A003500 -id:A003500 - cp's OEIS frontend

A321118 T(n,k) = A321119(n) - (-1)^kA321119(n-2k)/2 for 0 < k < n, with T(0,0) = 0 and T(n,0) = T(n,n) = A002530(n+1) for n > 0, triangle read by rows; unreduced numerator of the weights of Holladay-Sard's quadrature formula.

0, 1, 1, 3, 10, 3, 4, 11, 11, 4, 11, 32, 26, 32, 11, 15, 43, 37, 37, 43, 15, 41, 118, 100, 106, 100, 118, 41, 56, 161, 137, 143, 143, 137, 161, 56, 153, 440, 374, 392, 386, 392, 374, 440, 153, 209, 601, 511, 535, 529, 529, 535, 511, 601, 209

Offset: 0

Franck Maminirina Ramaharo, Nov 01 2018

The n-th row common denominator is factorized out and is given by A321119(n).

Given a continuous function f over the interval [0,n], the best quadrature formula in the sense of Holladay-Sard is given by Integral_{x=0..n} f(x) dx = Sum_{k=0..n} T(n,k)*f(k)/A321119(n). The formula is exact if f belongs to the class of natural cubic splines.

			Triangle begins (denominator is factored out):
    0;                                                 1/4
    1,   1;                                            1/2
    3,  10,   3;                                       1/8
    4,  11,  11,   4;                                  1/10
   11,  32,  26,  32,  11;                             1/28
   15,  43,  37,  37,  43,  15;                        1/38
   41, 118, 100, 106, 100, 118,  41;                   1/104
   56, 161, 137, 143, 143, 137, 161,  56;              1/142
  153, 440, 374, 392, 386, 392, 374, 440, 153;         1/388
  209, 601, 511, 535, 529, 529, 535, 511, 601, 209;    1/530
  ...
If f is a continuous function over the interval [0,3], then the quadrature formula yields Integral_{x=0..3} f(x) d(x) = (1/10)*(4*f(0) + 11*f(1) + 11*f(2) + 4*f(3)).

Harold J. Ahlberg, Edwin N. Nilson and Joseph L. Walsh, The Theory of Splines and Their Applications, Academic Press, 1967. See p. 47, Table 2.5.2.

Franck Maminirina Ramaharo, Rows n = 0..150 of triangle, flattened
Harold J. Ahlberg, Edwin N. Nilson and Joseph L. Walsh, Chapter II The Cubic Spline, Mathematics in Science and Engineering Volume 38 (1967), p. 9-74.
John C. Holladay, A smoothest curve approximation, Math. Comp. Vol. 11 (1957), 233-243.
Peter Köhler, On the weights of Sard's quadrature formulas, CALCOLO Vol. 25 (1988), 169-186.
Leroy F. Meyers and Arthur Sard, Best approximate integration formulas, J. Math. Phys. Vol. 29 (1950), 118-123.
Arthur Sard, Best approximate integration formulas; best approximation formulas, American Journal of Mathematics Vol. 71 (1949), 80-91.
Isaac J. Schoenberg, Spline interpolation and best quadrature formulae, Bull. Amer. Math. Soc. Vol. 70 (1964), 143-148.
Frans Schurer, On natural cubic splines, with an application to numerical integration formulae, EUT report. WSK, Dept. of Mathematics and Computing Science Vol. 70-WSK-04 (1970), 1-32.

Cf. A093735, A093736, A100641, A002176, A100640, A321119, A321120, A321121, A321122.

alpha = (Sqrt[2] + Sqrt[6])/2; T[0,0] = 0;
T[n_, k_] := If[n > 0 && k == 0 || k == n, (alpha^(n + 1) - (-alpha)^(-(n + 1)))/(2*Sqrt[6]*(alpha^n + (-alpha)^(-n))), 1 - (-1)^k*(alpha^(n - 2*k) + (-alpha)^(2*k - n))/(2*(alpha^n + (-alpha)^(-n)))];
a321119[n_] := 2^(-Floor[(n - 1)/2])*((1 - Sqrt[3])^n + (1 + Sqrt[3])^n);
Table[FullSimplify[a321119[n]*T[n, k]],{n, 0, 10}, {k, 0, n}] // Flatten

(b[0] : 0, b[1] : 1, b[2] : 1, b[3] : 3, b[n] := 4*b[n-2] - b[n-4])$ /* A002530 */
d(n) := 2^(-floor((n - 1)/2))*((1 - sqrt(3))^n + (1 + sqrt(3))^n) $ /* A321119 */
T(n, k) := if n = 0 and k = 0 then 0 else if n > 0 and k = 0 or k = n then b[n + 1] else d(n) - (-1)^k*d(n - 2*k)/2$
create_list(ratsimp(T(n, k)), n, 0, 10, k, 0, n);

T(n,k)/A321119(n) = (alpha^(n + 1) - (-alpha)^(-(n + 1)))/(2*sqrt(6)*(alpha^n + (-alpha)^(-n))) if k = 0 or k = n, and 1 - (-1)^k*(alpha^(n - 2*k) + (-alpha)^(2*k - n))/(2*(alpha^n + (-alpha)^(-n))) if 0 < k < n, where alpha = (sqrt(2) + sqrt(6))/2.
T(n,k) = T(n,n-k).
T(n,k) = 4*T(n-2,k) - T(n-4,k), n >= k + 4.
T(2*n+2,k)*A001834(n+1) = A001834(n)*T(2*n,k) + 2*A003500(n)*T(2*n+1,k) for k < 2*n.
T(2*n+3,k)*A003500(n+1) = A003500(n)*T(2*n+1,k) + 2*A001834(n+1)*T(2*n+2,k) for k < 2*n + 1.
Sum_{k=0..n} T(n,k)/A321119(n) = n.

