A057721 -id:A057721 - cp's OEIS frontend

A073133 Table by antidiagonals of T(n,k) = n*T(n,k-1) + T(n,k-2) starting with T(n,1) = 1.

1, 1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 4, 10, 12, 5, 1, 5, 17, 33, 29, 8, 1, 6, 26, 72, 109, 70, 13, 1, 7, 37, 135, 305, 360, 169, 21, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89

Offset: 1

Henry Bottomley, Jul 16 2002

Columns of the array are generated from Fibonacci polynomials f(x). They are: (1), (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 +1), ... If column headings start 0, 1, 2, ... then the terms in the n-th column are generated from the n-th degree Fibonacci polynomial. For example, column 5 (8, 70, 360, ...) is generated from f(x), x = 1,2,3,...; fifth-degree polynomial x^5 + 4x^3 + 3x; e.g., f(2) = 70 = 2^5 + 4*8 + 3*2. - Gary W. Adamson, Apr 02 2006
The ratio of two consecutive entries of the sequence in the n-th row approaches (n + sqrt(n^2 + 4))/2. Example: The sequence beginning (1, 3, 10, 33, ...) tends to 3.302775... = (3 + sqrt(13))/2. - Gary W. Adamson, Aug 12 2013
As to the array sequences, (n+1)-th sequence is the INVERT transform of the n-th sequence. - Gary W. Adamson, Aug 20 2013
The array can be extended infinitely above the Fibonacci row by taking successive INVERTi transforms, resulting in:
  ...
  1,  -2,   5, -12,  29, -70, ...
  1,  -1,   2,  -3,   5,  -8, ...
  l,   0,   1,   0,   1,   0, ...
  1,   1,   2,   3,   5,   8, ...
  1,   2,   5,  12,  29,  70, ...
  ...
  This results in an infinite array in which sequences above the (1, 0, 1, 0, ...) are reflections of the sequences below, except for the alternate signs. Any sequence in the (+ sign) row starting (1, n, ...) is the (2*n-th) INVERT transform of the same sequence but with alternate signs.  Example: (1, 2, 5, 12, ...) is the (2*2) = fourth INVERT transform of (1, -2, 5, -12, ...) by inspection. Conjecture:  This "reflection" principle will result from taking successive INVERT transforms of any aerated sequence starting 1, ... and with positive signs. Likewise, the rows above the aerated sequence are successive INVERTi transforms of the aerated sequence. - Gary W. Adamson, Jul 14 2019
From Michael A. Allen, Feb 21 2023: (Start)
Row n is the n-metallonacci sequence.
T(n,k) is the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			Table begins:
  1, 1,  2,  3,   5,    8,   13, ...
  1, 2,  5, 12,  29,   70,  169, ...
  1, 3, 10, 33, 109,  360, 1189, ...
  1, 4, 17, 72, 305, 1292, 5473, ... etc.

G. C. Greubel, Antidiagonals n = 1..100, flattened
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Rows are A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A041041, A049666, A041061, A140455, A041085, etc.
Columns are A000012, A000027, A002522, A054602, A057721, A124152, etc.
Different from A081572.

T:= function(n,k)
    if k<0 then return 0;
    elif k=1 then return 1;
    else return n*T(n,k-1) + T(n,k-2);
    fi;
  end;
Flat(List([1..15], n-> List([1..n], k-> T(n-k+1,k) ))); # G. C. Greubel, Aug 12 2019

A073133 := proc(n,k)
    option remember;
    if k <= 1 then
        k;
    else
        n*procname(n,k-1)+procname(n,k-2) ;
    end if;
end proc:
seq(seq( A073133(d-k,k),k=1..d-1),d=2..13) ; # R. J. Mathar, Aug 16 2019

T[n_, 1]:= 1; T[n_, k_]:= T[n, k] = If[k<0, 0, n*T[n, k-1] + T[n, k-2]]; Table[T[n-k+1, k], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Aug 12 2019 *)

T(n,k) = if(k==1, 1, k<0, 0, n*T(n,k-1)+T(n,k-2));
for(n=1,15, for(k=1,n, print1(T(n-k+1,k), ", "))) \\ G. C. Greubel, Aug 12 2019

def T(n, k):
    if (k<0): return 0
    elif (k==1): return 1
    else: return n*T(n, k-1) + T(n, k-2)
[[T(n-k+1, k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Aug 12 2019

