A022998 -id:A022998 - cp's OEIS frontend

A345082 Number of elements of order n in R/Z X Z/2Z.

1, 3, 2, 4, 4, 6, 6, 8, 6, 12, 10, 8, 12, 18, 8, 16, 16, 18, 18, 16, 12, 30, 22, 16, 20, 36, 18, 24, 28, 24, 30, 32, 20, 48, 24, 24, 36, 54, 24, 32, 40, 36, 42, 40, 24, 66, 46, 32, 42, 60, 32, 48, 52, 54, 40, 48, 36, 84, 58, 32, 60, 90, 36, 64, 48, 60, 66, 64

Offset: 1

Michel Marcus, Jul 30 2021

From Peter Bala, Dec 30 2023: (Start)
Denoted phi_2(n) in van der Kamp.
The number of solutions of the congruence x*y == 2 (mod n), 1 <= x, y <= n.
Can be regarded as a generalization of Euler's totient function phi(n) = Sum_{k = 1..n, gcd(k,n) = 1} gcd(k,n) since a(n) = Sum_{k = 1..n, gcd(k,n) divides 2} gcd(k,n). (End)

N. Anghel, Heron triangles with constant area and perimeter, Rev. Roumaine Math. Pures Appl. 65 (2020), 4, 403-422.
Peter H. van der Kamp, On the Fourier transform of the greatest common divisor, arXiv:1201.3139 [math.NT], 2012.

Cf. A000010, A088245.

Cf. A319998, A319997.

with(numtheory):
seq(add(d*phi(n/d), d in divisors(igcd(2, n))), n = 1..70); # Peter Bala, Dec 30 2023

Table[If[OddQ[n],EulerPhi[n],If[Mod[n,4]==0,2EulerPhi[n],2EulerPhi[n]+EulerPhi[n/2]]],{n,68}] (* Stefano Spezia, Jul 30 2021 *)

a(n) = if (n%2, eulerphi(n), if (n%4, 2*eulerphi(n) + eulerphi(n/2), 2*eulerphi(n)));

from sympy import totient as phi
def a(n): return phi(n) if n%2 else 2*phi(n)+phi(n//2) if n%4 else 2*phi(n)
print([a(n) for n in range(1, 69)]) # Michael S. Branicky, Jul 30 2021

a(n) = phi(n) if n is odd; 2*phi(n) if n == 0 (mod 4); 2*phi(n) + phi(n/2) if n == 2 (mod 4).
From Ridouane Oudra, Oct 17 2021: (Start)
a(n) = A000010(n) + A319998(n);
a(n) = 2*A000010(n) - A319997(n);
a(n) = Sum_{j = 1..n} gcd(n,j)*cos(4*Pi*j/n). (End)
From Peter Bala, Dec 30 2023: (Start)
a(n) = Sum_{d divides gcd(2,n)} d*phi(n/d), where phi(n) = A000010(n) denotes Euler's totient function.
Sum_{d divides n} a(d) = 2*n for n even, else equals n (van der Kamp, equation 26).
Dirichlet g.f.: zeta(s-1)*(1 + 2^(1-s))/zeta(s).
The Lambert series Sum_{n >= 1} a(n)*x^n/(1 - x^n) = x*(1 + 4*x + x^2)/(1 - x^2)^2. See A022998.
Multiplicative with a(2) = 3, a(2^k) = 2^k for k >= 2 and a(p^k) = p*k - p^(k-1) for odd primes p.
If n divides m then a(n) divides 3*a(m). (End)
Sum_{k=1..n} a(k) ~ c * n^2, where c = 9/(2*Pi^2) = 0.455945... (A088245). - Amiram Eldar, Jan 18 2024

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.

A022997 Numerator of n(n-2)(2n-1)/(2(n-1)).