A321119 a(n) = ((1 - sqrt(3))^n + (1 + sqrt(3))^n)/2^floor((n - 1)/2); n-th row common denominator of A321118.

Original entry on oeis.org

4, 2, 8, 10, 28, 38, 104, 142, 388, 530, 1448, 1978, 5404, 7382, 20168, 27550, 75268, 102818, 280904, 383722, 1048348, 1432070, 3912488, 5344558, 14601604, 19946162, 54493928, 74440090, 203374108, 277814198, 759002504, 1036816702, 2832635908, 3869452610

Offset: 0

Franck Maminirina Ramaharo, Nov 01 2018

			a(0) = ((1 - sqrt(3))^0 + (1 + sqrt(3))^0)/2^floor((0 - 1)/2) = 2*(1 + 1) = 4.

Harold J. Ahlberg, Edwin N. Nilson and Joseph L. Walsh, The Theory of Splines and Their Applications, Academic Press, 1967. See p. 47, Table 2.5.2.

Encyclopedia of Mathematics, Quadrature formula
John C. Holladay, A smoothest curve approximation, Math. Comp. Vol. 11 (1957), 233-243.
Peter Köhler, On the weights of Sard's quadrature formulas, CALCOLO Vol. 25 (1988), 169-186.
Leroy F. Meyers and Arthur Sard, Best approximate integration formulas, J. Math. Phys. Vol. 29 (1950), 118-123.
Arthur Sard, Best approximate integration formulas; best approximation formulas, American Journal of Mathematics Vol. 71 (1949), 80-91.
Isaac J. Schoenberg, Spline interpolation and best quadrature formulae, Bull. Amer. Math. Soc. Vol. 70 (1964), 143-148.
Frans Schurer, On natural cubic splines, with an application to numerical integration formulae, EUT report. WSK, Dept. of Mathematics and Computing Science Vol. 70-WSK-04 (1970), 1-32.
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).

Cf. A002176 (common denominators of Cotesian numbers).
Cf. A321118, A321120.
Cf. A001834, A002530, A002531, A003500, A005246, A048788, A083336, A125905.

LinearRecurrence[{0, 4, 0, -1}, {4, 2, 8, 10}, 50]

a(n) := ((1 - sqrt(3))^n + (1 + sqrt(3))^n)/2^floor((n - 1)/2)$
makelist(ratsimp(a(n)), n, 0, 50);

a(n) = (((sqrt(2) - sqrt(6))/2)^n + ((sqrt(6) + sqrt(2))/2)^n)*((2 - sqrt(2))*(-1)^n + 2 + sqrt(2))/2.
a(-n) = (-1)^n*a(n).
a(n) = 2*A000034(n+1)*A002531(n).
a(2*n) = 2*A001834(n).
a(2*n+1) = 2*A003500(n).
a(n) = 4*a(n-2) - a(n-4) with a(0) = 4, a(1) = 2, a(2) = 8, a(3) = 10.
a(2*n+3) = a(2*n+1) + a(2*n+2).
a(2*n+2) = a(2*n) + 2*a(2*n+1).
G.f.: 2*(1 - x)*(2 + 3*x - x^2)/(1 - 4*x^2 + x^4).
E.g.f.: (1 + exp(-sqrt(6)*x))*((2 - sqrt(2))*exp(sqrt(2 - sqrt(3))*x) + (2 + sqrt(2))*exp(sqrt(2 + sqrt(3))*x))/2.
Lim_{n->infinity} a(2*n+1)/a(2*n) = (1 + sqrt(3))/2.

A001352 a(0) = 1, a(1) = 6, a(2) = 24; for n>=3, a(n) = 4a(n-1) - a(n-2).

Original entry on oeis.org

1, 6, 24, 90, 336, 1254, 4680, 17466, 65184, 243270, 907896, 3388314, 12645360, 47193126, 176127144, 657315450, 2453134656, 9155223174, 34167758040, 127515808986, 475895477904, 1776066102630, 6628368932616, 24737409627834, 92321269578720, 344547668687046

Offset: 0

N. J. A. Sloane

Also the coordination sequence of a {4,6} tiling of the hyperbolic plane, where there are 6 squares (with vertex angles Pi/3) around every vertex. - toen (tca110(AT)rsphysse.anu.edu.au), May 16 2005

a(n) is related to the almost-equilateral Heronian triangles because it is the area of the Heronian triangle with edge lengths A003500(n)-1, A003500(n)+1 and 4. - Herbert Kociemba, Mar 19 2021

Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
D. Fortin, B-spline Toeplitz inverse under corner perturbations, International Journal of Pure and Applied Mathematics, Volume 77, No. 1, 2012, 107-118. - From _N. J. A. Sloane_, Oct 22 2012
T. N. E. Greville, Table for third-degree spline interpolations with equally spaced arguments, Math. Comp., 24 (1970), 179-183.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (4,-1).

First differences of A082841. Pairwise sums of A001834.