T(n, k) = A073134(n, k) + 2*A073135(n, k-2) = Sum_{j=0..k-1} abs(A049310(k-1, j)*n^j).
T(n,k) = [[0,1; 1,n]^{k+1}]{1,1}, n,k in {1,2,...}. - _L. Edson Jeffery, Sep 23 2012
G.f. for row n: x/(1-n*x-x^2). - L. Edson Jeffery, Aug 28 2013

A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 3, 0, 1, 4, 10, 12, 5, 0, 1, 5, 17, 33, 29, 8, 0, 1, 6, 26, 72, 109, 70, 13, 0, 1, 7, 37, 135, 305, 360, 169, 21, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 0, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89, 0, 1, 11, 101, 747, 4289, 18200, 53353, 98145, 98209, 42837, 5741, 144

Offset: 1

Roger L. Bagula, Jan 29 2010

Equals A073133 with an additional column A(.,0).
If the first column and top row are deleted, antidiagonal reading yields A118243.
Adding a top row of 1's and antidiagonal reading downwards yields A157103.
Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, ....
From Jianing Song, Jul 14 2018: (Start)
All rows have strong divisibility, that is, gcd(A(n,k_1), A(n,k_2)) = A(n,gcd(k_1,k_2)) holds for all k_1, k_2 >= 0.
Let E(n,m) be the smallest number l such that m divides A(n,l), we have: for odd primes p that are not divisible by n^2 + 4, E(n,p) divides p - ((n^2+4)/p) if p == 3 (mod 4) and (p - ((n^2+4)/p))/2 if p == 1 (mod 4). E(n,p) = p for odd primes p that are divisible by n^2 + 4. E(n,2) = 2 for even n and 3 for odd n. Here ((n^2+4)/p) is the Legendre symbol. A prime p such that p^2 divides T(n,E(n,p)) is called an n-Wall-Sun-Sun prime.
E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.
Let pi(n,m) be the Pisano period of A(n, k) modulo m, i.e, the smallest number l such that A(n, k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p - 1 if ((n^2+4)/p) = 1 and 2(p+1) if ((n^2+4)/p) = -1. pi(n,p) = 4p for odd primes p that are divisible by n^2 + 4. pi(n,2) = 2 even n and 3 for odd n.
pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n, p), so pi(n,p^e) = 4p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1), pi(n,m_2)) if gcd(m_1,m_2) = 1.
If n != 2, the largest possible value of pi(n,m)/m is 4 for even n and 6 for odd n. For even n, pi(n,p^e) = 4p^e; for odd n, pi(n,2p^e) = 12p^e, where p is any odd prime factor of n^2 + 4. For n = 2 it is 8/3, obtained by m = 3^e.
Let z(n,m) be the number of zeros in a period of A(n, k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: z(n,p) = 4 for odd primes p that are divisible by n^2 + 4. For other odd primes p, z(n,p) = 4 if E(n,p) is odd; 1 if E(n,p) is even but not divisible by 4; 2 if E(n,p) is divisible by 4; see the table below. z(n,2) = z(n,4) = 1.
Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have:
((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4
......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)*
......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)*
......1.........3,7.............1...................0...................0
.....-1.........1,5.............0...................0...................1
.....-1.........3,7.............0...................1...................0
* The result is that among all odd primes that are not divisible by n^2 + 4, 7/24 of them are with z(n,p) = 1, 5/12 are with z(n,p) = 2 and 7/24 are with z(n,p) = 4 if n^2 + 4 is a twice a square; 1/3 of them are with z(n,p) = 1, 1/3 are with z(n,p) = 2 and 1/3 are with z(n,p) = 4 otherwise. [Corrected by Jianing Song, Jul 06 2019]
z(n,p^e) = z(n,p) for all odd primes p; z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.
(End)
From Michael A. Allen, Mar 06 2023: (Start)
Removing the first (n=0) row of A352361 gives this sequence.
Row n is the n-metallonacci sequence.
A(n,k) is (for k>0) the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			The array, A(n, k), starts in row n = 1 with columns k >= 0 as
  0      1      1      2      3      5      8
  0      1      2      5     12     29     70
  0      1      3     10     33    109    360
  0      1      4     17     72    305   1292
  0      1      5     26    135    701   3640
  0      1      6     37    228   1405   8658
  0      1      7     50    357   2549  18200
  0      1      8     65    528   4289  34840
  0      1      9     82    747   6805  61992
  0      1     10    101   1020  10301 104030
  0      1     11    122   1353  15005 166408
Antidiagonal triangle, T(n, k), begins as:
  0;
  0, 1;
  0, 1, 1;
  0, 1, 2,  2;
  0, 1, 3,  5,   3;
  0, 1, 4, 10,  12,   5;
  0, 1, 5, 17,  33,  29,    8;
  0, 1, 6, 26,  72, 109,   70,   13;
  0, 1, 7, 37, 135, 305,  360,  169,  21;
  0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34;