Original entry on oeis.org

0, 15, 28, 135, 132, 455, 360, 1071, 760, 2079, 1380, 3575, 2268, 5655, 3472, 8415, 5040, 11951, 7020, 16359, 9460, 21735, 12408, 28175, 15912, 35775, 20020, 44631, 24780, 54839, 30240, 66495, 36448, 79695, 43452, 94535, 51300, 111111, 60040, 129519, 69720, 149855

Offset: 2

N. J. A. Sloane

			Fractions begins with 0, 15/4, 28/3, 135/8, 132/5, 455/12, 360/7, 1071/16, 760/9, 2079/20, 1380/11, 3575/24, ...

Amiram Eldar, Table of n, a(n) for n = 2..10000
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Cf. A022998 (denominators, with an offset shift).

a[n_] := Numerator[n*(n - 2)*(2*n - 1)/(2*(n - 1))]; Array[a, 50, 2] (* Amiram Eldar, Sep 21 2023 *)

a(n) = numerator(n*(n-2)*(2*n-1)/(2*(n-1))); \\ Amiram Eldar, Sep 21 2023

G.f.: x^3*(x^6+5x^4+20x^3+75x^2+28x+15)/((x-1)^4*(x+1)^4). - Ralf Stephan, Sep 03 2003

Sum_{n>=3} 1/a(n) = 11/9 + Pi/6 - 7*log(2)/3. - Amiram Eldar, Sep 21 2023

More terms from Amiram Eldar, Sep 21 2023

A118583 Numerator of sum of first p reciprocals of p-simplex numbers divided by p^4, where p = prime(n) for n > 2.

Original entry on oeis.org

1, 5, 53, 789, 237493, 2576561, 338350897, 616410400171, 2603853251291, 5745400286707685, 3081677433937346539, 41741941495866750557, 7829195555633964779233, 21066131970056662377432067

Offset: 3

Alexander Adamchuk, May 09 2007

nonn

			Prime(3) = 5.
a(3) = 1 because A118431(5)/5^4 = 1, where A118431(5) = Numerator[ 1/C(4+1,5) + 1/C(4+2,5) + 1/C(4+3,5) + 1/C(4+4,5) +1/C(4+5,5) ] = Numerator[ 1/1 + 1/6 + 1/21 + 1/56 + 1/126 ] = 625.

G. C. Greubel, Table of n, a(n) for n = 3..250
Eric Weisstein's World of Mathematics, Composition.
Eric Weisstein's World of Mathematics, Tetrahedral Number.
Eric Weisstein's World of Mathematics, Triangular Number.

Cf. A022998 = Numerator of sum of reciprocals of first n triangular numbers
Cf. A118391 = Numerator of sum of reciprocals of first n tetrahedral numbers A000292.
Cf. A118431 = Numerator of sum of reciprocals of first n 5-simplex numbers A000389.

Table[Numerator[Sum[1 /Binomial[ n+Prime[k]-1, Prime[k]], {n,1,Prime[k]} ]]/ Prime[k]^4, {k,3,25}]

for(n=3,10, print1(numerator(sum(k=1,prime(n), 1/(binomial(k+ prime(n)-1, prime(n)))))/prime(n)^4, ", ")) \\ G. C. Greubel, Nov 25 2017

a(n) = numerator( Sum_{k=1..prime(n)} ( 1/binomial( k + prime(n) - 1, prime(n) ) ))/prime(n)^4 for n > 2.

A244840 Denominators of the triangle T(n,k) = (n*(n+1)/2+k+1)/(k+1) for n >= k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 5, 2, 1, 1, 1, 3, 1, 5, 3, 1, 2, 1, 1, 1, 1, 5, 1, 7, 2, 1, 1, 2, 1, 4, 1, 2, 7, 8, 1, 2, 1, 2, 3, 4, 1, 6, 7, 8, 9, 2, 1, 1, 1, 1, 2, 5, 1, 7, 4, 3, 5, 1, 2

Offset: 0

Paul Curtz, Jul 07 2014

Numerators: A244734(n,k).

See A244734 for the first entries of the rational triangle T(n,k).

