A228564 -id:A228564 - cp's OEIS frontend

A306367 a(n) = numerator of (n^2 + 2)/(n + 2).

1, 1, 3, 11, 3, 27, 19, 17, 33, 83, 17, 123, 73, 57, 99, 227, 43, 291, 163, 121, 201, 443, 81, 531, 289, 209, 339, 731, 131, 843, 451, 321, 513, 1091, 193, 1227, 649, 457, 723, 1523, 267, 1683, 883, 617, 969, 2027, 353, 2211, 1153, 801, 1251, 2603, 451

Offset: 0

Peter Bala, Feb 14 2019

If P(x) and Q(x) are coprime integral polynomials such that Q(n) > 0 for n >= 0 then the sequence of numerators of the rational numbers P(n)/Q(n) for n >= 0 and the sequence of denominators of P(n)/Q(n) for n >= 0 are both quasi-polynomial in n. In fact, there exists a purely periodic sequence b(n) such that numerator(P(n)/Q(n)) = P(n)/b(n) and denominator(P(n)/Q(n)) = Q(n)/b(n).

Here we take P(n) = n^2 + 2 and Q(n) = n + 2. Cf. A228564 (case P(n) = n^2 + 1, Q(n) = n + 1).

Muniru A Asiru, Table of n, a(n) for n = 0..10000
P. Bala, A note on the sequence of numerators of a rational function
Wikipedia, Quasi-polynomial

Cf. A060789, A228564, A306368.

List([0..100],n->NumeratorRat((n^2+2)/(n+2)); # Muniru A Asiru, Feb 25 2019

seq(numer( (n^2 + 2)/(n + 2) ), n = 0..100);

a(n) = numerator((n^2 + 2)/(n + 2)); \\ Michel Marcus, Feb 26 2019

O.g.f.: (3*x^17 + x^16 + 11*x^15 + 9*x^14 + 9*x^13 + 19*x^12 + 42*x^11 + 8*x^10 + 50*x^9 + 24*x^8 + 14*x^7 + 16*x^6 + 27*x^5 + 3*x^4 + 11*x^3 + 3*x^2 + x + 1)/(1 - x^6)^3.
a(n) = (n^2 + 2)/b(n), where (b(n))n>=0 is the purely periodic sequence [2, 3, 2, 1, 6, 1, 2, 3, 2, 1, 6, 1, ...] with period 6.
a(n) is a quasi-polynomial in n: a(6*n) = 18*n^2 + 1; a(6*n+1) = 12*n^2 + 4*n + 1; a(6*n+2) = 18*n^2 + 12*n + 3; a(6*n+3) = 36*n^2 + 36*n + 11; a(6*n+4) = 6*n^2 + 8*n + 3; a(6*n+5) = 36*n^2 + 60*n + 27.
a(n) = (n^2 + 2)/( gcd(n+2,2)*gcd(n+2,3) ).
a(n) = (n^2 + 2)/(n + 2) * A060789(n+2).

A382252 Triangle T(n,k) = numerator of (n+k)/(1+n*k), 0 <= k <= n >= 0, read by rows.

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 3, 1, 5, 3, 4, 1, 2, 7, 8, 5, 1, 7, 1, 3, 5, 6, 1, 8, 9, 2, 11, 12, 7, 1, 3, 5, 11, 1, 13, 7, 8, 1, 10, 11, 4, 13, 2, 5, 16, 9, 1, 11, 3, 13, 7, 3, 1, 17, 9, 10, 1, 4, 13, 14, 5, 16, 17, 2, 19, 20, 11, 1, 13, 7, 1, 2, 17, 3, 19, 1, 7, 11, 12, 1, 14, 15, 16, 17, 18, 19, 20, 21, 2, 23, 24

Offset: 0

M. F. Hasler, Apr 15 2025

Since the operation n @ k := (n + k)/(1 + n*k) is commutative, it is sufficient to list only the lower half of the "multiplication table", which would otherwise be an infinite square array. This triangle lists the numerators, and A382253 lists the denominators.