Jianing Song, Antidiagonals n = 1..100, flattened (the first 30 antidiagonals from Vincenzo Librandi)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Rows n include: A000045 (n=1), A000129 (n=2), A006190 (n=3), A001076 (n=4), A052918 (n=5), A005668 (n=6), A054413 (n=7), A041025 (n=8), A099371 (n=9), A041041 (n=10), A049666 (n=11), A041061 (n=12), A140455 (n=13), A041085 (n=14), A154597 (n=15), A041113 (n=16), A178765 (n=17), A041145 (n=18), A243399 (n=19), A041181 (n=20). (Note that there are offset shifts for rows n = 5, 7, 8, 10, 12, 14, 16..20.)
Columns k include: A000004 (k=0), A000012 (k=1), A000027 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
Entry points for A(n,k) modulo m: A001177 (n=1), A214028 (n=2), A322907 (n=3).
Pisano period for A(n,k) modulo m: A001175 (n=1), A175181 (n=2), A175182 (n=3), A175183 (n=4), A175184 (n=5), A175185 (n=6).
Number of zeros in a period for A(n,k) modulo m: A001176 (n=1), A214027 (n=2), A322906 (n=3).
Cf. A084844, A084845, A118243, A157103, A271782, A316269.
Sums include: A304357, A304359.
Similar to: A073133.

A172236:= func< n,k | k le 1 select k else Evaluate(DicksonSecond(k-1,-1), n-k) >;
[A172236(n,k): k in [0..n-1], n in [1..13]]; // G. C. Greubel, Sep 29 2024

A172236[n_,k_]:=Fibonacci[k, n-k];
Table[A172236[n, k], {n,15}, {k,0,n-1}]//Flatten

A(n, k) = if (k==0, 0, if (k==1, 1, n*A(n, k-1) + A(n, k-2)));
tabl(nn) = for(n=1, nn, for (k=0, nn, print1(A(n, k), ", ")); print); \\ Jianing Song, Jul 14 2018 (program from Michel Marcus; see also A316269)

A(n, k) = ([n, 1; 1, 0]^k)[2, 1] \\ Jianing Song, Nov 10 2018

def A172236(n,k): return sum(binomial(k-j-1,j)*(n-k)^(k-2*j-1) for j in range(1+(k-1)//2))
flatten([[A172236(n,k) for k in range(n)] for n in range(1,14)]) # G. C. Greubel, Sep 29 2024

A(n,k) = (((n + sqrt(n^2 + 4))/2)^k - ((n-sqrt(n^2 + 4))/2)^k)/sqrt(n^2 + 4), n >= 1, k >= 0. - Jianing Song, Jun 27 2018
For n >= 1, Sum_{i=1..k} 1/A(n,2^i) = ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/A(n,2^k)), where u = (n + sqrt(n^2 + 4))/2, v = (n - sqrt(n^2 + 4))/2 are the two roots of the polynomial x^2 - n*x - 1. As a result, Sum_{i>=1} 1/A(n,2^i) = (n^2 + 4 - n*sqrt(n^2 + 4))/(2*n). - Jianing Song, Apr 21 2019
From G. C. Greubel, Sep 29 2024: (Start)
A(n, k) = F_{k}(n) (Fibonacci polynomials F_{n}(x)) (array).
T(n, k) = F_{k}(n-k) (antidiagonal triangle).
Sum_{k=0..n-1} T(n, k) = A304357(n) - (1-(-1)^n)/2.
Sum_{k=0..n-1} (-1)^k*T(n, k) = (-1)*A304359(n) + (1-(-1)^n)/2.
T(2*n, n) = A084844(n).
T(2*n+1, n+1) = A084845(n). (End)