A001352 := proc(n) coeftayl(1+6*x/(1-4*x+x^2),x=0,n) ; end: for n from 0 to 30 do printf("%d,",A001352(n)) ; od ; # R. J. Mathar, Jun 06 2007
A001352:=(z+1)**2/(1-4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Join[{1},LinearRecurrence[{4,-1},{6,24},30]] (* Harvey P. Dale, Jul 20 2011 *)

Vec((x+1)^2/(x^2-4*x+1) + O(x^40)) \\ Colin Barker, Oct 12 2015

G.f.: 1+6x/(1-4x+x^2). - R. J. Mathar, Jun 06 2007

a(n) = sqrt(3)*(-(2-sqrt(3))^n+(2+sqrt(3))^n) for n>0. - Colin Barker, Oct 12 2015

More terms from R. J. Mathar, Jun 06 2007

A151961 Semiperimeter of the n-th Heronian triangle.

Original entry on oeis.org

3, 6, 21, 78, 291, 1086, 4053, 15126, 56451, 210678, 786261, 2934366, 10951203, 40870446, 152530581, 569251878, 2124476931, 7928655846, 29590146453, 110431929966, 412137573411, 1538118363678, 5740335881301, 21423225161526, 79952564764803, 298387033897686

Offset: 1

Juri-Stepan Gerasimov, Jul 13 2009

The side lengths are consecutive integers (A016064) and the area is an integer (A011945).
Except for the first term, positive values of x (or y) satisfying  x^2 - 4*x*y + y^2 + 27 = 0. - Colin Barker, Feb 08 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 432 = 0. - Colin Barker, Feb 16 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
G. Jacob Martens, Rational right triangles and the Congruent Number Problem, arXiv:2112.09553 [math.GM], 2021, see section 10.1 The Brahmaguptra triangles, equation (99).
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075, A003500, A011945, A016064.

I:=[3,6]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 11 2014

CoefficientList[Series[3(1-2x)/(1-4x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Feb 11 2014 *)
3*ChebyshevT[Range[0, 40], 2] (* G. C. Greubel, Oct 10 2022 *)
LinearRecurrence[{4,-1},{3,6},30] (* Harvey P. Dale, Dec 21 2022 *)

Vec(3*x*(1-2*x)/(1-4*x+x^2) + O(x^40)) \\ Colin Barker, Oct 12 2015

[3*chebyshev_T(n, 2) for n in range(41)] # G. C. Greubel, Oct 10 2022

a(n) = 3 * A001075(n-1). - Joerg Arndt, Oct 10 2022
a(n) = 3*(A016064(n-1) + 1)/2 = 3*A003500(n-1)/2. - R. J. Mathar, Jul 27 2009
From Colin Barker, Mar 30 2012: (Start)
a(n) = 4*a(n-1) - a(n-2).
G.f.: 3*x*(1-2*x)/(1-4*x+x^2). (End)
a(n) = 3*( (2+sqrt(3))*(2-sqrt(3))^n + (2-sqrt(3))*(2+sqrt(3))^n )/2. - Colin Barker, Oct 12 2015

More terms from R. J. Mathar, Jul 27 2009

A191348 Array read by antidiagonals: ((ceiling(sqrt(n)) + sqrt(n))^k + (ceiling(sqrt(n)) - sqrt(n))^k)/2 for columns k >= 0 and rows n >= 0.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 6, 2, 1, 0, 8, 20, 7, 2, 1, 0, 16, 68, 26, 8, 3, 1, 0, 32, 232, 97, 32, 14, 3, 1, 0, 64, 792, 362, 128, 72, 15, 3, 1, 0, 128, 2704, 1351, 512, 376, 81, 16, 3, 1, 0

Offset: 0

Charles L. Hohn, May 31 2011

			1, 0,  0,   0,    0,     0,      0,      0,       0,        0,         0, ...
1, 1,  2,   4,    8,    16,     32,     64,     128,      256,       512, ...
1, 2,  6,  20,   68,   232,    792,   2704,    9232,    31520,    107616, ...
1, 2,  7,  26,   97,   362,   1351,   5042,   18817,    70226,    262087, ...
1, 2,  8,  32,  128,   512,   2048,   8192,   32768,   131072,    524288, ...
1, 3, 14,  72,  376,  1968,  10304,  53952,  282496,  1479168,   7745024, ...
1, 3, 15,  81,  441,  2403,  13095,  71361,  388881,  2119203,  11548575, ...
1, 3, 16,  90,  508,  2868,  16192,  91416,  516112,  2913840,  16450816, ...
1, 3, 17,  99,  577,  3363,  19601, 114243,  665857,  3880899,  22619537, ...
1, 3, 18, 108,  648,  3888,  23328, 139968,  839808,  5038848,  30233088, ...
1, 4, 26, 184, 1316,  9424,  67496, 483424, 3462416, 24798784, 177615776, ...
1, 4, 27, 196, 1433, 10484,  76707, 561236, 4106353, 30044644, 219825387, ...
1, 4, 28, 208, 1552, 11584,  86464, 645376, 4817152, 35955712, 268377088, ...
1, 4, 29, 220, 1673, 12724,  96773, 736012, 5597777, 42574180, 323800109, ...
1, 4, 30, 232, 1796, 13904, 107640, 833312, 6451216, 49943104, 386642400, ...
...

Row 1 is A000007, row 2 is A011782, row 3 is A006012, row 4 is A001075, row 5 is A081294, row 6 is A098648, row 7 is A084120, row 8 is A146963, row 9 is A001541, row 10 is A081341, row 11 is A084134, row 13 is A090965.
Row 3*2 is A056236, row 4*2 is A003500, row 5*2 is A155543, row 9*2 is A003499.
Cf. A191347 which uses floor() in place of ceiling().

