A100536 -id:A100536 - cp's OEIS frontend

A239331 Square array, read by antidiagonals: column k has g.f. (1+(k-1)*x)^2/(1-x)^3.

1, 1, 1, 1, 3, 1, 1, 5, 6, 1, 1, 7, 13, 10, 1, 1, 9, 22, 25, 15, 1, 1, 11, 33, 46, 41, 21, 1, 1, 13, 46, 73, 79, 61, 28, 1, 1, 15, 61, 106, 129, 121, 85, 36, 1, 1, 17, 78, 145, 191, 201, 172, 113, 45, 1, 1, 19, 97, 190, 265, 301, 289, 232, 145, 55, 1, 1, 21

Offset: 0

Philippe Deléham, Mar 16 2014

			Square array begins:
n\k : 0......1......2......3......4......5......6......7......8......9
======================================================================
.0||  1......1......1......1......1......1......1......1......1......1
.1||  1......3......5......7......9.....11.....13.....15.....17.....19
.2||  1......6.....13.....22.....33.....46.....61.....78.....97....118
.3||  1.....10.....25.....46.....73....106....145....190....241....298
.4||  1.....15.....41.....79....129....191....265....351....449....559
.5||  1.....21.....61....121....201....301....421....561....721....901
.6||  1.....28.....85....172....289....436....613....820...1057...1324
.7||  1.....36....113....232....393....596....841...1128...1457...1828
.8||  1.....45....145....301....513....781...1105...1485...1921...2413
.9||  1.....55....181....379....649....991...1405...1891...2449...3079
10||  1.....66....221....466....801...1226...1741...2346...3041...3826
11||  1.....78....265....562....969...1486...2113...2850...3697...4654

Cf. Columns: A000012, A000217, A001844, A038764, A081585, A081587, A081589, A081591, A081593.

Cf. Rows: A000012, A005408, A028872, A100536, A239325, A069133.

T(n,k) = 3*T(n-1,k) - 3*T(n-2,k) + T(n-3,k).
T(n,k) = 3*T(n,k-1) - 3*T(n,k-2) + T(n,k-3).
T(n,k) = (T(n,k-1) + T(n,k+1))/2 - A161680(n).
T(n,k) = (T(n-1,k) + T(n+1,k) - A000290(n))/2.

A271723 Numbers k such that 3*k - 8 is a square.

Original entry on oeis.org

3, 4, 8, 11, 19, 24, 36, 43, 59, 68, 88, 99, 123, 136, 164, 179, 211, 228, 264, 283, 323, 344, 388, 411, 459, 484, 536, 563, 619, 648, 708, 739, 803, 836, 904, 939, 1011, 1048, 1124, 1163, 1243, 1284, 1368, 1411, 1499, 1544, 1636, 1683, 1779, 1828, 1928, 1979, 2083, 2136, 2244, 2299

Offset: 1

Juri-Stepan Gerasimov, Apr 13 2016

Square roots of resulting squares gives A001651. - Ray Chandler, Apr 14 2016

			a(1) = 3 because 3*3 - 8 = 1^2.

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A001651.

Cf. numbers n such that 3*n + k is a square: this sequence (k=-8), A120328 (k=-6), A271713 (k=-5), A056107 (k=-3), A257083 (k=-2), A033428 (k=0), A001082 (k=1), A080663 (k=3), A271675 (k=4), A100536 (k=6), A271741 (k=7), A067725 (k=9).

Magma
```
[n: n in [1..2400] | IsSquare(3*n-8)];
```

seq(seq(((3*m+k)^2+8)/3, k=1..2),m=0..50); # Robert Israel, Dec 05 2016

Select[Range@ 2400, IntegerQ@ Sqrt[3 # - 8] &] (* Bruno Berselli, Apr 14 2016 *)
LinearRecurrence[{1,2,-2,-1,1},{3,4,8,11,19},60] (* Harvey P. Dale, Oct 02 2020 *)

from gmpy2 import is_square
[n for n in range(3000) if is_square(3*n-8)] # Bruno Berselli, Dec 05 2016

[(6*(n-1)*n-(2*n-1)*(-1)**n+23)/8 for n in range(1, 60)] # Bruno Berselli, Dec 05 2016

From Ilya Gutkovskiy, Apr 13 2016: (Start)
G.f.: x*(3 + x - 2*x^2 + x^3 + 3*x^4)/((1 - x)^3*(1 + x)^2).
a(n) = (6*(n - 1)*n - (2*n - 1)*(-1)^n + 23)/8. (End)