More terms from Jianing Song, Jul 14 2018

A352361 Array read by ascending antidiagonals. A(n, k) = Fibonacci(k, n), where Fibonacci(n, x) are the Fibonacci polynomials.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 2, 2, 0, 0, 1, 3, 5, 3, 1, 0, 1, 4, 10, 12, 5, 0, 0, 1, 5, 17, 33, 29, 8, 1, 0, 1, 6, 26, 72, 109, 70, 13, 0, 0, 1, 7, 37, 135, 305, 360, 169, 21, 1, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 1

Offset: 0

Peter Luschny, Mar 18 2022

From Michael A. Allen, Mar 26 2023: (Start)
Row n is the n-metallonacci sequence for n>0.
A(n,k), for n > 0 and k > 0, is the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			Array, A(n,k), starts:
  n\k 0, 1, 2,  3,   4,    5,     6,      7,       8,        9, ...
  -------------------------------------------------------------------------
  [0] 0, 1, 0,  1,   0,    1,     0,      1,       0,        1, ... A000035;
  [1] 0, 1, 1,  2,   3,    5,     8,     13,      21,       34, ... A000045;
  [2] 0, 1, 2,  5,  12,   29,    70,    169,     408,      985, ... A000129;
  [3] 0, 1, 3, 10,  33,  109,   360,   1189,    3927,    12970, ... A006190;
  [4] 0, 1, 4, 17,  72,  305,  1292,   5473,   23184,    98209, ... A001076;
  [5] 0, 1, 5, 26, 135,  701,  3640,  18901,   98145,   509626, ... A052918;
  [6] 0, 1, 6, 37, 228, 1405,  8658,  53353,  328776,  2026009, ... A005668;
  [7] 0, 1, 7, 50, 357, 2549, 18200, 129949,  927843,  6624850, ... A054413;
  [8] 0, 1, 8, 65, 528, 4289, 34840, 283009, 2298912, 18674305, ... A041025;
  [9] 0, 1, 9, 82, 747, 6805, 61992, 564733, 5144589, 46866034, ... A099371;
      |  |  |  | A054602 |   A124152;
      |  |  |  A002522   A057721;
      |  |  A001477;
      |  A000012;
      A000004;
Antidiagonals, T(n, k), begin as:
  0;
  0, 1;
  0, 1, 0;
  0, 1, 1,  1;
  0, 1, 2,  2,   0;
  0, 1, 3,  5,   3,   1;
  0, 1, 4, 10,  12,   5,   0;
  0, 1, 5, 17,  33,  29,   8,   1;
  0, 1, 6, 26,  72, 109,  70,  13,  0;
  0, 1, 7, 37, 135, 305, 360, 169, 21, 1;

G. C. Greubel, Antidiagonals n = 0..50, flattened
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Other versions of this array are A073133, A157103, A172236.
Rows n: A000035 (n=0), A000045 (n=1), A000129 (n=2), A006190 (n=3), A001076 (n=4), A052918 (n=5), A005668 (n=6), A054413 (n=7), A041025 (n=8), A099371 (n=9).
Columns k: A000004 (k=0), A000012 (k=1), A001477 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
Cf. A084844 (main diagonal), A352362 (Lucas polynomials), A350470 (Jacobsthal polynomials).
Sums include: A304357 (row sums), A304359.
Cf. A084845.

