A300401 -id:A300401 - cp's OEIS frontend

A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4x + 3)(x + 1)^(2n) + (x^2 - 1)(x^2 + 3*x + 3).

0, 1, 3, 3, 1, 0, 7, 14, 9, 2, 0, 13, 37, 43, 26, 8, 1, 0, 19, 72, 129, 141, 98, 42, 10, 1, 0, 25, 119, 291, 463, 504, 378, 192, 63, 12, 1, 0, 31, 178, 553, 1156, 1716, 1848, 1452, 825, 330, 88, 14, 1, 0, 37, 249, 939, 2432, 4576, 6435, 6864, 5577, 3432, 1573

Offset: 0

Franck Maminirina Ramaharo, Jul 18 2018

The triangle is related to the Kauffman bracket polynomial evaluated at the shadow diagram of the two-bridge knot with Conway's notation C(2n,3).

			The triangle T(n,k) begins:
n\k| 0   1    2    3     4     5     9     7     8     9    10   11   12  13 14
-------------------------------------------------------------------------------
0  | 0   1    3    3     1
1  | 0   7   14    9     2
2  | 0  13   37   43    26     8     1
3  | 0  19   72  129   141    98    42    10     1
4  | 0  25  119  291   463   504   378   192    63    12     1
5  | 0  31  178  553  1156  1716  1848  1452   825   330    88   14    1
6  | 0  37  249  939  2432  4576  6435  6864  5577  3432  1573  520  117  16  1
...

Ji-Young Ham and Joongul Lee, An explicit formula for the A-polynomial of the knot with Conway’s notation C(2n,3), Journal of Knot Theory and Its Ramifications 25 (2016), 1-9.
Ryo Hanaki, On scannable properties of the original knot from a knot shadow, Topology and its Applications 194 (2015), 296-305.
Bin Lu and Jianyuan K. Zhong, The Kauffman Polynomials of 2-bridge Knots, arXiv:math/0606114 [math.GT], 2006.

Row sums: 4*A004171.

Cf. A093560, A137396, A299989, A300184, A300192, A300453, A300454, A316659.

T := proc (n, k) if k = 1 then 6*n + 1 else binomial(2*n + 3, k + 1) + (binomial(2*n + 1, k)*(2*k - 2*n) + binomial(4, k)*(2*k - 3))/(k + 1) end if end proc:
for n from 0 to 12 do seq(T(n, k), k = 0 .. max(4, 2*(n + 1))) od;

row[n_] := CoefficientList[(x^2 + 4*x + 3)*(x + 1)^(2*n) + (x^2 - 1)*(x^2 + 3*x + 3), x];
Array[row, 12, 0] // Flatten

T(n, k) := binomial(2*n + 3, k + 1) + (binomial(2*n + 1, k)*(2*k - 2*n) + binomial(4, k)*(2*k - 3))/(k + 1) - kron_delta(1, k)$
for n:0 thru 12 do print(makelist(T(n, k), k, 0, max(4, 2*(n + 1))));

T(n,1) = A016921(n) and T(n,k) = C(2*n+3,k+1) + (C(2*n+1,k)*(2*k - 2*n) + C(4,k)*(2*k - 3))/(k + 1) for k > 1.
T(n,2) = A173247(2*n+1) = A300401(2*n,3).
T(n,3) = 2*A099721(n) + 3.
T(n,4) = A244730(n) - A002412(n) + 1.
T(n,k) = A093560(2*n,k) for n > 2 and k > 4.
G.f.: (x^2 + 4*x + 3)/(1 - y*(x + 1)^2) + (x^4 + 3*x^3 + 2*x^2 - 3*x - 3)/(1 - y).

A321127 Irregular triangle read by rows: row n gives the coefficients in the expansion of ((x + 1)^(2n) + (x^2 - 1)(2*(x + 1)^n - 1))/x.

Original entry on oeis.org

0, 1, 0, 2, 2, 0, 5, 8, 3, 0, 10, 24, 21, 8, 1, 0, 17, 56, 80, 64, 30, 8, 1, 0, 26, 110, 220, 270, 220, 122, 45, 10, 1, 0, 37, 192, 495, 820, 952, 804, 497, 220, 66, 12, 1, 0, 50, 308, 973, 2030, 3059, 3472, 3017, 2004, 1001, 364, 91, 14, 1

Offset: 0

Franck Maminirina Ramaharo, Nov 19 2018

These are the coefficients of the Kauffman bracket polynomial evaluated at the shadow diagram of the two-bridge knot with Conway's notation C(n,n). Hence, T(n,k) gives the corresponding number of Kauffman states having exactly k circles.