A193516 T(n,k) = number of ways to place any number of 4X1 tiles of k distinguishable colors into an nX1 grid.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 5, 4, 1, 1, 1, 5, 7, 7, 5, 1, 1, 1, 6, 9, 10, 9, 7, 1, 1, 1, 7, 11, 13, 13, 15, 10, 1, 1, 1, 8, 13, 16, 17, 25, 25, 14, 1, 1, 1, 9, 15, 19, 21, 37, 46, 39, 19, 1, 1, 1, 10, 17, 22, 25, 51, 73, 76, 57, 26, 1, 1, 1, 11, 19, 25, 29, 67, 106, 125

Offset: 1

R. H. Hardin, with proof and formula from Robert Israel in the Sequence Fans Mailing List, Jul 29 2011

Table starts:
..1...1...1...1....1....1....1....1....1....1....1.....1.....1.....1.....1
..1...1...1...1....1....1....1....1....1....1....1.....1.....1.....1.....1
..1...1...1...1....1....1....1....1....1....1....1.....1.....1.....1.....1
..2...3...4...5....6....7....8....9...10...11...12....13....14....15....16
..3...5...7...9...11...13...15...17...19...21...23....25....27....29....31
..4...7..10..13...16...19...22...25...28...31...34....37....40....43....46
..5...9..13..17...21...25...29...33...37...41...45....49....53....57....61
..7..15..25..37...51...67...85..105..127..151..177...205...235...267...301
.10..25..46..73..106..145..190..241..298..361..430...505...586...673...766
.14..39..76.125..186..259..344..441..550..671..804...949..1106..1275..1456
.19..57.115.193..291..409..547..705..883.1081.1299..1537..1795..2073..2371
.26..87.190.341..546..811.1142.1545.2026.2591.3246..3997..4850..5811..6886
.36.137.328.633.1076.1681.2472.3473.4708.6201.7976.10057.12468.15233.18376

			Some solutions for n=9 k=3; colors=1, 2, 3; empty=0
..0....3....0....0....3....3....0....0....0....0....2....2....0....0....1....2
..1....3....0....2....3....3....3....0....0....0....2....2....1....0....1....2
..1....3....0....2....3....3....3....2....0....0....2....2....1....0....1....2
..1....3....3....2....3....3....3....2....1....0....2....2....1....0....1....2
..1....0....3....2....0....3....3....2....1....0....2....0....1....0....0....0
..2....3....3....2....0....3....3....2....1....3....2....2....0....0....0....3
..2....3....3....2....0....3....3....0....1....3....2....2....0....0....0....3
..2....3....0....2....0....3....3....0....0....3....2....2....0....0....0....3
..2....3....0....2....0....0....3....0....0....3....0....2....0....0....0....3

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 1 is A003269(n+1),
Column 2 is A052942,
Column 3 is A143454(n-3),
Row 8 is A082111,
Row 9 is A100536(n+1),
Row 10 is A051866(n+1).

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<4 or k=0, 1, k*T(n-4, k) +T(n-1, k)))
    end:
seq(seq(T(n, d+1-n), n=1..d), d=1..13); # Alois P. Heinz, Jul 29 2011

T[n_, k_] := T[n, k] = If[n < 0, 0, If[n < 4 || k == 0, 1, k*T[n-4, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 13}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. So T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n=0,1,...,z-1. The solution is T(n,k) = sum_r r^(-n-1)/(1 + z k r^(z-1)) where the sum is over the roots of the polynomial k x^z + x - 1.
T(n,k) = sum {s=0..[n/4]} (binomial(n-3*s,s)*k^s).
For z X 1 tiles, T(n,k,z) = sum{s=0..[n/z]} (binomial(n-(z-1)*s,s)*k^s). - R. H. Hardin, Jul 31 2011

A196507 a(n) = n(3n^2 + 6*n + 1).