T(n, k) = if (k==0, 1, if (k==1, ceil(sqrt(n)), T(n,k-2)*(n-T(n,1)^2) + T(n,k-1)*T(n,1)*2));
matrix(9, 9, n, k, T(n-1, k-1)) \\ Charles L. Hohn, Aug 23 2019

For each row n >= 0 let T(n,0)=1 and T(n,1) = ceiling(sqrt(n)), then for each column k >= 2: T(n,k) = T(n,k-2)*(n-T(n,1)^2) + T(n,k-1)*T(n,1)*2. - Charles L. Hohn, Aug 23 2019

A219163 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 4.

Original entry on oeis.org

4, 194, 1416317954, 4023861667741036022825635656102100994

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A003010.

a(4) has 147 digits and a(5) has 586 digits. - Harvey P. Dale, Mar 03 2020

Cf. A001999, A003010, A003500, A219162, A219164, A219165.

NestList[#^4-4#^2+2&,4,5] (* Harvey P. Dale, Mar 03 2020 *)

a(n)={if(n==0,4,a(n-1)^4-4*a(n-1)^2+2)} \\ Edward Jiang, Sep 11 2014

Let alpha = 2 + sqrt(3). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003010(2*n) = A003500(4^n).
Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(3).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 4. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

A237250 Values of x in the solutions to x^2 - 4xy + y^2 + 11 = 0, where 0 < x < y.

Original entry on oeis.org

2, 3, 5, 10, 18, 37, 67, 138, 250, 515, 933, 1922, 3482, 7173, 12995, 26770, 48498, 99907, 180997, 372858, 675490, 1391525, 2520963, 5193242, 9408362, 19381443, 35112485, 72332530, 131041578, 269948677, 489053827, 1007462178, 1825173730, 3759900035

Offset: 1

Colin Barker, Feb 05 2014

The corresponding values of y are given by a(n+2).

Positive values of x (or y) satisfying x^2 - 14xy + y^2 + 176 = 0.

			10 is in the sequence because (x, y) = (10, 37) is a solution to x^2 - 4xy + y^2 + 11 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).

Cf. A001075, A001835, A003500, A082841.

Vec(-x*(x-1)*(x+2)*(2*x+1)/(x^4-4*x^2+1) + O(x^100))

a(n) = 4*a(n-2)-a(n-4).

G.f.: -x*(x-1)*(x+2)*(2*x+1) / (x^4-4*x^2+1).

A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

Original entry on oeis.org

2, 2, 2, 2, 3, 2, 2, 4, 7, 2, 2, 5, 14, 18, 2, 2, 6, 23, 52, 47, 2, 2, 7, 34, 110, 194, 123, 2, 2, 8, 47, 198, 527, 724, 322, 2, 2, 9, 62, 322, 1154, 2525, 2702, 843, 2, 2, 10, 79, 488, 2207, 6726, 12098, 10084, 2207, 2, 2, 11, 98, 702, 3842, 15127, 39202, 57965, 37634, 5778, 2

Offset: 0

William W. Collier, Feb 18 2018

Note the similarity in form of the recursive steps in the array definition above and the polynomial definition under FORMULA.

			i\j |0  1   2    3      4       5        6          7           8            9
----+-------------------------------------------------------------------------
   0|2  2   2    2      2       2        2          2           2            2
   1|2  3   7   18     47     123      322        843        2207         5778
   2|2  4  14   52    194     724     2702      10084       37634       140452
   3|2  5  23  110    527    2525    12098      57965      277727      1330670
   4|2  6  34  198   1154    6726    39202     228486     1331714      7761798
   5|2  7  47  322   2207   15127   103682     710647     4870847     33385282
   6|2  8  62  488   3842   30248   238142    1874888    14760962    116212808
   7|2  9  79  702   6239   55449   492802    4379769    38925119    345946302
   8|2 10  98  970   9602   95050   940898    9313930    92198402    912670090
   9|2 11 119 1298  14159  154451  1684802   18378371   200477279   2186871698
  10|2 12 142 1692  20162  240252  2862862   34114092   406506242   4843960812
  11|2 13 167 2158  27887  360373  4656962   60180133   777684767  10049721838
  12|2 14 194 2702  37634  524174  7300802  101687054  1416317954  19726764302
  13|2 15 223 3330  49727  742575 11088898  165590895  2472774527  36926027010
  14|2 16 254 4048  64514 1028176 16386302  261152656  4162056194  66331746448
  15|2 17 287 4862  82367 1395377 23639042  400468337  6784322687 114933017342
  16|2 18 322 5778 103682 1860498 33385282  599074578 10749957122 192900153618
  17|2 19 359 6802 128879 2441899 46267202  876634939 16609796639 314709501202
  18|2 20 398 7940 158402 3160100 63043598 1257711860 25091193602 500566160180
  19|2 21 439 9198 192719 4037901 84603202 1772629341 37140612959 778180242798

William W. Collier, a(i,j) = f(i+2,j)
William W. Collier, Experimental Mathematics on Wisteria Tables, Talk to Poughkeepsie ACM Chapter.
OEIS Wiki, The (1,2) Pascal Triangle.

The array first appeared in A298675.
Rows 1 through 29 of the array appear in these OEIS entries: A005248, A003500, A003501, A003499, A056854, A086903, A056918, A087799, A057076, A087800, A078363, A067902, A078365, A090727, A078367, A087215, A078369, A090728, A090729, A090730, A090731, A090732, A090733, A090247, A090248, A090249, A090251. Also entries occur for rows 45, 121, and 320:  A087265, A065705, A089775. Each of these entries asserts that a(i,j)=f(i+2,j) is true for that row.
A few of the columns appear in the OEIS: A008865 (for column 2), A058794 and A007754 (for column 3), and A230586 (for column 5).
Main diagonal gives A343261.