			T(0,0) = 1/1, T(1,0) = 2/1, T(1,1) = 3/2,... .
The triangle a(n,k) begins:
n/k  0 1 2 3 4 5 6 7 8  9 10 11 12 13 14 15 16 17 18 19 20 ...
0:   1
1:   1 2
2:   1 2 1
3:   1 1 1 2
4:   1 1 3 2 1
5:   1 2 1 4 1 2
6:   1 2 1 4 5 2 1
7:   1 1 3 1 5 3 1 2
8:   1 1 1 1 5 1 7 2 1
9:   1 2 1 4 1 2 7 8 1  2
10:  1 2 3 4 1 6 7 8 9  2  1
11:  1 1 1 2 5 1 7 4 3  5  1  2
12:  1 1 1 2 5 1 7 4 3  5 11  2 1
13:  1 2 3 4 5 6 1 8 9 10 11 12 1   2
14:  1 2 1 4 1 2 1 8 3  2 11  4 13  2  1
15:  1 1 1 1 1 1 7 1 3  1 11  1 13  7  1  2
16:  1 1 3 1 5 3 7 1 9  5 11  3 13  7 15  2  1
17:  1 2 1 4 5 2 7 8 1 10 11  4 13 14  5 16  1  2
18:  1 2 1 4 5 2 7 8 1 10 11  4 13 14  5 16 17  2  1
19:  1 1 3 2 1 3 7 4 9  1 11  6 13  7  3  8 17  9  1  2
20:  1 1 1 2 1 1 1 4 3  1 11  2 13  1  1  8 17  3 19  2  1
n/k  0 1 2 3 4 5 6 7 8  9 10 11 12 13 14 15 16 17 18 19 20 ...
.. reformatted - _Wolfdieter Lang_, Jul 28 2014 .
The second column is of period 4: repeat 2, 2, 1, 1. From A014695 or A130658.
The third column is of period 3: repeat 1, 1, 3. From A109007.
The fourth column is of period 8: repeat 2, 2, 4, 4, 1, 1, 4, 4.
The fifth column is of period 5: repeat 1, 1, 5, 5, 5.
The sixth column is of period 12: repeat 2, 2, 3, 1, 2, 6, 1, 1, 6, 2, 1, 3 .
The seventh column is of period 7: repeat 1, 1, 7, 7, 7, 7, 7.
Hence the positive terms of A022998.
Main diagonal: A000034(n).
Alternate main and second diagonal: A130658(n).
Common denominator by row: 1, 2, 2, 2, 6, 4, 20, 30, 70, ... .

Cf. A014695, A130658, A109007, A002260, A022998, A244734, A000034.

Table[(n*(n+1)/2+k+1)/(k+1) // Denominator, {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 08 2014 *)

a(n,k) = denominator((n*(n+1)/2 + k + 1)/(k+1)) for n >= k >= 0.

Editse: Name reformulated, comment with T(n,k) reference added. - Wolfdieter Lang, Jul 28 2014

A257942 a(n) = (n+1)*(n+2)/A014695(n+1), where A014695 is repeat (1, 2, 2, 1).

Original entry on oeis.org

1, 3, 12, 20, 15, 21, 56, 72, 45, 55, 132, 156, 91, 105, 240, 272, 153, 171, 380, 420, 231, 253, 552, 600, 325, 351, 756, 812, 435, 465, 992, 1056, 561, 595, 1260, 1332, 703, 741, 1560, 1640, 861, 903, 1892, 1980, 1035, 1081, 2256, 2352, 1225, 1275, 2652