Original entry on oeis.org

0, 10, 50, 138, 292, 530, 870, 1330, 1928, 2682, 3610, 4730, 6060, 7618, 9422, 11490, 13840, 16490, 19458, 22762, 26420, 30450, 34870, 39698, 44952, 50650, 56810, 63450, 70588, 78242, 86430, 95170, 104480, 114378, 124882

Offset: 0

R. J. Mathar, Oct 03 2011

Jolley, Summation of Series, Dover (1961), eq. 45 on page 8.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

[n*(3*n^2+6*n+1): n in [0..30]]; // Vincenzo Librandi, Oct 05 2011

a(n) = 2*5 + 5*8 + 8*11 + ... + (3*k-1)*(3*k+2) + ... (n terms) = n*A100536(n+1).
G.f.: -2*x*(-5 - 5*x + x^2) / (x-1)^4.
E.g.f.: exp (x)*(10*x + 15*x^2 + 3*x^3). - Franck Maminirina Ramaharo, Nov 22 2018

A361731 Array read by descending antidiagonals. A(n, k) = hypergeom([-k, -3], [1], n).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 10, 7, 1, 1, 20, 25, 10, 1, 1, 35, 63, 46, 13, 1, 1, 56, 129, 136, 73, 16, 1, 1, 84, 231, 307, 245, 106, 19, 1, 1, 120, 377, 586, 593, 396, 145, 22, 1, 1, 165, 575, 1000, 1181, 1011, 595, 190, 25, 1, 1, 220, 833, 1576, 2073, 2076, 1585, 848, 241, 28, 1

Offset: 0

Peter Luschny, Mar 22 2023

			Array A(n, k) starts:
 [0] 1,  1,   1,   1,    1,    1,    1,     1, ...  A000012
 [1] 1,  4,  10,  20,   35,   56,   84,   120, ...  A000292
 [2] 1,  7,  25,  63,  129,  231,  377,   575, ...  A001845
 [3] 1, 10,  46, 136,  307,  586, 1000,  1576, ...  A081583
 [4] 1, 13,  73, 245,  593, 1181, 2073,  3333, ...  A081586
 [5] 1, 16, 106, 396, 1011, 2076, 3716,  6056, ...  A081588
 [6] 1, 19, 145, 595, 1585, 3331, 6049,  9955, ...  A081590
 [7] 1, 22, 190, 848, 2339, 5006, 9192, 15240, ...
.
Table T(n, k) starts:
 [0] 1;
 [1] 1,   1;
 [2] 1,   4,   1;
 [3] 1,  10,   7,    1;
 [4] 1,  20,  25,   10,    1;
 [5] 1,  35,  63,   46,   13,    1;
 [6] 1,  56, 129,  136,   73,   16,   1;
 [7] 1,  84, 231,  307,  245,  106,  19,   1;
 [8] 1, 120, 377,  586,  593,  396, 145,  22,  1;
 [9] 1, 165, 575, 1000, 1181, 1011, 595, 190, 25, 1;

Rows: A000012, A000292, A001845, A081583, A081586, A081588, A081590.
Columns: A000012, A016777, A100536.
Hypergeometric family: A000012 (m=0), A077028 (m=1), A361682 (m=2), this array (m=3).

A := (n, k) -> 1 + (((k*n - 3*n + 9)*n*k + (2*n - 9)*n + 18)*n*k)/6;
seq(print(seq(A(n, k), k = 0..7)), n = 0..7);
# Alternative:
ogf := n -> (1 + (n - 1) * x)^3 / (1 - x)^4:
ser := n -> series(ogf(n), x, 12):
row := n -> seq(coeff(ser(n), x, k), k = 0..9):
seq(print(row(n)), n = 0..9);

A(n, k) = [x^k] (1 + (n - 1) * x)^3 / (1 - x)^4.
A(n, k) = 1 + (((k*n - 3*n + 9)*n*k + (2*n - 9)*n + 18)*n*k)/6.
T(n, k) = 1 + (((k*(n - k) - 3*k + 9)*k*(n - k) + (2*k - 9)*k + 18)*k*(n - k))/6.

A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

Original entry on oeis.org

1, 1, -3, 1, -5, 5, 1, -7, 25, -5, 1, -9, 23, -35, 49, 1, -11, 73, -27, 112, -49, 1, -13, 53, -77, 629, -91, 58, 1, -15, 145, -130, 1399, -451, 753, -58, 1, -17, 95, -135, 2699, -2301, 8573, -869, 341, 1, -19, 241