A:= proc(i, j) option remember; `if`(min(i, j)=0, 2,
      `if`(j=1, i+2, (i+2)*A(i, j-1)-A(i, j-2)))
    end:
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Mar 05 2019

a[, 0] = a[0, ] = 2; a[i_, 1] := i + 2;
a[i_, j_] := a[i, j] =(i + 2) a[i, j - 1] - a[i, j - 2];
Table[a[i - j, j], {i, 0, 10}, {j, 0, i}] // Flatten (* Jean-François Alcover, Dec 07 2019 *)

Let k be an integer, and let r1 and r2 be the roots of x + 1/x = k. Then f(k,n) = r1^n + r2^n is an integer, for integer n >= 0. Theorem: a(i,j) = f(i+2,j), for i,j >= 0. Proof: See the Collier link.
Define polynomials recursively by:
    p[0](n) = 2, for n >= 0 ( [ and ] demark subscripts).
    p[1](n) = n + 2, for n >= 0.
    p[j](n) = p[j-1](n) * p[1](n) - p[j-2](n), for j > 1, n >= 0. The coefficients of these polynomials occur as the even numbered, upward diagonals in the OEIS Wiki link. Conjecture: a(i,j) = p[j](i), i,j >= 0.

Edited by N. J. A. Sloane, Apr 04 2018

A363348 Turn sequence of a non-Eulerian path for drawing an infinite aperiodic tiling based on the "hat" monotile. See the comments section for details.

Original entry on oeis.org

3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, -2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, -2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, -2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, 2, -3, -2, 3, 2, -3, 2, 3, -2, 3, -2

Offset: 1

Thomas Scheuerle, May 28 2023

The curve can be drawn using turtle graphics rules. Each term of the sequence encodes an angle of rotation in units of (1/6)*Pi. For example, a(k) = 3 would mean a turn of 90 degrees to the left, a(k) = -2 a turn of 60 degrees to the right. To draw the tiling we draw a line of length l and then take a term of the sequence to determine the direction of further drawing by rotation relative to the current drawing orientation. The length of the line segments between terms of the sequence is either sqrt(3) or 1 units. We start by drawing with sqrt(3) units of length; every time we reach a term with 3 or -3 in the sequence we toggle the selected line length from sqrt(3) to 1, or back again from 1 to sqrt(3).

The drawing process works by recursion into the H8 metatile and its supertiles; this means a(1..14) draws a single "hat" monotile. Then the terms a(1..140) draw the H8 metatile and a(1..1588) and so forth (see formula section) draw the next larger supertile of the H8 metatile. (For details regarding H8 see page 18 in arXiv:2303.10798.) The number of "hat" tiles visible after k recursions is Fibonacci(4*k + 2) (A033890); however, tiles and line segments will be overdrawn multiple times in this process.

			We start by drawing a line of length sqrt(3):
___
We take then the first term of the sequence a(1) = 3 this means
we turn our drawing turtle 90 degrees to the left and also switch to a length unit of 1.
___|
We take the second term from the sequence a(2) = -2 this means
we turn our drawing turtle 60 degrees to the right, and we keep the selected line length of 1 unit.
    /
___|
(In this ASCII representation, angles and length units are only symbolically represented and do not match the exact values in the description.)

Thomas Scheuerle, Table of n, a(n) for n = 1..1588
Thomas Scheuerle, MATLAB program
Thomas Scheuerle, 1st iteration: drawing of a(1..140) (results in 8 "hat" tiles).
Thomas Scheuerle, 2nd iteration: drawing of a(1..1588) (results in 55 "hat" tiles).
Thomas Scheuerle, 3rd iteration: drawing of a(1..206104) (results in 377 "hat" tiles).
Thomas Scheuerle, 4th iteration: drawing of a(1..2462272) (results in 2584 "hat" tiles).
David Smith, Joseph Samuel Myers, Craig S. Kaplan, and Chaim Goodman-Strauss, An aperiodic monotile, arXiv:2303.10798 [math.CO], 2023.

Cf. A363445 describes a curve around the perimeter of this tiling.

Cf. A003500, A003699, A033890, A052530, A061278, A108946.