A352361:= func< n, k | k le 1 select k else Evaluate(DicksonSecond(k-1, -1), n-k) >;
[A352361(n, k): k in [0..n], n in [0..13]]; // G. C. Greubel, Sep 29 2024

seq(seq(combinat:-fibonacci(k, n - k), k = 0..n), n = 0..11);

Table[Fibonacci[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten
(* or *)
A[n_, k_] := With[{s = Sqrt[n^2 + 4]}, ((n + s)^k - (n - s)^k) / (2^k*s)];
Table[Simplify[A[n, k]], {n, 0, 9}, {k, 0, 9}] // TableForm

A(n, k) = ([1, k; 1, k-1]^n)[2, 1] ;
export(A)
for(k = 0, 9, print(parvector(10, n, A(n - 1, k))))

def A352361(n, k): return lucas_number1(k,n-k,-1)
flatten([[A352361(n, k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Sep 29 2024

A(n, k) = Sum_{j=0..floor((k-1)/2)} binomial(k-j-1, j)*n^(k-2*j-1).
A(n, k) = ((n + s)^k - (n - s)^k) / (2^k*s) where s = sqrt(n^2 + 4).
A(n, k) = [x^k] (x / (1 - n*x - x^2)).
A(n, k) = n^(k-1)*hypergeom([1 - k/2, 1/2 - k/2], [1 - k], -4/n^2) for n,k >= 1.
A(n, n) = T(2*n, n) = A084844(n).
From G. C. Greubel, Sep 29 2024: (Start)
T(n, k) = A(n-k, k) (antidiagonal triangle).
T(2*n+1, n+1) = A084845(n).
Sum_{k=0..n} T(n, k) = A304357(n) (row sums).
Sum_{k=0..n} (-1)^k*T(n, k) = (-1)*A304359(n). (End)

A120062 Number of triangles with integer sides a <= b <= c having integer inradius n.

Original entry on oeis.org

1, 5, 13, 18, 15, 45, 24, 45, 51, 52, 26, 139, 31, 80, 110, 89, 33, 184, 34, 145, 185, 103, 42, 312, 65, 96, 140, 225, 36, 379, 46, 169, 211, 116, 173, 498, 38, 123, 210, 328, 44, 560, 60, 280, 382, 134, 64, 592, 116, 228, 230, 271, 47, 452, 229, 510, 276, 134, 54

Offset: 1

Hugo Pfoertner, Jun 11 2006

It is conjectured that the longest possible side c of a triangle with integer sides and inradius n is given by A057721(n) = n^4 + 3*n^2 + 1.
For n >= 1, a(n) >= 1 because triangle (a, b, c) = (n^2 + 2, n^4 + 2*n^2 + 1, n^4 + 3*n^2 + 1) has inradius n. - David W. Wilson, Jun 17 2006
Previous name was "Number of triangles with integer sides a<=bA362669); so, now effectively, a(10) = 52. - Bernard Schott, Apr 24 2023

			a(1)=1: {3,4,5} is the only triangle with integer sides and inradius 1.
a(2)=5: {5,12,13}, {6,8,10}, {6,25,29}, {7,15,20}, {9,10,17} are the only triangles with integer sides and inradius 2.
a(4)=A120252(1)+A120252(2)+A120252(4)=1+4+13 because 1, 2 and 4 are the factors of 4. The 1 primitive triangle with inradius n=1 is (3,4,5). The 4 primitive triangles with n=2 are (5,12,13), (9,10,17), (7,15,20), (6,25,29). The 13 primitive triangles with n=4 are (13,14,15), (15,15,24), (11,25,30), (15,26,37), (10,35,39), (9,40,41), (33,34,65), (25,51,74), (9,75,78), (11,90,97), (21,85,104), (19,153,170), (18,289,305). (Primitive means GCD(a, b, c, n)=1.)

David W. Wilson, Table of n, a(n) for n = 1..10000
Alan F. Beardon and Paul Stephenson, The Heron parameters of a triangle, Mathematical Gazette May 8, 2014.
Frank M Jackson, Mathematica program
Thomas Mautsch, Additional terms

