A152947 -id:A152947 - cp's OEIS frontend

A302537 a(n) = (n^2 + 13*n + 2)/2.

1, 8, 16, 25, 35, 46, 58, 71, 85, 100, 116, 133, 151, 170, 190, 211, 233, 256, 280, 305, 331, 358, 386, 415, 445, 476, 508, 541, 575, 610, 646, 683, 721, 760, 800, 841, 883, 926, 970, 1015, 1061, 1108, 1156, 1205, 1255, 1306, 1358, 1411, 1465, 1520, 1576

Offset: 0

Franck Maminirina Ramaharo, Apr 09 2018

Binomial transform of [1, 7, 1, 0, 0, 0, ...].

Numbers m > 0 such that 8*m + 161 is a square.

			Illustration of initial terms (by the formula a(n) = A052905(n) + 3*n):
.                                                                    o
.                                                                  o o
.                                                    o           o o o
.                                                  o o         o o o o
.                                      o         o o o       o o o o o
.                                    o o       o o o o     o o o o o o
.                          o       o o o     o o o o o   o . . . . . o
.                        o o     o o o o   o . . . . o   o . . . . . o
.                o     o o o   o . . . o   o . . . . o   o . . . . . o
.              o o   o . . o   o . . . o   o . . . . o   o . . . . . o
.        o   o . o   o . . o   o . . . o   o . . . . o   o . . . . . o
.      o o   o . o   o . . o   o . . . o   o . . . . o   o . . . . . o
.  o   o o   o o o   o o o o   o o o o o   o o o o o o   o o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
----------------------------------------------------------------------
.  1     8      16        25          35            46              58

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

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Sequences whose n-th terms are of the form binomial(n, 2) + n*k + 1:
A152947 (k = 0); A000124 (k = 1); A000217 (k = 2); A034856 (k = 3);
A052905 (k = 4); A051936 (k = 5); A246172 (k = 6).
Cf. A000578, A002939, A004120, A008589, A056119, A060544, A162261.

A302537:= func< n | ((n+1)^2 +12*n +1)/2 >;
[A302537(n): n in [0..50]]; // G. C. Greubel, Jan 21 2025

a := n -> (n^2 + 13*n + 2)/2;
seq(a(n), n = 0 .. 100);

Table[(n^2 + 13 n + 2)/2, {n, 0, 100}]
CoefficientList[ Series[(5x^2 - 5x - 1)/(x - 1)^3, {x, 0, 50}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 8, 16}, 51] (* Robert G. Wilson v, May 19 2018 *)

makelist((n^2 + 13*n + 2)/2, n, 0, 100);

a(n) = (n^2 + 13*n + 2)/2; \\ Altug Alkan, Apr 12 2018

def A302537(n): return (n**2 + 13*n + 2)//2
print([A302537(n) for n in range(51)]) # G. C. Greubel, Jan 21 2025

a(n) = binomial(n + 1, 2) + 6*n + 1 = binomial(n, 2) + 7*n + 1.
a(n) = a(n-1) + n + 6.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, where a(0) = 1, a(1) = 8 and a(2) = 16.
a(n) = 2*a(n-1) - a(n-2) + 1.
a(n) = A004120(n+1) for n > 1.
a(n) = A056119(n) + 1.
a(n) = A152947(n+1) + A008589(n).
a(n) = A060544(n+1) - A002939(n).
a(n) = A000578(n+1) - A162261(n) for n > 0.
G.f.: (1 + 5*x - 5*x^2)/(1 - x)^3.
E.g.f.: (1/2)*(2 + 14*x + x^2)*exp(x).
Sum_{n>=0} 1/a(n) = 24097/45220 + 2*Pi*tan(sqrt(161)*Pi/2) / sqrt(161) = 1.4630922534498496... - Vaclav Kotesovec, Apr 11 2018

A371417 Triangle read by rows: T(n,k) is the number of complete compositions of n with k parts.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 3, 1, 0, 0, 0, 3, 4, 1, 0, 0, 0, 6, 6, 5, 1, 0, 0, 0, 0, 16, 10, 6, 1, 0, 0, 0, 0, 12, 30, 15, 7, 1, 0, 0, 0, 0, 12, 35, 50, 21, 8, 1, 0, 0, 0, 0, 24, 50, 75, 77, 28, 9, 1, 0, 0, 0, 0, 0, 90, 126, 140, 112, 36, 10, 1

Offset: 0

John Tyler Rascoe, Mar 23 2024

A composition (ordered partition) is complete if the set of parts both covers an interval (is gap-free) and contains 1.