MATLAB
```
% See Scheuerle link.
```

L(k) = { my(v = [0, 14, 140, 1588]); if(k > 3, return(12*L(k-1) - 7*L(k-2) + L(k-3)), return(v[k+1])) }
r1(k) = if(k > 1, return(r5(k-1) + r1(k-1) + r7(k-1)), return(6))
r2(k) = if(k > 1, return(r2(k-1) + r7(k-1)), return(6))
r3(k) = if(k > 1, return(2*r5(k-1) + r3(k-1) + r5(k) + r7(k-1)), return(6))
r5(k) = if(k > 1, return(r5(k-1) + r3(k-1)), return(2))
r7(k) = if(k > 1, return(r5(k-1) + 2*r3(k-1)), return(4))
r8(k) = if(k > 1, return(r12(k-1) + r8(k-1) + r14(k-1)), return(1))
r9(k) = if(k > 1, return(r9(k-1) + r14(k-1)), return(1))
r10(k) = if(k > 1, return(2*r13(k-1) + r10(k-1) + r11(k-1) + r14(k-1)), return(1))
r11(k) = if(k > 1, return(2*r13(k-1) + 3*r10(k-1) + r11(k-1)), return(1))
r12(k) = if(k > 1, return(r13(k-1) + r10(k-1)), return(1))
r13(k) = if(k > 1, return(r12(k-1) + r13(k-1) + r14(k-1)), return(1))
r14(k) = if(k > 1, return(r13(k-1) + 2*r10(k-1)), return(1))
c1(k) = r2(k) + sum(m=1, k-1, r9(k+1-m)*L(m))
c2(k) = c1(k) - sum(m=1, k-1, L(m))
c3(k) = r2(k) + r3(k) + sum(m=1, k-1, (r9(k+1-m) + r10(k+1-m) - 1)*L(m))
c4(k) = r2(k) + r5(k+1) + sum(m=1, k-1, (r9(k+1-m) + r11(k+1-m) - 1)*L(m))
c5(k) = r2(k) + r7(k) + sum(m=1, k-1, (r9(k+1-m) + r14(k+1-m) - 2)*L(m))
c6(k) = c4(k) - sum(m=1, k-1, L(m))
a(NumIter) = { my(a = [3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, 2, -3, 2]); for(k = 1, NumIter, a = concat([a, a[1..(c1(k)-1)], -a[c1(k)], a[(c2(k)+1)..L(k)], a[1..(c3(k)-1)], -a[c3(k)], a[(c2(k)+1)..L(k)], a[1..(c3(k)-1)], -a[c3(k)], a[(c2(k)+1)..L(k)], a[1..(c4(k)-1)], -a[c4(k)], a[(c5(k)+1)..L(k)], a[1..(c3(k)-1)], -a[c3(k)], a[(c2(k)+1)..L(k)], a[1..(c3(k)-1)], -a[c3(k)],  a[(c2(k)+1)..L(k)], a[1..(c4(k)-1)], -a[c4(k)], a[(c6(k)+1)..L(k)]]) ); return(a) }
draw(NumIter) = {my(p = [0, sqrt(3)]); my(dl = [1]); my(s = a(NumIter)); for(j=2, length(s), dl = concat(dl, ((dl[j-1]+(abs(s[j-1])==3))%2)); p = concat(p, p[j]+sqrt(1+2*dl[j])*exp(I*Pi*vecsum(s[1..j-1])*(1/6)) )); plothraw(apply(real, p), apply(imag, p), 1); }

a(1..14) = {3,-2, 3,-2, 3, 2, 0, 2, -3, 2, 3, 2,-3, 2} = a(1..L(1)) and for k > 0:
a(1..L(k+1)) = {a(1..L(k)), a(1..c1(k)-1), -a(c1(k)), a(c2(k)+1..L(k)), a(1..c3(k)-1), -a(c3(k)), a(c2(k)+1..L(k)), a(1..c3(k)-1), -a(c3(k)), a(c2(k)+1..L(k)), a(1..c4(k)-1), -a(c4(k)), a(c5(k)+1..L(k)), a(1..c3(k)-1), -a(c3(k)), a(c2(k)+1..L(k)), a(1..c3(k)-1), -a(c3(k)), a(c2(k)+1..L(k)), a(1..c4(k)-1), -a(c4(k)), a(c6(k)+1..L(k))}. With:
L(k) = 12*L(k-1) - 7*L(k-2) + L(k-3) for k > 3 with L(1..3) = {14, 140, 1588}.
r1(k) = r5(k-1) + r1(k-1) + r7(k-1), with r1(1) = 6.
r2(k) = r2(k-1) + r7(k-1), with r2(1) = 6.
r3(k) = 2*r6(k-1) + r3(k-1) + r4(k-1) + r7(k-1), with r3(1) = 6 (A003699).
r4(k) = r5(k+1) = 2*r5(k-1) + 3*r3(k-1) + r4(k-1), with r4(1) = 8 (A052530).
r5(k) = r5(k-1) + r3(k-1), with r5(1) = 2. r4, r5, r6 are in the case of this tiling accidentally essentially the same recurrence.
r6(k) = r5(k) = r5(k-1) + r6(k-1) + r7(k-1), with r6(1) = 2 (A052530).
r7(k) = r6(k-1) + 2*r3(k-1), with r7(1) = 4 (A003500).
r8(k) = r12(k-1) + r8(k-1) + r14(k-1), with r8(1) = 1
r9(k) = r9(k-1) + r14(k-1), with r9(1) = 1.
r10(k) = 2*r13(k-1) + r10(k-1) + r11(k-1) + r14(k-1), with r10(1) = 1 (A061278).
r11(k) = 2*r13(k-1) + 3*r10(k-1) + r11(k-1), with r11(1) = 1.
r12(k) = r13(k-1) + r10(k-1), with r12(1) = 1.
r13(k) = r12(k-1) + r13(k-1) + r14(k-1), with r13(1) = 1.
r14(k) = r13(k-1) + 2*r10(k-1), with r14(1) = 1 (A108946 unsigned).
c1(k) = r2(k) + Sum_{m=1..k-1} (r9(k+1-m)*L(m)) = {6, 38, 374, 4204, ...}.
c2(k) = c1(k) - Sum_{m=1..k-1} L(m) = {6, 24, 220, 2462, ...}.
c3(k) = r2(k) + r3(k) + Sum_{m=1..k-1} ((r9(k+1-m) + r10(k+1-m) - 1)*L(m)) = {12, 116, 1282, 14572, ...}.
c4(k) = r2(k) + r4(k) + Sum_{m=1..k-1} ((r9(k+1-m) + r11(k+1-m) - 1)*L(m)) = {14, 138, 1550, 17630, ...}.
c5(k) = r2(k) + r7(k) + Sum_{m=1..k-1} ((r9(k+1-m) + r14(k+1-m) - 2)*L(m)) = {10, 66, 720, 8170, ...}.
c6(k) = c4(k) - Sum_{m=1..k-1} L(m) = {14, 124, 1396, 15888, ...}.
Description of curve position:
  OrientationAngle(n) = Sum_{k = 1..n-1} a(k)*Pi*(1/6).
Xcoordinate(n) = Sum_{k = 1..n} cos(OrientationAngle(n))*sqrt(1 + 2*((1 + Sum_{k = 1..n-1} [abs(a(k)) = 3]) mod 2)).
Ycoordinate(n) = Sum_{k = 1..n} sin(OrientationAngle(n))*sqrt(1 + 2*((1 + Sum_{k = 1..n-1} [abs(a(k)) = 3]) mod 2)). [] is the Iverson bracket here.