Offset: 0

Paul Curtz, Jul 14 2015

Consider, for n >= 0, a sequence s(n). A useful transform is wi(n) = s(0), s(2), s(3), ..., i.e., s(n) without s(1).
For s(n) = 1/(n+1), wi(n)= 1, 1/3, 1/4, 1/5, ..., whose inverse binomial transform is f(n) = 1, -2/3, 7/12, -11/20, 8/15, -11/21, 29/56, -37/72, 23/45, -28/55, 67/132, -79/156, 46/91, -53/105, 121/240, -137/272, ...
The denominator of f(n) is a(n), for n >= 0.
If the numerator of f(n) is b(n), then it can be seen that b(n+1) = -(-1)^n* A226089(n).
Alternating a(n) - b(n) with a(n) + b(n) yields 0, 1, 5, 9, ... = A160050(n+1).
a(4n+1) is linked to the Rydberg-Ritz spectra of hydrogen.
h(n) = 0, 0, 1, 1, 4, 4, 3, 3, 8, 8, 5, 5, ... = duplicated A022998(n).
A022998(n) is linked to the Balmer series (see A246943(n)).
With an initial 0 and offset=0, a(-n) = a(n). Then (a(n+10) - a(n-10))/10 = 1, 6, 10, 7, 9, 22, 26, ... = g(n). a(n) mod 9 is of period 20.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A002378(n+1), A014695(n+1)=A130658(n+2), A014634, A033567(n+1), A104188(n+1), 4*A007742(n+1), A160050 (in A226089), A022998, A109613, A000217(n+1), A246943.

A257942:=n->(n+1)*(n+2)/(3/2+(-1)^((2*n+7+(-1)^n)/4)/2): seq(A257942(n), n=0..100); # Wesley Ivan Hurt, Jul 18 2015

CoefficientList[Series[-(x^6 + 9 x^4 - 8 x^3 + 9 x^2 + 1)/((x - 1)^3 (x^2 + 1)^3), {x, 0, 50}], x] (* Michael De Vlieger, Jul 14 2015 *)
(* Using inverse binomial transform *) s[0]=1; s[n_] := 1/(n+2); f[n_] := Sum[(-1)^(n-k)*Binomial[n, k]*s[k], {k, 0, n}]; Table[f[n]//Denominator, {n, 0, 50}] (* Jean-François Alcover, Jul 14 2015 *)
LinearRecurrence[{3, -6, 10, -12, 12, -10, 6, -3, 1}, {1, 3, 12, 20, 15, 21, 56, 72, 45}, 55] (* Vincenzo Librandi, Jul 15 2015 *)

Vec(-(x^6+9*x^4-8*x^3+9*x^2+1)/((x-1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jul 14 2015

a(n)=(n+1)*(n+2)/if(n%4<2,2,1) \\ Charles R Greathouse IV, Jul 14 2015

a(4n)    = (2*n+1)*(4*n+1).
a(4n+1)  = (2*n+1)*(4*n+3).
a(4n+2)  = (4*n+3)*(4*n+4).
a(4n+3)  = (4*n+4)*(4*n+5).
a(n) = A064038(n+2) * (period 4: repeat 1, 1, 4, 4).
From Colin Barker, Jul 14 2015: (Start)
a(n) = (-1/8+i/8)*(((-3-i*3)+i*(-i)^n+i^n)*(2+3*n+n^2)) where i=sqrt(-1).
G.f.: -(x^6+9*x^4-8*x^3+9*x^2+1) / ((x-1)^3*(x^2+1)^3). (End)
a(n) = h(n+2) * A109613(n+1).
a(n) = (n+1)*(n+2) * period 4:repeat (1, 1, 2, 2) /2.
From Wesley Ivan Hurt, Jul 18 2015: (Start)
a(n) = (n+1)*(n+2)/(3/2+(-1)^((2*n+7+(-1)^n)/4)/2).
a(n) = 3*a(n-1)-6*a(n-2)+10*a(n-3)-12*a(n-4)+12*a(n-5)-10*a(n-6)+6*a(n-7)-3*a(n-8)+a(n-9), n>9. (End)
Sum_{n>=0} 1/a(n) = Pi/4 + 1. - Amiram Eldar, Aug 14 2022

A281098 a(n) is the GCD of the sequence d(n) = A261327(k+n) - A261327(k) for all k.