			The triangle begins:
    k=0  1  2  3   4   5   6   7   8   9  10
n=0:  1;
n=1:  0, 1;
n=2:  0, 0, 1;
n=3:  0, 0, 2, 1;
n=4:  0, 0, 0, 3,  1;
n=5:  0, 0, 0, 3,  4,  1;
n=6:  0, 0, 0, 6,  6,  5,  1;
n=7:  0, 0, 0, 0, 16, 10,  6,  1;
n=8:  0, 0, 0, 0, 12, 30, 15,  7,  1;
n=9:  0, 0, 0, 0, 12, 35, 50, 21,  8,  1;
n=10: 0, 0, 0, 0, 24, 50, 75, 77, 28,  9,  1;
...
For n = 5 there are a total of 8 complete compositions:
  T(5,3) = 3: (221), (212), (122)
  T(5,4) = 4: (2111), (1211), (1121), (1112)
  T(5,5) = 1: (11111)

Alois P. Heinz, Rows n = 0..200, flattened

A107428 counts gap-free compositions.
A251729 counts gap-free but not complete compositions.
Cf. A107429 (row sums give complete compositions of n), A000670 (column sums), A152947 (number of nonzero terms per column).
Cf. A066099, A124774, A373118.

b:= proc(n, i, t) option remember; `if`(n=0,
     `if`(i=0, t!, 0), `if`(i<1 or n (p-> seq(coeff(p, x, i), i=0..n))(add(b(n, i, 0), i=0..n)):
seq(T(n), n=0..12);  # Alois P. Heinz, Apr 03 2024

G(N)={ my(z='z+O('z^N)); Vec(sum(i=1,N,z^(i*(i+1)/2)*t^i*prod(j=1,i,sum(k=0,N, (z^(j*k)*t^k)/(k+1)!))))}
my(v=G(10)); for(n=0, #v, if(n<1,print([1]), my(p=v[n], r=vector(n+1)); for(k=0, n, r[k+1] =k!*polcoeff(p, k)); print(r)))

T(n,k) = k!*[z^n*t^k] Sum_{i>0} z^(i*(i+1)/2)*t^i * Product_{j=1..i} Sum_{k>=0} (z^(j*k)*t^k)/(k+1)!.

A092365 Coefficient of X^2 in expansion of (1 + nX + nX^2)^n.

Original entry on oeis.org

1, 8, 36, 112, 275, 576, 1078, 1856, 2997, 4600, 6776, 9648, 13351, 18032, 23850, 30976, 39593, 49896, 62092, 76400, 93051, 112288, 134366, 159552, 188125, 220376, 256608, 297136, 342287, 392400, 447826, 508928, 576081, 649672, 730100, 817776

Offset: 1

Jon Perry, Mar 19 2004

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000290, A152947.

[n^2*(Binomial(n, 2)+1): n in [1..40]]; // Vincenzo Librandi, Aug 15 2017

Coefficient[Table[Expand[(1+n x+n x^2)^n],{n,60}],x,2]  (* Harvey P. Dale, Mar 13 2011 *)
Table[n^2 (Binomial[n, 2] + 1), {n, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 36, 112, 275}, 40] (* Vincenzo Librandi, Aug 15 2017 *)

q(n)=(1+n*X+n*X^2)^n; for(i=1,40,print1(","polcoeff(q(i),2)))

a(n) = n^2*(binomial(n, 2) + 1).
G.f.: x*(1 + 3*x + 6*x^2 + 2*x^3)/(1-x)^5. [Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009; corrected by R. J. Mathar, Sep 16 2009]
From Stefano Spezia, Oct 08 2022: (Start)
E.g.f.: exp(x)*x*(2 + 6*x + 5*x^2 + x^3)/2.
a(n) = A000290(n)*A152947(n+1). (End)

A159938 The number of homogeneous trisubstituted linear alkanes.

Original entry on oeis.org

2, 6, 16, 36, 70, 122, 196, 296, 426, 590, 792, 1036, 1326, 1666, 2060, 2512, 3026, 3606, 4256, 4980, 5782, 6666, 7636, 8696, 9850, 11102, 12456, 13916, 15486, 17170, 18972, 20896, 22946, 25126, 27440, 29892, 32486

Offset: 2

Parthasarathy Nambi, Apr 26 2009

See the paper by Valentin Vankov Iliev for details.

This sequence is related to A152947 by a(n) = (n-1)*A152947(n) + sum( A152947(i), i=1..n-1 ). - Bruno Berselli, Dec 19 2013

			The number of homogeneous trisubstituted linear alkane with ten carbon atoms is 426.

