74893491693
Offset: 0
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..3296
- Robert Davis, Greg Simay, Further Combinatorics and Applications of Two-Toned Tilings, arXiv:2001.11089 [math.CO], 2020.
- Index entries for linear recurrences with constant coefficients, signature (9, -33, 63, -66, 36, -8).
Programs
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Mathematica
CoefficientList[Series[1/(((1 - 2 x)^3) (1 - x)^3), {x, 0, 25}], x] (* Michael De Vlieger, Apr 23 2020 *)
Formula
G.f.: 1/(((1-2*x)^3)*(1-x)^3).
a(n)= (n^2 - 3*n + 8)*(2^(n+3) -1)/2 - 9*(n+3). [Yahia Kahloune, Aug 11 2013]
A055250 Seventh column of triangle A055249.
1, 9, 47, 187, 630, 1898, 5282, 13866, 34831, 84575, 199977, 462973, 1053804, 2365704, 5250660, 11543700, 25177005, 54539205, 117456115, 251676495, 536892146, 1140875254, 2415947382, 5100306062, 10737455195, 22548620283
Offset: 0
Programs
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Maple
a:= n-> (Matrix(7, (i,j)-> if (i=j-1) then 1 elif j=1 then [9,-34,70,-85,61,-24,4][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 + 5, k + 5], {k, 0, n}], {n, 1, 26}] (* Zerinvary Lajos, Jul 08 2009 *)
Formula
G.f.: 1/(((1-2*x)^2)*(1-x)^5).
a(n) = A055249(n+6, 6).
For n >= 1, a(n) = A035038(n+6) + Sum_{j=0..n-1} a(j).
a(n) = Sum_{k=0..n+5} Sum_{i=0..n+5} (i-k) * C(n-k+5,i+3). - Wesley Ivan Hurt, Sep 19 2017
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; ...
Comments