Cf. A078644 [Pythagorean triangles with inradius n], A057721 [n^4+3*n^2+1].
Let S(n) be the set of triangles with integer sides a<=b<=c and inradius n. Then:
A120062(n) gives number of triangles in S(n).
A120261(n) gives number of triangles in S(n) with gcd(a, b, c) = 1.
A120252(n) gives number of triangles in S(n) with gcd(a, b, c, n) = 1.
A005408(n) = 2n+1 gives shortest short side a of triangles in S(n).
A120064(n) gives shortest middle side b of triangles in S(n).
A120063(n) gives shortest long side c of triangles in S(n).
A120570(n) gives shortest perimeter of triangles in S(n).
A120572(n) gives smallest area of triangles in S(n).
A058331(n) = 2n^2+1 gives longest short side a of triangles in S(n).
A082044(n) = n^4+2n^2+1 gives longest middle side b of triangles in S(n).
A057721(n) = n^4+3n^2+1 gives longest long side c of triangles in S(n).
A120571(n) = 2n^4+6n^2+4 gives longest perimeter of triangles in S(n).
A120573(n) = gives largest area of triangles in S(n).
Cf. A120252 [primitive triangles with integer inradius], A120063 [minimum of longest sides], A057721 [maximum of longest sides], A120064 [minimum of middle sides], A082044 [maximum of middle sides], A005408 [minimum of shortest sides], A058331 [maximum of shortest sides], A007237 [number of triangles with integer sides and area = n times perimeter].
Cf. A362669, A362670.

Mathematica
```
(* See link above. *)
```

The even-numbered terms are given by a(2*n)=A007237(n).

a(n) = Sum_{k|n} A120252(k).

More terms from Graeme McRae and Hugo Pfoertner, Jun 12 2006

Name corrected by Bernard Schott, Apr 24 2023

A331040 Numerator of squared radius of inscribed circle of a triangle with integer sides i <= j <= k, such that the number of triangles with this radius sets a new record. Denominators are A331041.

Original entry on oeis.org

1, 35, 3, 7, 3, 15, 8, 35, 55, 63, 95, 119, 135, 56, 231, 255, 80, 351, 455, 495, 855, 216, 224, 1071, 360

Offset: 1

Hugo Pfoertner, Jan 11 2020

The radius rho of the inscribed circle of a triangle (a,b,c) is rho = sqrt((s-a)*(s-b)*(s-c)/s), with s=(a+b+c)/2. For given integer values of a <= b and a rational target value r2 of the squared incircle radius, c is given by the two positive real roots of the polynomial P(a,b,x,r2) = x^3 - x^2 * (a+b) + x * (4*r2-(b-a)^2) + (a+b)^3 + 4*(a+b)*(r2-a*b). P(a,b,x,r2) = 0 may have 0, 1 or 2 positive integer solutions.

The potential ranges of the side lengths of the triangles can be determined in analogy to the ranges for the case of integer radii of the incircles, see A120062 for the relevant formulas and sequences.

			b(1) = a(1)/A331041(1) = 1/12: Triangle (1,1,1) has the least possible radius of incircle = sqrt(1/12).
b(2) = a(2)/A331041(2) = 35/52: Triangles (2,18,19) and (3,4,6) are the first occurrence of more than one triangle with the same radius of their incircles. rho = sqrt(35/52) in both cases.
b(3) = a(3)/A331041(3) = 3/4: Triangles are (2,7,7), (3,3,3), and (3,5,7).
b(4) = a(4)/A331041(4) = 7/4: (3,12,12), (3,22,23), (4,5,6), (5,18,22), (6,11,16) are the A331043(4) = 5 triangles with rho^2 = b(4).
b(15) = 231/4 includes the rare case, where two distinct integer solutions for the same pair of sides a and b exist: (20,37,38) and (20,37,39), both with rho^2=231/4 and thus contributing 2 of the A331043(15)=84 triangles with this squared radius of the incircle.

Hugo Pfoertner, Illustration of side lengths for terms a(13)-a(25).

Cf. A057721, A082044, A120062, A120572, A331012.

Cf. A331041 (corresponding denominators), A331042 (4*a(n)/A331041(n)), A331043 (records of numbers of triangles).