Original entry on oeis.org

0, 1, 1, 3, 4, 1, 3, 1, 8, 3, 5, 1, 12, 1, 7, 3, 16, 1, 9, 1, 20, 3, 11, 1, 24, 1, 13, 3, 28, 1, 15, 1, 32, 3, 17, 1, 36, 1, 19, 3, 40, 1, 21, 1, 44, 3, 23, 1, 48, 1, 25, 3, 52, 1, 27, 1, 56, 3, 29, 1, 60, 1, 31, 3, 64, 1, 33, 1, 68, 3, 35, 1, 72, 1, 37, 3, 76, 1, 39, 1

Offset: 0

Paul Curtz, Jan 14 2017

Successive sequences:
   0,   0,  0,    0, ...    = 0 * ( )
   4,  -3,  11,  -8, ...    = 1 * ( )
   1,   8,   3,  16, ...    = 1 * ( )                 A195161
  12,   0,  27,  -3, ...    = 3 * (4, 0, 9, -1, ...)
   4,  24,   8,  40, ...    = 4 * (1, 6, 2, 10, ...)  A064680
5;   28,   5,  51,   4, ...    = 1 * ( )
   9,  48,  15,  72, ...    = 3 * (3, 16, 5, 24, ...) A195161
  52,  12,  83,  13, ...    = 1 * ( )
  16,  80,  24, 112, ...    = 8 * (2, 10, 3, 14, ...) A064080
  84   21, 123,  24, ...    = 3 * (28, 7, 41, 8, ...)
 25, 120,  35, 160, ...    = 5 * (5, 24, 7, 32, ...) A195161

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for linear recurrences with constant coefficients, signature (0,-1,0,1,0,2,0,1,0,-1,0,-1).

Cf. A064680, A144433 or A195161.

Cf. A000012, A005408, A008586, A010701, A109007 (bisection), A016825, A165988 (via A022998), A166138, A166304, A280579.

CoefficientList[Series[(-x (-1 - x - 4 x^2 - 5 x^3 - 3 x^4 - 6 x^5 + 3 x^6 - 5 x^7 + 4 x^8 - x^9 + x^10))/((x^2 - x + 1) (1 + x + x^2) (x - 1)^2*(1 + x)^2*(1 + x^2)^2), {x, 0, 79}], x] (* Michael De Vlieger, Feb 02 2017 *)

f(n) = numerator((4 + n^2)/4);
a(n) = gcd(vector(1000, k, f(k+n) - f(k))); \\ Michel Marcus, Jan 15 2017

A281098(n) = if(n%2, gcd((n\2)-1,3), n>>(bitand(n,2)/2)); \\ Antti Karttunen, Feb 15 2023

G.f.: -x*( -1 - x - 4*x^2 - 5*x^3 - 3*x^4 - 6*x^5 + 3*x^6 - 5*x^7 + 4*x^8 - x^9 + x^10 )/( (x^2 - x + 1)*(1 + x + x^2)*(x - 1)^2*(1 + x)^2*(1 + x^2)^2 ). - R. J. Mathar, Jan 31 2017
a(2*k)   = A022998(k).
a(2*k+1) = A109007(k-1).
a(3*k)   = interleave 3*k*(3 +(-1)^k)/2, 3.
a(3*k+1) = interleave 1, A166304(k).
a(3*k+2) = interleave A166138(k), 1.
a(4*k)   = 4*k.
a(4*k+1) = period 3: repeat [1, 1, 3].
a(4*k+2) = 1 + 2*k.
a(4*k+3) = period 3: repeat [3, 1, 1].
a(n+12) - a(n) = 6*A131743(n+3).
a(n) = (18*n + 40 - 16*cos(n*Pi/3) + 9*n*cos(n*Pi/2) + 32*cos(2*n*Pi/3) + (18*n - 40)*cos(n*Pi) + 3*n*cos(3*n*Pi/2) - 16*cos(5*n*Pi/3))/48. - Wesley Ivan Hurt, Oct 04 2018

Corrected and extended by Michel Marcus, Jan 15 2017

A290168 If n is even then a(n) = n^2(n+2)/8, otherwise a(n) = (n-1)n*(n+1)/8.