			Triangle begins:
n\k | 0   1    2    3    4    5    6    7    8   9  11 12
----+----------------------------------------------------
  0 | 0   1
  1 | 0   2    2
  2 | 0   5    8    3
  3 | 0  10   24   21    8    1
  4 | 0  17   56   80   64   30    8    1
  5 | 0  26  110  220  270  220  122   45   10   1
  6 | 0  37  192  495  820  952  804  497  220  66  12  1
  ...

Louis H. Kauffman, Formal Knot Theory, Princeton University Press, 1983.

Michael De Vlieger, Table of n, a(n) for n = 0..14282 (rows 0 <= n <= 120, flattened).
Louis H. Kauffman, State models and the Jones polynomial, Topology Vol. 26 (1987), 395-407.
Kelsey Lafferty, The three-variable bracket polynomial for reduced, alternating links, Rose-Hulman Undergraduate Mathematics Journal Vol. 14 (2013), 98-113.
Matthew Overduin, The three-variable bracket polynomial for two-bridge knots, California State University REU, 2013.
Franck Ramaharo, A generating polynomial for the two-bridge knot with Conway's notation C(n,r), arXiv:1902.08989 [math.CO], 2019.
Wikipedia, 2-bridge knot
Wikipedia, Bracket polynomial

Row sums: A000302.
Row 1 is row 2 in A300453.
Row 2 is also row 2 in A300454 and A316659.
Cf. A299989, A300184, A300192, A316989.

row[n_] := CoefficientList[Expand[((x + 1)^(2*n) + (x^2 - 1)*(2*(x + 1)^n - 1))/x], x]; Array[row, 12, 0] // Flatten

T(n, k) := if k = 1 then n^2 + 1 else  ((4*k - 2*n)/(k + 1))*binomial(n + 1, k) + binomial(2*n, k + 1)$
create_list(T(n, k), n, 0, 12, k, 0, max(2*n - 1, n + 1));

T(n,k) = 0 if k = 0, n^2 + 1 if k = 1, and C(2*n, k + 1) - 2*(C(n, k + 1) + C(n, k - 1)) otherwise.
T(n,1) = A002522(n).
T(n,2) = A300401(n,n).
T(n,n) = A001791(n) + A005843(n) - A063524(n).
T(n,k) = A094527(n,k-n+1) if  n + 1 < k < 2*n  and n > 2.
G.f.: x*(1 - (1 + x + x^2)*y + (1 + x)*(2 - x^2)*y^2)/((1 - y)*(1 - y - x*y)*(1 - (1 + x)^2*y)).
E.g.f.: (exp((1 + x)^2*y) - (exp(x) + 2*exp((1 + x)*y))*(1 - x^2))/x.

A320530 T(n,k) = k^n + k^(n - 2)n(n - 1)(k(k - 1) + 1)/2 for 0 < k <= n and T(n,0) = A154272(n+1), square array read by antidiagonals upwards.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 7, 3, 1, 0, 7, 26, 16, 4, 1, 0, 11, 88, 90, 29, 5, 1, 0, 16, 272, 459, 220, 46, 6, 1, 0, 22, 784, 2133, 1504, 440, 67, 7, 1, 0, 29, 2144, 9234, 9344, 3775, 774, 92, 8, 1, 0, 37, 5632, 37908, 54016, 29375, 7992, 1246, 121, 9

Offset: 0

Franck Maminirina Ramaharo, Oct 14 2018

Construct a length n ternary word over the alphabet {a, b, c} as follows: letters from the set {a, b} are only used in pairs of at most one, and consist of either (a,b), (b,a) or (b,b). Next, replace each occurrence of a, b and c with a length k binary word such that 'a' has exactly two letters 1, 'b' contains no 0's and 'c' has exactly one letter 0 (empty words otherwise, respectively). Then T(n,k) gives the number of length n*k binary words resulting from this substitution. First column follows from the next definition.
In Kauffman's language, T(n,k) is the number of ways of splitting the crossings of the Pretzel knot shadow P(k, k, ..., k) having n tangles, of k half-twists respectively, such that the final diagram consists of two Jordan curves. This result can be achieved by assigning each tangle of the Pretzel knot a length k binary words in a way that letters 1 and 0 indicate the adequate choice for splitting the crossings.
Columns are linear recurrence sequences with signature (3*k, -3*k^2, k^3).