\\ Only suitable for demonstration of initial terms
rh2(a,b,c)={my(s=(a+b+c)/2);(s-a)*(s-b)*(s-c)/s};
lim_a(x)=ceil(4*(x^2+2));
lim_b(x)=ceil(4*(x^4+2*x^2+1));
target=35/4; v=vector(333938); n=0;
for(a=1,lim_a(sqrt(target)), for(b=a,lim_b(sqrt(target)), for(c=b,a+b-1, f=rh2(a,b,c);v[n++]=f)));
v=vecsort(v); print("A331040 A331041 A331043"); print(numerator(v[1])," ",denominator(v[1])," ",1); m=0; mm=0; for(k=2,#v, if(v[k]>target,break); if(v[k]==v[k-1], m++; if(m>mm&&v[k+1]>v[k], print(numerator(v[k])," ",denominator(v[k])," ",m);mm=m),m=1));

A082046 Square array, A(n, k) = (kn)^2 + 3k*n + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 11, 11, 1, 1, 19, 29, 19, 1, 1, 29, 55, 55, 29, 1, 1, 41, 89, 109, 89, 41, 1, 1, 55, 131, 181, 181, 131, 55, 1, 1, 71, 181, 271, 305, 271, 181, 71, 1, 1, 89, 239, 379, 461, 461, 379, 239, 89, 1, 1, 109, 305, 505, 649, 701, 649, 505, 305, 109, 1

Offset: 0

Paul Barry, Apr 03 2003

			Array, A(n, k), begins as:
  1,  1,   1,   1,   1,    1,    1,    1, ... A000012;
  1,  5,  11,  19,  29,   41,   55,   71, ... A028387;
  1, 11,  29,  55,  89,  131,  181,  239, ... A082108;
  1, 19,  55, 109, 181,  271,  379,  505, ... A069131;
  1, 29,  89, 181, 305,  461,  649,  869, ... ;
  1, 41, 131, 271, 461,  701,  991, 1331, ... ;
  1, 55, 181, 379, 649,  991, 1405, 1891, ... ;
  1, 71, 239, 505, 869, 1331, 1891, 2549, ... ;
Antidiagonals, T(n, k), begin as:
  1;
  1,  1;
  1,  5,   1;
  1, 11,  11,   1;
  1, 19,  29,  19,   1;
  1, 29,  55,  55,  29,   1;
  1, 41,  89, 109,  89,  41,   1;
  1, 55, 131, 181, 181, 131,  55,  1;
  1, 71, 181, 271, 305, 271, 181, 71,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A028387, A057721, A069131, A072025, A082039, A082043, A082047, A082105, A082108.

[(k*(n-k))^2 + 3*(k*(n-k)) + 1: k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 22 2022

T[n_, k_]:= (k*(n-k))^2 + 3*(k*(n-k)) + 1;
Table[T[n,k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 22 2022 *)

def A082046(n,k): return (k*(n-k))^2 + 3*(k*(n-k)) + 1
flatten([[A082046(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 22 2022

A(n, k) = (k*n)^2 + 3*k*n + 1 (square array).
A(k, n) = A(n, k).
A(n, n) = T(2*n, n) = A057721(n).
A(n, n+1) = A072025(n).
T(n, k) = (k*(n-k))^2 + 3*k*(n-k) + 1 (antidiagonals).
Sum_{k=0..n} T(n, k) = A082047(n) (antidiagonal sums).
From G. C. Greubel, Dec 22 2022: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*(1 + (-1)^n)*(1 - 2*n).
T(2*n+1, n-1) = T(2*n-1, n-1) = A072025(n-1). (End)

A117715 Triangle, read by rows, T(n, k) = Fibonacci(n, k), where Fibonacci(n, x) is the Fibonacci polynomial.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 1, 2, 5, 10, 0, 3, 12, 33, 72, 1, 5, 29, 109, 305, 701, 0, 8, 70, 360, 1292, 3640, 8658, 1, 13, 169, 1189, 5473, 18901, 53353, 129949, 0, 21, 408, 3927, 23184, 98145, 328776, 927843, 2298912, 1, 34, 985, 12970, 98209, 509626, 2026009, 6624850, 18674305, 46866034

Offset: 0

Roger L. Bagula, Apr 13 2006

			Triangle begins as:
  0;
  1,  1;
  0,  1,   2;
  1,  2,   5,   10;
  0,  3,  12,   33,   72;
  1,  5,  29,  109,  305,   701;
  0,  8,  70,  360, 1292,  3640,  8658;
  1, 13, 169, 1189, 5473, 18901, 53353, 129949;

Steven Wolfram, The Mathematica Book, Cambridge University Press, 3rd ed. 1996, page 728

Alois P. Heinz, Rows n = 0..140, flattened
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
Wikipedia, Fibonacci Polynomial

Cf. A000045, A117716, A049310, A073133, A157103 (antidiagonals).
Main diagonal and first lower diagonal give: A084844, A084845.
Cf. A352361.