			The table for the operation n @ k := (n + k)/(1 + n*k) starts as follows:
(0 is the neutral element for the operation: n @ 0 = n = 0 @ n, therefore row and column 0 give the column and row headers.)
  0    1    2     3     4     5     6     7     8    Numerators of  0;
  1    1    1     1     1     1     1     1     1     lower left    1, 1:
  2    1   4/5   5/7   2/3   7/11  8/13  3/5  10/17    triangle:    2, 1, 4;
  3    1   5/7   3/5   7/13  1/2   9/19  5/11 11/25                 3, 1, 5, 3
  4    1   2/3   7/13  8/17  3/7   2/5  11/29  4/11                 4, 1, 2, 7, 8;
  5    1   7/11  1/2   3/7   5/13 11/31  1/3  13/41                 etc.
  6    1   8/13  9/19  2/5  11/31 12/37 13/43  2/7
  7    1   3/5   5/11 11/29  1/3  13/43  7/25  5/19
  8    1  10/17 11/25  4/11 13/41  2/7   5/19 16/65
This sequence lists the numerators of the values, where numerator(x) = x for integers, and only for the lower left triangle of the table, by rows.

Cf. A382253 (denominators), A382257 (related); A228564 (main diagonal), A001477 (row & col. 0), A000012 (row & col. 1).

apply( {A382252(n,k=-1)= k<0&& k=n-(1+n=(sqrtint(8*n+1)-1)\2)*n/2; numerator((n+k)/(1+n*k))}, [0..30])

T(n,k) = T(k,n) for all n, k >= 0; therefore only k <= n is considered here.
T(n,0) = T(0,n) = n and T(n,1) = T(1,n) = 1 for all n >= 0.
T(n,n) = A022998(n) = n if odd, else 2*n.

A382253 Triangle T(n,k) = denominator of (n+k)/(1+n*k), 0 <= k <= n >= 0, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 5, 1, 1, 7, 5, 1, 1, 3, 13, 17, 1, 1, 11, 2, 7, 13, 1, 1, 13, 19, 5, 31, 37, 1, 1, 5, 11, 29, 3, 43, 25, 1, 1, 17, 25, 11, 41, 7, 19, 65, 1, 1, 19, 7, 37, 23, 11, 4, 73, 41, 1, 1, 7, 31, 41, 17, 61, 71, 9, 91, 101, 1

Offset: 0

M. F. Hasler, Apr 15 2025

Since the operation n @ k := (n + k)/(1 + n*k) is commutative, it is sufficient to list only the lower half of the "multiplication table", which would otherwise be an infinite square array. This triangle lists the denominators, and A382252 lists the numerators.

			The table for the operation n @ k := (n + k)/(1 + n*k) starts as follows:
(0 is the neutral element for the operation: n @ 0 = n = 0 @ n, therefore the elements in row 0 and column 0 equal the column and row index.)
  0    1    2     3     4     5     6     7     8    Denominators of lower left
  1    1    1     1     1     1     1     1     1    triangle:  1;
  2    1   4/5   5/7   2/3   7/11  8/13  3/5  10/17             1, 1
  3    1   5/7   3/5   7/13  1/2   9/19  5/11 11/25             1, 1, 5;
  4    1   2/3   7/13  8/17  3/7   2/5  11/29  4/11             1, 1, 7, 5;
  5    1   7/11  1/2   3/7   5/13 11/31  1/3  13/41             1, 1, 3, 13, 17;
  6    1   8/13  9/19  2/5  11/31 12/37 13/43  2/7                 etc.
  7    1   3/5   5/11 11/29  1/3  13/43  7/25  5/19
  8    1  10/17 11/25  4/11 13/41  2/7   5/19 16/65
The sequence lists the denominators of the values, where denominator(x) = 1 for integers, and only for the lower left triangle of the table, by rows.

Cf. A382252, A382257; A228564 (main diagonal), A001477 (row & col. 0 of the '@' table), A000012 (row & col. 0 & 1 of the table of denominators).

apply( {A382253(n,k=-1)= k<0&& k=n-(1+n=(sqrtint(8*n+1)-1)\2)*n/2; denominator((n+k)/(1+n*k))}, [0..66])

T(n,k) = T(k,n) for all n, k >= 0;
T(n,0) = T(0,n) = T(n,1) = T(1,n) = 1 for all n >= 0;
T(n,n) = denominator(2*n/(1+n^2)) = numerator((1+n^2)/2) = A228564(n).

A386824 Triangle read by rows: T(n,k) = denominator((n^2 - k^2)/(n^2 + k^2)), where 0 <= k < n.