			Square array begins:
    1,  1,     1,     1,      1,       1,       1, ...
    0,  1,     2,     3,      4,       5,       6, ...
    1,  2,     7,    16,     29,      46,      67, ...
    0,  4,    26,    90,    220,     440,     774, ...
    0,  7,    88,   459,   1504,    3775,    7992, ...
    0, 11,   272,  2133,   9344,   29375,   74736, ...
    0, 16,   784,  9234,  54016,  212500,  649296, ...
    0, 22,  2144, 37908, 295936, 1456250, 5342112, ...
    ...
T(3,2) = 2^3 + 2^(3 - 2)*3*(3 - 1)*(2*(2 - 1) + 1)/2 = 26. The corresponding ternary words are abc, acb, cab, bac, bca, cba, bbc, bcb, cbb, ccc.  Next, let a = {00}, b = {11} and c = {01, 10}. The resulting binary words are
    abc: 001101, 001110;
    acb: 000111, 001011;
    cab: 010011, 100011;
    bac: 110001, 110010;
    bca: 110100, 111000;
    cba: 011100, 101100;
    bbc: 111101, 111110;
    bcb: 110111, 111011;
    cbb: 011111, 101111;
    ccc: 010101, 101010, 010110, 011001, 100101, 101001, 100110, 011010.

Louis H. Kauffman, Formal Knot Theory, Princeton University Press, 1983.

Louis H. Kauffman, State models and the Jones polynomial, Topology, Vol. 26 (1987), 395-407.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
Alexander Stoimenow, Everywhere Equivalent 2-Component Links, Symmetry Vol. 7 (2015), 365-375.
Wikipedia, Pretzel link

Column 1 is column 2 of A300453.
Column 2 is column 2 of A300184.
Cf. A300401, A303273, A320530.

T[n_, k_] = If[k > 0, k^n + k^(n - 2)*n*(n - 1)*(k*(k - 1) + 1)/2, If[k == 0 && (n == 0 || n == 1), 1, 0]];
Table[Table[T[n - k, k], {k, 0, n}], {n, 0, 10}]//Flatten

t(n, k) := k^n + k^(n - 2)*binomial(n, 2)*(2*binomial(k, 2) + 1)$
u(n) := if n = 0 or n = 1 then 1 else 0$
T(n, k) := if k = 0 then u(n) else t(n,k)$
tabl(nn) := for n:0 thru 10 do print(makelist(T(n, k), k, 0, nn))$

T(n,k) = k^n + k^(n - 2)*binomial(n, 2)*(2*binomial(k, 2) + 1), k > 0.
T(n,k) = (3*k)*T(n-1,k) - (3*k^2)*T(n-2,k) + (k^3)*T(n-3,k), n > 3.
T(n,1) = A152947(n+1).
T(n,2) = A300451(n).
T(2,n) = A130883(n).
G.f. for columns: (1 - 2*k*x + (1 - k + 2*k^2)*x^2 )/(1 - k*x)^3.
E.g.f. for columns: ((1 - k + k^2)*x^2 + 2)*exp(k*x)/2.

A303273 Array T(n,k) = binomial(n, 2) + k*n + 1 read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 4, 4, 1, 4, 6, 7, 7, 1, 5, 8, 10, 11, 11, 1, 6, 10, 13, 15, 16, 16, 1, 7, 12, 16, 19, 21, 22, 22, 1, 8, 14, 19, 23, 26, 28, 29, 29, 1, 9, 16, 22, 27, 31, 34, 36, 37, 37, 1, 10, 18, 25, 31, 36, 40, 43, 45, 46, 46, 1, 11, 20, 28, 35, 41

Offset: 0

Franck Maminirina Ramaharo, Apr 20 2018

Columns are linear recurrence sequences with signature (3,-3,1).
8*T(n,k) + A166147(k-1) are squares.
Columns k are binomial transforms of [1, k, 1, 0, 0, 0, ...].
Antidiagonals sums yield A116731.