Original entry on oeis.org

0, 0, 2, 3, 12, 15, 36, 42, 80, 90, 150, 165, 252, 273, 392, 420, 576, 612, 810, 855, 1100, 1155, 1452, 1518, 1872, 1950, 2366, 2457, 2940, 3045, 3600, 3720, 4352, 4488, 5202, 5355, 6156, 6327, 7220, 7410, 8400

Offset: 0

Jean-François Alcover and Paul Curtz, Jul 23 2017

nonn

Bisection of a(n) [0, 2, 12, 36, 80, 150, 252, ...] is A011379.
Bisection [0, 3, 15, 42, 90, 165, 273, ...] is A059270.
Considering s(n) = [0, 0, 0, 0, 1, 1, 3, 3, 6, 6, 10, 10, 15, 15, ...] (triangular numbers repeated - see A008805), a(n) = n*s(n+2) holds.
Considering the first differences of a(n), b(n) = [0, 2, 1 , 9, 3, 21, 6, 38, 10, 60, 15, 87, ...], b(n) shows bisections A000217 and A005476. In addition, b(n) begins like A249264 up to 12th term, and is an alternation of 4 multiples of 3 and 2 not multiples; b(n) is also such that b(2n) + b(2n+1) = A049450(n).
Considering the second differences c(n), c(n) shows bisections A001105(n+1) and -A000384(n+1), c(n) has 3 consecutive terms multiples of 3 alternating with 3 not multiples; in addition, c(2n) + c(2n+1) = A000027(n).
Considering a(n)/c(n) = [0, 0, 1/4, -1/2, 2/3, -1, 9/8, -3/2, 8/5, -2, 25/12, -5/2, ...], it appears that it is A129194(n)/A022998(n+1) and -A026741(n)/A000034(n) alternating.

Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A000027, A000034, A000217, A000384, A001105, A005476, A008805, A011379, A022998, A026741, A049450, A059270, A129194, A135713, A161680, A249264.

a[n_] := If[EvenQ[n], n^2*(n + 2)/8, (n - 1)*n*(n + 1)/8]; Table[a[n], {n, 0, 40}]

a(n) = if(n%2==0, n^2*(n+2)/8, (n-1)*n*(n+1)/8) \\ Felix Fröhlich, Jul 23 2017

G.f.: x^2*(2 + x + 3*x^2)/((x-1)^4*(x+1)^3).
a(n) = (1/16)*(-1)^n*n*(1 + (-1)^(n+1) + 2*(1 + (-1)^n)*n + 2*(-1)^n*n^2).
Sum_{n>=2} 1/a(n) = 5 + Pi^2/6 - 8*log(2). - Amiram Eldar, Sep 17 2022

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

A369304 Numbers k for which the polynomial (x-1)^3*(x+1)^k has more than one zero coefficient.

Original entry on oeis.org

3, 6, 14, 19, 31, 38, 54, 63, 83, 94, 118, 131, 159, 174, 206, 223, 259, 278, 318, 339, 383, 406, 454, 479, 531, 558, 614, 643, 703, 734, 798, 831, 899, 934, 1006, 1043, 1119, 1158, 1238, 1279, 1363, 1406, 1494, 1539, 1631, 1678, 1774, 1823, 1923, 1974, 2078, 2131, 2239, 2294, 2406, 2463

Offset: 1

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Vladimir Petrov Kostov, Jan 19 2024

Keywords

nonn
easy

Comments

In this sequence, pairs of consecutive even numbers (excluding the leading term) alternate with pairs of consecutive odd numbers. When in the sequence a(n) is even (resp. when a(n) is odd), the polynomial (x-1)^3*(x+1)^a(n) has two (resp. three) vanishing coefficients.