Valentin Vankov Iliev, A mathematical characterization of the groups of substitution isomerism of the linear alkanes, J. Math. Chem. 47 (2010), 52-61.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002522, A033816, A081489, A152947, A159940, A159941.

a(n) = (1/3)*(2*n^3 - 9*n^2 + 19*n - 12), where n is the number of carbons.
a(n) = 2*A081489(n-1) = (n-1)*(2*n^2-7*n+12)/3. - R. J. Mathar, Apr 28 2009
G.f.: 2*x^2*(1-x+2*x^2)/(1-x)^4. - Colin Barker, Aug 06 2012

More terms from Colin Barker, Aug 06 2012

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.

A342939 a(n) is the Skolem number of the triangular grid graph T_n.

Original entry on oeis.org

1, 2, 5, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79, 92, 106, 121, 137, 154, 172, 191, 211, 232, 254, 277, 301, 326, 352, 379, 407, 436, 466, 497, 529, 562, 596, 631, 667, 704, 742, 781, 821, 862, 904, 947, 991, 1036, 1082, 1129, 1177, 1226, 1276, 1327, 1379, 1432, 1486

Offset: 1

Stefano Spezia, Mar 30 2021

For the meaning of Skolem number of a graph, see Definitions 1.4 and 1.5 in Carrigan and Green.

Braxton Carrigan and Garrett Green, Skolem Number of Subgraphs on the Triangular Lattice, Communications on Number Theory and Combinatorial Theory 2 (2021), Article 2.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A152947, A161680, A247476, A342938, A342940.

For n > 1, 3*A002061(n) gives the Skolem number of the hexagonal grid graph H_n.

LinearRecurrence[{3,-3,1},{1,2,5,7,11,16},55]

O.g.f.: x*(1 - x + 2*x^2 - 3*x^3 + 3*x^4 - x^5)/(1 - x)^3.
E.g.f.: exp(x)*(2 + x^2)/2 - 1 + x^3/6.
a(n) = 3*a(n-1) - 3*a(n-2) - a(n-3) for n > 6.
Except for a(3) = 5:
  a(n) = 1 + n*(n - 1)/2 (see Theorem 2.5 in Carrigan and Green).
  a(n) = 1 + A161680(n).
  a(n) = A152947(n-1).

A240168 T(n,k) is the number of unlabeled graphs of n vertices and k edges that have endpoints, where an endpoint is a vertex with degree 1.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 5, 4, 2, 1, 1, 2, 4, 8, 13, 15, 16, 11, 5, 2, 1, 1, 2, 4, 9, 19, 35, 55, 75, 83, 72, 51, 29, 13, 5, 2, 1, 1, 2, 4, 10, 22, 50, 105, 196, 338, 511, 649, 695, 627, 473, 304, 172, 83, 35, 14, 5, 2, 1

Offset: 1

Chai Wah Wu, Aug 02 2014

The length of the rows are 1,1,2,4,7,11,16,22,...: (n-1)*(n-2)/2 + 1 = A152947(n).

T(n,k) = 0 if k > (n-1)*(n-2)/2 + 1. (Cf. A245796)

			First few rows of irregular triangle are:
..0
..1
..1....1
..1....2....2....1
..1....2....3....5....4....2....1
..1....2....4....8...13...15...16...11....5....2....1
..1....2....4....9...19...35...55...75...83...72...51...29...13....5....2....1
...

Cf. A245796. Sum of n-th row is equal to A141580(n).

A282283 Recursive 2-parameter sequence allowing calculation of the Euler Totient function.

Original entry on oeis.org

0, 1, -1, 1, 2, -4, 2, -4, 10, -6, -2, 2, 6, -16, 10, 4, -6, 8, -10, 4, -10, 28, -18, -8, 10, -10, 10, -2, 8, -10, 0, 2, 12, -34, 22, 10, -12, 12, -22, 30, -30, 6, 10, -10, 8, 0, 6, -14, 6, -18, 52, -34, -16, 18, -18, 34, -36, 20, 10, -6, -2, 4, -28, 18, 8

Offset: 0

Gevorg Hmayakyan, Feb 11 2017

The a(n,m) forms a table where each row has (n*(n-3)+4)/2 = A152947(n) elements.
The index of the first row is n=1 and the index of the first column is m=0.
The right diagonal a(n, A152947(n)) = A000010(n), Euler Totient function.

			The first few rows are:
0, 1;
-1, 1;
2, -4, 2;
-4, 10, -6, -2, 2;
6, -16, 10, 4, -6, 8, -10, 4;
-10, 28, -18, -8, 10, -10, 10, -2, 8, -10, 0, 2;
12, -34, 22, 10, -12, 12, -22, 30, -30, 6, 10, -10, 8, 0, 6, -14, 6;

Cf. A000010, A152947, A231599.