A363445 Turn sequence of a fractal-like curve which is also the perimeter around an aperiodic tiling based on the "hat" monotile. See the comments section for details.

Original entry on oeis.org

3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, -2, 0, 2, -3, 2, 3, -2, 0, 2, -3, 2, 3, -2, 0, 2, -3, 2, 3, 2, -3, -2, 3, -2, 0, 2, -3, 2, 3, -2, 0, 2, -3, 2, 3, 2, -3, -2, 3, -2, 3, -2, 3, -2, 0, 2, -3, -2, 3, -2, 0, 2, -3, 2, 3, -2, 0, 2, -3, 2, 3, 2, -3, -2, 3, -2, 0, 2, -3, -2

Offset: 1

Thomas Scheuerle, Jul 09 2023

sign

The curve can be drawn by turtle graphics rules. Each term of the sequence encodes an angle of rotation in units of (1/6)*Pi. For example, a(k) = 3 would mean a turn of 90 degrees to the left, a(k) = -2 a turn of 60 degrees to the right. To draw the tiling we draw a line of length l and then take a term of the sequence to determine the direction of further drawing by rotation relative to the current drawing orientation. The length of the line segments between each term of the sequence is either sqrt(3) or 1 units. We start by drawing with sqrt(3) units of length; every time we reach a term with 3 or -3 in the sequence we toggle the selected line length from sqrt(3) in 1, or back again from 1 in sqrt(3).
The curve is defined by recursion; this means a(1..14) draws a single "hat" monotile. Then the interval a(15..56) draws the perimeter around the H8 metatile and a(57..202) will be the perimeter around the next higher composition of these tiles. (For details regarding H8 see page 18 in arXiv:2303.10798.) The number of "hat" tiles enclosed by this curve after k recursions is Fibonacci(4*k + 2) (A033890).
The number of new terms added after each iteration can be calculated as m(k) = 5*m(k-1) - 5*m(k-2) + m(k-3) with m(1..3) = {14, 42, 146, ...}. After each such iteration the curve will be closed with an enclosed area equivalent to A033890(k+1) "hat" tiles.

			We start by drawing a line of length sqrt(3):
___
We then take the first term of the sequence, a(1) = 3: this means
we turn our drawing turtle 90 degrees to the left and also switch to a length unit of 1.
___|
We take the second term from the sequence, a(2) = -2: this means
we turn our drawing turtle 60 degrees to the right, and we keep the selected line length of 1 unit.
    /
___|
(In this ASCII representation, angles and length units are only symbolically represented and do not match the exact values in the description.)

Thomas Scheuerle, Table of n, a(n) for n = 1..2718
David Smith, Joseph Samuel Myers, Craig S. Kaplan, and Chaim Goodman-Strauss, An aperiodic monotile, arXiv:2303.10798 [math.CO], 2023.
Thomas Scheuerle, MATLAB program
Thomas Scheuerle, a(1..140480) drawn out by turtle graphics. This is the result of the seventh iteration.

Cf. A363348 describes how to draw this curve together with all "hat" monotiles enclosed by it.

Cf. A003500, A003699, A033890, A052530, A061278, A108946.

MATLAB
```
% See Scheuerle link.
```

L(k) = { my(v = [0, 14, 56, 202]); if(k > 3,return(6*L(k-1) - 10*L(k-2) + 6*L(k-3) - L(k-4)),return(v[k+1])) }
r1(k) = if(k > 1, return(r5(k-1) + r1(k-1) + r7(k-1)), return(6))
r2(k) = if(k > 1, return(r2(k-1) + r7(k-1)), return(6))
r3(k) = if(k > 1, return(2*r5(k-1) + r3(k-1) + r5(k) + r7(k-1)), return(6))
r5(k) = if(k > 1, return(r5(k-1) + r3(k-1)), return(2))
r7(k) = if(k > 1, return(r5(k-1) + 2*r3(k-1)), return(4))
c1(k) = r2(k) + L(k-1)
c2(k) = r2(k) + r3(k) + L(k-1)
c3(k) = r2(k) + r5(k+1) + L(k-1)
c4(k) = r2(k) + r7(k) + L(k-1)
a(NumIter) = { my(a = [3,-2, 3,-2, 3, 2, 0, 2, -3, 2, 3, 2,-3, 2]); for(k = 1, NumIter, a = concat([a, a[(L(k-1)+1)..(c1(k)-1)], -a[c1(k)], a[(c1(k)+1)..(c2(k)-1)], -a[c1(k)], a[(c1(k)+1)..(c2(k)-1)], -a[c1(k)], a[(c1(k)+1)..(c3(k)-1)], -a[c4(k)], a[(c4(k)+1)..(c2(k)-1)], -a[c1(k)], a[(c1(k)+1)..(c2(k)-1)], -a[c1(k)],  a[(c1(k)+1)..(c3(k)-1)], -a[c3(k)], a[(c3(k)+1)..length(a)] ]) ); return(a) }
draw(NumIter) = {my(p = [0, sqrt(3)]); my(dl = [1]); my(s = a(NumIter)); for(j=2,length(s), dl = concat(dl, ((dl[j-1]+(abs(s[j-1])==3))%2)); p = concat(p, p[j]+sqrt(1+2*dl[j])*exp(I*Pi*vecsum(s[1..j-1])*(1/6)) )); plothraw(apply(real,p),apply(imag,p), 1);}