			The array T(n,k) begins
1    1    1    1    1    1    1    1    1    1    1    1    1  ...  A000012
1    2    3    4    5    6    7    8    9   10   11   12   13  ...  A000027
2    4    6    8   10   12   14   16   18   20   22   24   26  ...  A005843
4    7   10   13   16   19   22   25   28   31   34   37   40  ...  A016777
7   11   15   19   23   27   31   35   39   43   47   51   55  ...  A004767
11  16   21   26   31   36   41   46   51   56   61   66   71  ...  A016861
16  22   28   34   40   46   52   58   64   70   76   82   88  ...  A016957
22  29   36   43   50   57   64   71   78   85   92   99  106  ...  A016993
29  37   45   53   61   69   77   85   93  101  109  117  125  ...  A004770
37  46   55   64   73   82   91  100  109  118  127  136  145  ...  A017173
46  56   66   76   86   96  106  116  126  136  146  156  166  ...  A017341
56  67   78   89  100  111  122  133  144  155  166  177  188  ...  A017401
67  79   91  103  115  127  139  151  163  175  187  199  211  ...  A017605
79  92  105  118  131  144  157  170  183  196  209  222  235  ...  A190991
...
The inverse binomial transforms of the columns are
1    1    1    1    1    1    1    1    1    1    1    1    1  ...
0    1    2    3    4    5    6    7    8    9   10   11   12  ...
1    1    1    1    1    1    1    1    1    1    1    1    1  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
...
T(k,n-k) = A087401(n,k) + 1 as triangle
1
1   1
1   2   2
1   3   4   4
1   4   6   7   7
1   5   8  10  11  11
1   6  10  13  15  16  16
1   7  12  16  19  21  22  22
1   8  14  19  23  26  28  29  29
1   9  16  22  27  31  34  36  37  37
1  10  18  25  31  36  40  43  45  46  46
...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

Cf. A085475, A086270, A086271, A086272, A086273, A130154, A159798, A162609, A162610, A300401.

T := (n, k) -> binomial(n, 2) + k*n + 1;
for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;

Table[With[{n = m - k}, Binomial[n, 2] + k n + 1], {m, 0, 11}, {k, m, 0, -1}] // Flatten (* Michael De Vlieger, Apr 21 2018 *)

T(n, k) := binomial(n, 2)+ k*n + 1$
for n:0 thru 20 do
    print(makelist(T(n, k), k, 0, 20));

T(n,k) = binomial(n, 2) + k*n + 1;
tabl(nn) = for (n=0, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, May 17 2018