U[n_, m_] := U[n, m] = If[n > 1, U[n - 1, n*(n - 1)/2 - m]*(-1)^n - U[n - 1, m], 0]
U[1, m_] := U[1, m] = If[m == 0, 1, 0]
Q[n_, m_] := U[n, m - 2] - 2*U[n, m - 1] + U[n, m]
nu[n_]:=(n-1)*n/2+2-n
a[n_, m_] := a[n, m] = If[(m < 0) || (nu[n] < m), 0, a[n - 1, m - n + 1] - a[n - 1, m] - a[n - 1, nu[n - 1]]*Q[n - 1, m]]
a[1, m_] := a[1, m] = If[m == 1, 1, 0]
Table[Table[a[n, m], {m, 0, nu[n]}], {n, 1, 20}]
Table[a[n, nu[n]], {n, 1, 50}]

nu(n) = (n*(n-3)+4)/2
Q(n,m) = 2*A231599(n,m-1)-A231599(n,m-2)-A231599(n,m)
a(n, m) = a(n - 1, m - n + 1) - a(n - 1, m) - a(n - 1, nu(n - 1))*Q(n - 1, m) if (m < 0) or (nu(n) < m)
a(1,m)=1 if m=1 and 0 otherwise.
a(n,nu(n))= A000010(n)

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.

A330794 Inverse of the Jacobsthal triangle (A322942). Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, -1, 1, 1, -2, 1, -1, 1, -3, 1, 1, 4, 2, -4, 1, -1, -7, 10, 4, -5, 1, 1, -14, -25, 16, 7, -6, 1, -1, 65, -21, -55, 21, 11, -7, 1, 1, -24, 196, -8, -98, 24, 16, -8, 1, -1, -367, -204, 400, 42, -154, 24, 22, -9, 1, 1, 774, -963, -688, 666, 148, -222, 20, 29, -10, 1

Offset: 0

Peter Luschny, Jan 03 2020

The inverse matrix of the Riordan square (cf. A321620) generated by (1 - 2*x^2)/((1 + x)*(1 - 2*x)).

			Triangle starts:
[0]   1;
[1]  -1,    1;
[2]   1,   -2,    1;
[3]  -1,    1,   -3,    1;
[4]   1,    4,    2,   -4,    1;
[5]  -1,   -7,   10,    4,   -5,    1;
[6]   1,  -14,  -25,   16,    7,   -6,    1;
[7]  -1,   65,  -21,  -55,   21,   11,   -7,    1;
[8]   1,  -24,  196,   -8,  -98,   24,   16,   -8,    1;
[9]  -1, -367, -204,  400,   42, -154,   24,   22,   -9,    1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
G. C. Greubel, SageMath code

Cf. A152947, A256893, A321620, A322942.

m=30;
A322942:= CoefficientList[CoefficientList[Series[(1-2*t^2)/(1-(x+1)*t-2*t^2), {x,0,m}, {t,0,m}], t], x];
M:= M= Table[If[k<=n, A322942[[n+1,k+1]], 0], {n,0,m}, {k,0,m}];
g:= g= Inverse[M];
A330794[n_, k_]:= g[[n+1,k+1]];
Table[A330794[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 20 2023 *)

# uses[riordan_array from A256893]
Jacobsthal = (2*x^2 - 1)/((x + 1)*(2*x - 1))
riordan_array(Jacobsthal, Jacobsthal, 10).inverse()

From G. C. Greubel, Sep 15 2023: (Start)
T(n, 0) = (-1)^n.
T(n, n) = 1.
T(n, n-1) = -n.
T(n, n-2) = A152947(n-1). (End)

cp's OEIS Frontend

A302537 a(n) = (n^2 + 13*n + 2)/2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Maxima

PARI

Python

Formula

A371417 Triangle read by rows: T(n,k) is the number of complete compositions of n with k parts.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

A092365 Coefficient of X^2 in expansion of (1 + n*X + n*X^2)^n.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A159938 The number of homogeneous trisubstituted linear alkanes.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions

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.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Maxima

Formula

A342939 a(n) is the Skolem number of the triangular grid graph T_n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A240168 T(n,k) is the number of unlabeled graphs of n vertices and k edges that have endpoints, where an endpoint is a vertex with degree 1.

Views

Author

Keywords

Comments

Examples

A092365 Coefficient of X^2 in expansion of (1 + nX + nX^2)^n.

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.