A117715:= func< n, k | k eq 0 select (n mod 2) else Evaluate(DicksonSecond(n-1, -1), k) >;
[A117715(n, k): k in [0..n], n in [0..13]]; // G. C. Greubel, Oct 01 2024

with(combinat):for n from 0 to 9 do seq(fibonacci(n,m), m = 0 .. n) od; # Zerinvary Lajos, Apr 09 2008

Table[Fibonacci[n, k], {n,0,12}, {k,0,n}]//Flatten

from sympy import fibonacci
def T(n, m): return 0 if n==0 else fibonacci(n, m)
for n in range(21): print([T(n, m) for m in range(n + 1)]) # Indranil Ghosh, Aug 12 2017

def A117715(n,k): return lucas_number1(n, k, -1)
flatten([[A117715(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 01 2024

T(n, 1) = A000045(n).
T(n, 3) = A006190(n).
T(n, 4) = A001076(n).
T(n, 5) = A052918(n-1).
T(5, k) = A057721(k).
T(6, k) = A124152(k).
T(n, k) = (-1)^(n-1)*A352361(n-k, n). - G. C. Greubel, Oct 01 2024

Definition simplified by the Assoc. Editors of the OEIS, Nov 17 2009

A120063 Shortest side c of all integer-sided triangles with sides a<=b<=c and inradius n.

Original entry on oeis.org

5, 10, 12, 15, 25, 24, 35, 30, 36, 39, 55, 45, 65, 63, 53, 60, 85, 68, 95, 75, 77, 88, 115, 85, 125, 130, 108, 105, 145, 106, 155, 120, 132, 170, 137, 135, 185, 190, 156, 150, 205, 154, 215, 165, 159, 230, 235, 170, 245, 195, 204, 195, 265, 204, 200, 195, 228, 290

Offset: 1

Hugo Pfoertner, Jun 13 2006

nonn

Terms a(11),..., a(100) computed by Thomas Mautsch (mautsch(AT)ethz.ch).
Empirically, 2*sqrt(3) < a(n)/n <= 5. The lower bound is provably tight, the upper bound seems to be achieved infinitely often, e.g, for prime n >= 5. It appears that a(p) = 5p for prime p != 3. - David W. Wilson, Jun 17 2006
Minimum of longest side occurring among all A120062(n) triangles having integer sides with integer inradius n.

			a(1)=5 because the only triangle with integer sides and inradius 1 is {3,4,5}; its longest side is 5.
a(2)=10: The triangles with inradius 2 are {5,12,13}, {6,8,10}, {6,25,29}, {7,15,20}, {9,10,17}. The minimum of their longest sides is min(13,10,29,20,17)=10.

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.

David W. Wilson, Table of n, a(n) for n = 1..10000

See A120062 for sequences related to integer-sided triangles with integer inradius n.

Cf. A120062 [triangles with integer inradius], A120252 [primitive triangles with integer inradius], A057721 [maximum of longest sides], A058331 [maximum of shortest sides], A120064 [minimum of middle sides], A082044 [maximum of middle sides], A005408 [minimum of shortest sides], A007237.

cp's OEIS Frontend

A119672 Number of prime factors (with multiplicity) of n^4 + 3*n^2 + 1 (A057721).

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A073133 Table by antidiagonals of T(n,k) = n*T(n,k-1) + T(n,k-2) starting with T(n,1) = 1.

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A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

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A352361 Array read by ascending antidiagonals. A(n, k) = Fibonacci(k, n), where Fibonacci(n, x) are the Fibonacci polynomials.

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A120062 Number of triangles with integer sides a <= b <= c having integer inradius n.

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A331040 Numerator of squared radius of inscribed circle of a triangle with integer sides i <= j <= k, such that the number of triangles with this radius sets a new record. Denominators are A331041.

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A082046 Square array, A(n, k) = (k*n)^2 + 3*k*n + 1, read by antidiagonals.

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A082046 Square array, A(n, k) = (kn)^2 + 3k*n + 1, read by antidiagonals.