Original entry on oeis.org

1, 1, 5, 1, 5, 13, 1, 17, 5, 25, 1, 13, 29, 17, 41, 1, 37, 5, 5, 13, 61, 1, 25, 53, 29, 65, 37, 85, 1, 65, 17, 73, 5, 89, 25, 113, 1, 41, 85, 5, 97, 53, 13, 65, 145, 1, 101, 13, 109, 29, 5, 17, 149, 41, 181, 1, 61, 125, 65, 137, 73, 157, 85, 185, 101, 221

Offset: 1

Stefano Spezia, Aug 04 2025

			The triangle of the fractions begins as:
  1/1;
  1/1,   3/5;
  1/1,   4/5,  5/13;
  1/1, 15/17,   3/5,  7/25;
  1/1, 12/13, 21/29,  8/17,  9/41;
  1/1, 35/37,   4/5,   3/5,  5/13, 11/61;
  1/1, 24/25, 45/53, 20/29, 33/65, 12/37, 13/85;
  ...

Stefano Spezia, First 150 rows of the triangle, flattened

Cf. A000012 (k=0), A000290, A001844, A069011, A094728, A228564 (k=1), A243883 (k=2), A386823 (numerators).

T[n_,k_]:=Denominator[(n^2-k^2)/(n^2+k^2)]; Table[T[n,k],{n,11},{k,0,n-1}]//Flatten

T(n,n-1) = A001844(n-1).

A362218 Three-column array read by rows: row n gives the unique ordered primitive Pythagorean triple (a,b,c) with a

Original entry on oeis.org

3, 4, 5, 8, 15, 17, 5, 12, 13, 12, 35, 37, 7, 24, 25, 16, 63, 65, 9, 40, 41, 20, 99, 101, 11, 60, 61, 24, 143, 145, 13, 84, 85, 28, 195, 197, 15, 112, 113, 32, 255, 257, 17, 144, 145, 36, 323, 325, 19, 180, 181, 40, 399, 401

Offset: 3

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Author

Miguel-Ángel Pérez García-Ortega, Apr 11 2023

Keywords

nonn
tabf

Comments

Given an ordered primitive Pythagorean triple (a,b,c) with a
For n>=3 there exists a unique ordered primitive Pythagorean triple such that (b+c)/a = n.

For n odd, the triple is {n, (n^2-1)/2, (n^2+1)/2}.

For n even, the triple is { 2*n, n^2-1, n^2+1 }.

Examples

Irregular array begins: n=3: 3, 4, 5; n=4: 8, 15, 17; n=5: 5, 12, 13; n=6: 12, 35, 37; n=7: 7, 24, 25; ... Row n=3 is (3,4,5) and has (b+c)/a = (4+5)/3 = 3. Row n=4 is (8,15,17) and has (b+c)/a = (15+17)/8 = 4.

References

J. M. Blanco Casado, J. M. Sánchez Muñoz, and M. A. Pérez García-Ortega, El Libro de las Ternas Pitagóricas, Preprint 2023.

Links

Miguel-Ángel Pérez García-Ortega, Radial proportions (in Spanish).

Crossrefs

Cf. A022998 (short leg), A066830 (long leg), A228564 (hypotenuse).

Programs

Mathematica

k=50; ternas={{n," ",a,b,c," ",r," "," γ2 "," ",s," ",rb}};Do[If[Mod[t,2]==0,ternas=Join[ternas,{{t," ",2t,t^2-1,t^2+1," ",t-1," ",t," ",t(t+1)," ",t(t-1)}}],ternas=Join[ternas,{{t," ",t,(t^2-1)/2,(t^2+1)/2," ",(t-1)/2," ",t," ",(t(t+1))/2," ",(t(t-1))/2}}]],{t,3,k+2}] MatrixForm[Transpose[ternas]]

Formula

T(n,1) = A022998(n).

T(n,2) = A066830(n).

T(n,3) = A228564(n).

a(6*k-3) = 2*k+1;

a(6*k-2) = ((2*k+1)^2 - 1)/2;

a(6*k-1) = ((2*k+1)^2 + 1)/2;

a(6*k) = 4*(k+1);

a(6*k+1) = 4*(k+1)^2 - 1;

a(6*k+2) = 4*(k+1)^2 + 1.

Extensions

Edited by N. J. A. Sloane, Apr 30 2023

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cp's OEIS Frontend

A306367 a(n) = numerator of (n^2 + 2)/(n + 2).

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A382252 Triangle T(n,k) = numerator of (n+k)/(1+n*k), 0 <= k <= n >= 0, read by rows.

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A382253 Triangle T(n,k) = denominator of (n+k)/(1+n*k), 0 <= k <= n >= 0, read by rows.

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A386824 Triangle read by rows: T(n,k) = denominator((n^2 - k^2)/(n^2 + k^2)), where 0 <= k < n.

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A362218 Three-column array read by rows: row n gives the unique ordered primitive Pythagorean triple (a,b,c) with a

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