Offset: 0

Paul Curtz, Apr 27 2012

In A190339 we saw that (the second Bernoulli numbers) A164555/A027642 is an eigensequence (its inverse binomial transform is the sequence signed) of the second kind, see A192456/A191302. We consider this array preceded by 1 for the second row, by 1, -3/2, for the third one; 1 is chosen and is followed by the differences of successive rows.
Hence
                    1    1/2   1/6      0
            1    -1/2   -1/3  -1/6  -1/30
      1   -3/2    1/6    1/6  2/15   1/15
  1 -5/2   5/3      0  -1/30 -1/15 -8/105.
The second row is A051716/A051717.
The (reduced) triangle before the square array (T(n,m) in A190339) is a(n)/b(n)=
B(0)=    1 = 1                 Redbernou1li
B(1)= -1/2 = 1  -3/2
B(2)=  1/6 = 1  -5/2  5/3
B(3)=    0 = 1  -7/2 25/6  -5/3
B(4)=-1/30 = 1  -9/2 23/3 -35/6  49/30
B(5)=    0 = 1 -11/2 73/6 -27/2 112/15 -49/30.
For the main diagonal, see A165142.
Denominator b(n) will be submitted.
This transform is valuable for every eigensequence of the second kind. For instance Leibniz's 1/n (A003506).
With increasing exponents for coefficients, polynomials CB(n,x) create Redbernou1li. See the formula.
Triangle Bernou1li for A027641/A027642 with the same denominator A080326 for every column is
1
1  -3/2
1  -5/2 10/6
1  -7/2 25/6 -10/6
1  -9/2 46/6 -35/6  49/30
1 -11/2 73/6 -81/6 224/30 -49/30.
For numerator by columns,see A000012, -A144396, A100536, Q(n)=n*(2*n^2+9*n+9)/2 , new.
Triangle Checkbernou1 with the same denominator A080326 for every row is
1/1
(2    -3)/2
(6   -15  +10)/6
(6   -21  +25  -10)/6
(30 -135 +230 -175  +49)/30
(30 -165 +365 -405 +224 -49)/30;
Hence for numerator: 1, 2-3, 16-15, 31-31, 309-310, 619-619, 8171-8166.
Absolute sum: 1, 5, 31, 62, 619, 1238, 17337. Reduced division by A080326:
1, 5/2, 31/6, 31/3, 619/30, 619/15, 5779/70, = A172030(n+1)/A172031(n+1).

Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane), A051714/A051715 (Akiyama-Tanigawa), A192456/A191302 for other triangles that lead to the Bernoulli numbers.

CB(0,x) = 1,
CB(1,x) = 1 - 3*x/2,
CB(n,x) = (1-x)*CB(n-1,x) + B(n)*x^n , n > 1.

A374584 Numbers k such that 7*k + 2 is a square.

Original entry on oeis.org

1, 2, 14, 17, 41, 46, 82, 89, 137, 146, 206, 217, 289, 302, 386, 401, 497, 514, 622, 641, 761, 782, 914, 937, 1081, 1106, 1262, 1289, 1457, 1486, 1666, 1697, 1889, 1922, 2126, 2161, 2377, 2414, 2642, 2681, 2921, 2962, 3214, 3257, 3521, 3566, 3842, 3889

Offset: 1

Juri-Stepan Gerasimov, Aug 12 2024

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

The numbers k such that (m + (9-m)*k) is a square: A000217 (m = 1), this sequence (m = 2), A003154 (m = 3), A195162 (m = 4), A028387 (m = 5), A100536 (m = 6), A059993 (m = 7), A028884 (m = 8).

Cf. A047341.

[k: k in [0..4000] | IsSquare(7*k + 2)];

((Table[7*n + {3, 4}, {n, 0, 23}] // Flatten)^2 - 2)/7 (* Amiram Eldar, Aug 12 2024 *)

a(n) = (A047341(n)^2 - 2)/7. - Amiram Eldar, Aug 12 2024

cp's OEIS Frontend

A239331 Square array, read by antidiagonals: column k has g.f. (1+(k-1)*x)^2/(1-x)^3.

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A271723 Numbers k such that 3*k - 8 is a square.

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A193516 T(n,k) = number of ways to place any number of 4X1 tiles of k distinguishable colors into an nX1 grid.

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A196507 a(n) = n*(3*n^2 + 6*n + 1).

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A361731 Array read by descending antidiagonals. A(n, k) = hypergeom([-k, -3], [1], n).

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A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

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A374584 Numbers k such that 7*k + 2 is a square.

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Magma

Mathematica

Formula

A196507 a(n) = n(3n^2 + 6*n + 1).