These are the coefficients of x^j with j = (m(n) +- 2)*(m(n) +- 1)/6, where m(n) = (6*n - 3 - (-1)^n)/4, and for odd a(n), also with j = (a(n) + 3)/2.

The first differences are a(n) - a(n-1) = n+1 if n even, or 2*(n+1) if n odd, for n >= 2 (A022998).

a(n) = A001082(n+2)-2. Indeed, this formula is valid for n=1,...,20 and the even and odd terms of both sequences A001082 and A369304 are the values of quadratic polynomials in n.

The sequence terms are the exponents in the expansion of Sum_{n >= 1} (-1)^(n+1) * x^(3*n) * Product_{k = 2..n} (1 + x^(2*k-1)) = x^3 - x^6 + x^14 - x^19 + x^31 - x^38 + x^54 - x^63 + x^83 - x^94 + ... (set x = -q and replace q with q^2 in Andrews, equation 8). - Peter Bala, Nov 19 2024

Examples

For n=1, a(1)=3 and the polynomial (x-1)^3*(x+1)^3 = x^6 - 3*x^4 + 3*x^2 - 1 has three vanishing coefficients, those of x^5, x^3 and x. For n=2, a(2)=6 and the polynomial (x-1)^3*(x+1)^6 = x^9 + 3*x^8 - 8*x^6 - 6*x^5 + 6*x^4 + 8*x^3 - 3*x - 1 has two vanishing coefficients, those of x^7 and x^2.

Links

Michael De Vlieger, Table of n, a(n) for n = 1..10000

George E. Andrews, Euler's Pentagonal Number Theorem, Mathematics Magazine, Vol. 56, No. 5 (Nov., 1983), pp. 279-284.

Vladimir Petrov Kostov, On universal sign patterns, arXiv:2405.18895 [math.CA], 2024. See p. 5.

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

Crossrefs

Cf. A001651, A022998.

Programs

Mathematica

LinearRecurrence[{1, 2, -2, -1, 1}, {3, 6, 14, 19, 31}, 56] (* Hugo Pfoertner, Feb 12 2024 *)

PARI

isok(k) = #select(x->(x==0), Vec((x-1)^3*(x+1)^k)) > 1; \\ Michel Marcus, Jan 19 2024

Python

def A369304(n): return ((n+1<<1)-(n>>1))**2//3-2 # Chai Wah Wu, Mar 05 2024

Formula

a(n) = ((m(n) + 3)^2 - 7)/3 where m(n) = A001651(n) is the n-th natural number not divisible by 3.

G.f.: (x*(1+x+x^2)*(3-x^2))/((1-x)^3*(1+x)^2). - Joerg Arndt, Jan 19 2024

E.g.f.: (4 + (3*x^2 + 13*x - 4)*cosh(x) + (3*x^2 + 11*x - 1)*sinh(x))/4. - Stefano Spezia, Feb 13 2024

Sum_{n>=1} 1/a(n) = 3/2 + (tan((1+2*sqrt(7))*Pi/6) - cot((1+sqrt(7))*Pi/3)) * Pi/(2*sqrt(7)). - Amiram Eldar, Mar 07 2024

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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.
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cp's OEIS Frontend

A345082 Number of elements of order n in R/Z X Z/2Z.

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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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A022997 Numerator of n*(n-2)*(2*n-1)/(2*(n-1)).

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A118583 Numerator of sum of first p reciprocals of p-simplex numbers divided by p^4, where p = prime(n) for n > 2.

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A244840 Denominators of the triangle T(n,k) = (n*(n+1)/2+k+1)/(k+1) for n >= k >= 0.

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A257942 a(n) = (n+1)*(n+2)/A014695(n+1), where A014695 is repeat (1, 2, 2, 1).

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A281098 a(n) is the GCD of the sequence d(n) = A261327(k+n) - A261327(k) for all k.

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A022997 Numerator of n(n-2)(2n-1)/(2(n-1)).

A290168 If n is even then a(n) = n^2(n+2)/8, otherwise a(n) = (n-1)n*(n+1)/8.