A060789 a(n) = n / (gcd(n,2) * gcd(n,3)).

1, 1, 1, 2, 5, 1, 7, 4, 3, 5, 11, 2, 13, 7, 5, 8, 17, 3, 19, 10, 7, 11, 23, 4, 25, 13, 9, 14, 29, 5, 31, 16, 11, 17, 35, 6, 37, 19, 13, 20, 41, 7, 43, 22, 15, 23, 47, 8, 49, 25, 17, 26, 53, 9, 55, 28, 19, 29, 59, 10, 61, 31, 21, 32, 65, 11, 67, 34, 23, 35, 71, 12, 73, 37, 25, 38, 77

Offset: 1

Author

Len Smiley, Apr 26 2001

Keywords

• nonn
• easy
• mult

Comments

a(n+2) is absolute value of numerator of determinant of n X n matrix with M(i,j) = 2/(i(i+1)) if i=j otherwise 1. - Alexander Adamchuk, May 19 2006

Numerator of n/(n+6). - Gerry Martens, Aug 06 2015

In addition to being multiplicative, this sequence is also a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n, m)) for n, m >= 1. In particular, it follows that a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 20 2019

Connected to the Diophantine equation x^2 + (x + 1)^2 + … + (x + k)^2 = C, C an integer, x >= 0, k >= 0. Rewritten it is (k + 1)*x^2 + k*(k+1)*x + k*(k + 1)*(2*k + 1)/6 = C. The existence of its solutions depends on gcd(k + 1, k*(k + 1), k*(k + 1)*(2*k + 1)/6). - Ctibor O. Zizka, Oct 04 2023

Links

• Harry J. Smith, Table of n, a(n) for n=1..1000
• Wikipedia, Quasi-polynomial
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Crossrefs

Cf. A011782, A109047, A146535, A220466.

Cf. other sequences given by the formula n/gcd(n,k) = numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

Programs

GAP

List([1..80],n->n/(Gcd(n,2)*Gcd(n,3))); # Muniru A Asiru, Feb 20 2019

Magma

[n/(Gcd(n,2)*Gcd(n,3)) : n in [1..100]]; // Wesley Ivan Hurt, Aug 06 2015

Magma

I:=[1,1,1,2,5,1,7,4,3,5,11,2]; [n le 12 select I[n] else 2*Self(n-6)-Self(n-12): n in [1..80]]; // Vincenzo Librandi, Aug 07 2015

Maple

a := proc(n): if n = 1 then 1 else denom((4*n-6)/n) fi: end: seq(a(n), n=1..77); # Johannes W. Meijer, Dec 19 2012

Mathematica

Numerator[Table[(-1)^(n+1) Det[ DiagonalMatrix[ Table[ 2/(i(i+1)) - 1, {i, 1, n-2} ] ] + 1 ], {n, 30} ]] (* Alexander Adamchuk, May 19 2006 *)
Table[Numerator[(n+3)/(n+2)/(n+1)/n],{n,60}] (* Vladimir Joseph Stephan Orlovsky, Nov 17 2009 *)
Table[n/(GCD[n, 2] GCD[n, 3]), {n, 100}] (* Wesley Ivan Hurt, Aug 06 2015 *)
LinearRecurrence[{0,0,0,0,0,2,0,0,0,0,0,-1}, {1,1,1,2,5,1,7,4,3,5,11,2}, 80] (* Vincenzo Librandi, Aug 07 2015 *)

PARI

a(n) = { n / (gcd(n, 2) * gcd(n, 3)) } \\ Harry J. Smith, Jul 11 2009

Sage

[lcm(n,6)/6 for n in range(1, 78)] # Zerinvary Lajos, Jun 07 2009

Formula

G.f.: x*(1 + x + x^2 + 2*x^3 + 5*x^4 + x^5 + 5*x^6 + 2*x^7 + x^8 + x^9 + x^10)/(1 - x^6)^2.

Multiplicative with a(2^e)=2^(e-1), a(3^e)=3^(e-1), a(p^e)=p^e, p>3. - Vladeta Jovovic, Sep 09 2004

a(n) = Numerator[(-1)^(n+1)*Det[DiagonalMatrix[Table[2/(i(i+1))-1, {i,1,n-2}]]+1]], n>2. - Alexander Adamchuk, May 19 2006

a(n) divides n. a(6k) = k for integer k>0. a(p^k) = p^k for prime p>3 and integer k>0. - Alexander Adamchuk, Sep 20 2006

From R. J. Mathar, Apr 18 2011: (Start)

a(n) = A109047(n)/6.

Dirichlet g.f. zeta(s-1)*(1-1/2^s-2/3^s+2/6^s). (End)

a(n) = denominator((4*n-6)/n), n >= 2, with a(1) = 1. - Johannes W. Meijer, Dec 19 2012

a((2*n-1)*2^p) = A011782(p)*A146535(n), p >= 0. - Johannes W. Meijer, Feb 06 2013

a(n) = 2*a(n-6) - a(n-12) for n >= 12. - Robert Israel, Aug 06 2015

a(n) = gcd((n-1)*n*(n+1)/6, n). - Lechoslaw Ratajczak, Feb 16 2017

From Peter Bala, Feb 13 2019: (Start)

a(n) = n/gcd(n,n + 6) = n/gcd(n,6).

a(n) = n/b(n), where b(n) is the purely periodic sequence [1,2,3,2,1,6,...] with period 6.

a(n) is a quasi-polynomial in n: a(6*n+1) = 6*n + 1; a(6*n+2) = 3*n + 1; a(6*n+3) = 2*n + 1; a(6*n+4) = 3*n + 2; a(6*n+5) = 6*n + 5; a(6*n) = n.

a(n) = numerator(n/(n + 6)); a(n) = denominator((n + 6)/n).

(End)

Sum_{k=1..n} a(k) ~ 7*n^2/24. - Vaclav Kotesovec, Aug 09 2022

From Ctibor O. Zizka, Oct 04 2023: (Start)

For k >=0, a(k) = gcd(k + 1, k*(k + 1), k*(k + 1)*(2*k + 1)/6).

If (k mod 6) = 0 or 4 then a(k) = (k + 1).

If (k mod 6) = 1 or 3 then a(k) = (k + 1)/2.

If (k mod 6) = 2 then a(k) = (k + 1)/3.

If (k mod 6) = 5 then a(k) = (k + 1)/6. (End)