G.f.: (3*x^2*y - 3*x*y + y - 2*x^2 + 2*x - 1)/((x - 1)^3*(y - 1)^2).
E.g.f.: (1/2)*(2*x*y + x^2 + 2)*exp(y + x).
T(n,k) = 3*T(n-1,k) - 3*T(n-2,k) + T(n-3,k), with T(0,k) = 1, T(1,k) = k + 1 and T(2,k) = 2*k + 2.
T(n,k) = T(n-1,k) + n + k - 1.
T(n,k) = T(n,k-1) + n, with T(n,0) = 1.
T(n,0) = A152947(n+1).
T(n,1) = A000124(n).
T(n,2) = A000217(n).
T(n,3) = A034856(n+1).
T(n,4) = A052905(n).
T(n,5) = A051936(n+4).
T(n,6) = A246172(n+1).
T(n,7) = A302537(n).
T(n,8) = A056121(n+1) + 1.
T(n,9) = A056126(n+1) + 1.
T(n,10) = A051942(n+10) + 1, n > 0.
T(n,11) = A101859(n) + 1.
T(n,12) = A132754(n+1) + 1.
T(n,13) = A132755(n+1) + 1.
T(n,14) = A132756(n+1) + 1.
T(n,15) = A132757(n+1) + 1.
T(n,16) = A132758(n+1) + 1.
T(n,17) = A212427(n+1) + 1.
T(n,18) = A212428(n+1) + 1.
T(n,n) = A143689(n) = A300192(n,2).
T(n,n+1) = A104249(n).
T(n,n+2) = T(n+1,n) = A005448(n+1).
T(n,n+3) = A000326(n+1).
T(n,n+4) = A095794(n+1).
T(n,n+5) = A133694(n+1).
T(n+2,n) = A005449(n+1).
T(n+3,n) = A115067(n+2).
T(n+4,n) = A133694(n+2).
T(2*n,n) = A054556(n+1).
T(2*n,n+1) = A054567(n+1).
T(2*n,n+2) = A033951(n).
T(2*n,n+3) = A001107(n+1).
T(2*n,n+4) = A186353(4*n+1) (conjectured).
T(2*n,n+5) = A184103(8*n+1) (conjectured).
T(2*n,n+6) = A250657(n-1) = A250656(3,n-1), n > 1.
T(n,2*n) = A140066(n+1).
T(n+1,2*n) = A005891(n).
T(n+2,2*n) = A249013(5*n+4) (conjectured).
T(n+3,2*n) = A186384(5*n+3) = A186386(5*n+3) (conjectured).
T(2*n,2*n) = A143689(2*n).
T(2*n+1,2*n+1) = A143689(2*n+1) (= A030503(3*n+3) (conjectured)).
T(2*n,2*n+1) = A104249(2*n) = A093918(2*n+2) = A131355(4*n+1) (= A030503(3*n+5) (conjectured)).
T(2*n+1,2*n) = A085473(n).
a(n+1,5*n+1)=A051865(n+1) + 1.
a(n,2*n+1) = A116668(n).
a(2*n+1,n) = A054569(n+1).
T(3*n,n) = A025742(3*n-1), n > 1 (conjectured).
T(n,3*n) = A140063(n+1).
T(n+1,3*n) = A069099(n+1).
T(n,4*n) = A276819(n).
T(4*n,n) = A154106(n-1), n > 0.
T(2^n,2) = A028401(n+2).
T(1,n)*T(n,1) = A006000(n).
T(n*(n+1),n) = A211905(n+1), n > 0 (conjectured).
T(n*(n+1)+1,n) = A294259(n+1).
T(n,n^2+1) = A081423(n).
T(n,A000217(n)) = A158842(n), n > 0.
T(n,A152947(n+1)) = A060354(n+1).
floor(T(n,n/2)) = A267682(n) (conjectured).
floor(T(n,n/3)) = A025742(n-1), n > 0 (conjectured).
floor(T(n,n/4)) = A263807(n-1), n > 0 (conjectured).
ceiling(T(n,2^n)/n) = A134522(n), n > 0 (conjectured).
ceiling(T(n,n/2+n)/n) = A051755(n+1) (conjectured).
floor(T(n,n)/n) = A133223(n), n > 0 (conjectured).
ceiling(T(n,n)/n) = A007494(n), n > 0.
ceiling(T(n,n^2)/n) = A171769(n), n > 0.
ceiling(T(2*n,n^2)/n) = A046092(n), n > 0.
ceiling(T(2*n,2^n)/n) = A131520(n+2), n > 0.

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A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4*x + 3)*(x + 1)^(2*n) + (x^2 - 1)*(x^2 + 3*x + 3).

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A321127 Irregular triangle read by rows: row n gives the coefficients in the expansion of ((x + 1)^(2*n) + (x^2 - 1)*(2*(x + 1)^n - 1))/x.

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A320530 T(n,k) = k^n + k^(n - 2)*n*(n - 1)*(k*(k - 1) + 1)/2 for 0 < k <= n and T(n,0) = A154272(n+1), square array read by antidiagonals upwards.

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A303273 Array T(n,k) = binomial(n, 2) + k*n + 1 read by antidiagonals.

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A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4x + 3)(x + 1)^(2n) + (x^2 - 1)(x^2 + 3*x + 3).

A321127 Irregular triangle read by rows: row n gives the coefficients in the expansion of ((x + 1)^(2n) + (x^2 - 1)(2*(x + 1)^n - 1))/x.

A320530 T(n,k) = k^n + k^(n - 2)n(n - 1)(k(k - 1) + 1)/2 for 0 < k <= n and T(n,0) = A154272(n+1), square array read by antidiagonals upwards.