a(1..14) = {3,-2, 3,-2, 3, 2, 0, 2, -3, 2, 3, 2,-3, 2} = a(1..L(1)) and for k > 0:
a(1..L(k+1)) = {a(1..L(k-1)), a(L(k)+1..c1(k)-1), -a(c1(k)), a(c1(k)+1..c2(k)-1), -a(c1(k)), a(c1(k)+1..c2(k)-1), -a(c1(k)), a(c1(k)+1..c3(k)-1), -a(c4(k)), a(c4(k)+1..c2(k)-1), -a(c1(k)), a(c1(k)+1..c2(k)-1), -a(c1(k)),  a(c1(k)+1..c3(k)-1), -a(c3(k)), a(c3(k)+1..L(k))}. With:
L(k) = 6*L(k-1) - 10*L(k-2) + 6*L(k-3) - L(k-4), for k > 3 and L(0..3) = {0, 14, 56, 202}.
L(k) = L(k-1) + r1(k-1) + 3*r3(k-1) + 2*r4(k-1) + r6(k-1).
r1(k) = r5(k-1) + r1(k-1) + r7(k-1), with r1(1) = 6.
r2(k) = r2(k-1) + r7(k-1), with r2(1) = 6.
r3(k) = 2*r6(k-1) + r3(k-1) + r4(k-1) + r7(k-1), with r3(1) = 6 (A003699).
r4(k) = r6(k+1) = 2*r5(k-1) + 3*r3(k-1) + r4(k-1), with r4(1) = 8 (A052530).
r5(k) = r5(k-1) + r3(k-1), with r5(1) = 2.  r4, r5, r6 are in the case of this tiling accidentally essentially the same recurrence.
r6(k) = r5(k) = r5(k-1) + r6(k-1) + r7(k-1), with r6(1) = 2 (A052530).
r7(k) = r6(k-1) + 2*r3(k-1), with r7(1) = 4 (A003500).
c1(k) = r2(k) + L(k)         = {6, 24, 80, ...}.
c2(k) = r2(k) + r3(k) + L(k) = {12, 46, 162, ...}.
c3(k) = r2(k) + r4(k) + L(k) = {14, 54, 192, ...}.
c4(k) = r2(k) + r7(k) + L(k) = {10, 38, 132, ...}.
Description of curve position:
  OrientationAngle(n) = Sum_{k = 1..n-1} a(k)*Pi*(1/6).
Xcoordinate(n) = Sum_{k = 1..n} cos(OrientationAngle(n))*sqrt(1 + 2*((1 + Sum_{k = 1..n-1} [abs(a(k)) = 3]) mod 2)).
Ycoordinate(n) = Sum_{k = 1..n} sin(OrientationAngle(n))*sqrt(1 + 2*((1 + Sum_{k = 1..n-1} [abs(a(k)) = 3]) mod 2)). [] is the Iverson bracket here.
For some nonnegative integers b and c:
  OrientationAngle(L(b)) = OrientationAngle(L(c)).
  Xcoordinate(L(b)) = Xcoordinate(L(c)).
  Ycoordinate(L(b)) = Ycoordinate(L(c)).

cp's OEIS Frontend

A321118 T(n,k) = A321119(n) - (-1)^k*A321119(n-2*k)/2 for 0 < k < n, with T(0,0) = 0 and T(n,0) = T(n,n) = A002530(n+1) for n > 0, triangle read by rows; unreduced numerator of the weights of Holladay-Sard's quadrature formula.

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A321119 a(n) = ((1 - sqrt(3))^n + (1 + sqrt(3))^n)/2^floor((n - 1)/2); n-th row common denominator of A321118.

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A001352 a(0) = 1, a(1) = 6, a(2) = 24; for n>=3, a(n) = 4a(n-1) - a(n-2).

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Maple

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A151961 Semiperimeter of the n-th Heronian triangle.

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A191348 Array read by antidiagonals: ((ceiling(sqrt(n)) + sqrt(n))^k + (ceiling(sqrt(n)) - sqrt(n))^k)/2 for columns k >= 0 and rows n >= 0.

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A219163 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 4.

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A237250 Values of x in the solutions to x^2 - 4xy + y^2 + 11 = 0, where 0 < x < y.

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A321118 T(n,k) = A321119(n) - (-1)^kA321119(n-2k)/2 for 0 < k < n, with T(0,0) = 0 and T(n,0) = T(n,n) = A002530(n+1) for n > 0, triangle read by rows; unreduced numerator of the weights of Holladay-Sard's quadrature formula.