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a(n) = numerator of n/(n+20).
19

%I #57 Jan 01 2023 13:47:46

%S 0,1,1,3,1,1,3,7,2,9,1,11,3,13,7,3,4,17,9,19,1,21,11,23,6,5,13,27,7,

%T 29,3,31,8,33,17,7,9,37,19,39,2,41,21,43,11,9,23,47,12,49,5,51,13,53,

%U 27,11,14,57,29,59,3,61,31,63,16,13,33,67,17,69,7,71,18,73,37,15,19,77,39,79

%N a(n) = numerator of n/(n+20).

%C Contains as subsequences A026741, A017281, A017305, A005408, A017353, and A017377. - _Luce ETIENNE_, Nov 04 2018

%C Multiplicative and also a strong divisibility sequence: gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. - _Peter Bala_, Feb 24 2019

%H Vincenzo Librandi, <a href="/A106621/b106621.txt">Table of n, a(n) for n = 0..2000</a>

%H Peter Bala, <a href="/A306367/a306367.pdf">A note on the sequence of numerators of a rational function</a>, 2019.

%H <a href="/index/Rec#order_40">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1).

%F a(n) = lcm(20, n)/20. - _Zerinvary Lajos_, Jun 12 2009

%F a(n) = n/gcd(n, 20). - _Andrew Howroyd_, Jul 25 2018

%F From _Luce ETIENNE_, Nov 04 2018: (Start)

%F a(n) = 9*a(n-20) - 36*a(n-40) + 84*a(n-60) - 126*a(n-80) + 126*a(n-100) - 84*a(n-120) + 36*a(n-140) - 9*a(n-160) + a(n-180).

%F a(n) = (5*(119*m^9 - 4923*m^8 + 86250*m^7 - 832230*m^6 + 4807887*m^5 - 16882299*m^4 + 34770400*m^3 - 37855620m^2 + 16581744*m + 54432)*floor(n/10) + 72*m*(3*m^8 - 120*m^7 + 2030*m^6 - 18900*m^5 + 105329*m^4 - 356580*m^3 + 706220*m^2 - 733200*m + 300258) + ((19*m^9 - 855*m^8 + 15810*m^7 - 154350*m^6 + 849387*m^5 - 2597175*m^4 + 4037840*m^3 - 2600100*m^2 + 540144*m - 90720)*floor(n/10) - 72*m*(m^7 - 35*m^6 + 490*m^5 - 3500*m^4 + 13489*m^3 - 27335*m^2 + 26340*m - 9450))*(-1)^floor(n/10))/362880 where m = (n mod 10). (End)

%F From _Peter Bala_, Feb 24 2019: (Start)

%F a(n) = n/gcd(n,20) is a quasi-polynomial in n since gcd(n,20) is a purely periodic sequence of period 20.

%F O.g.f.: F(x) - F(x^2) - F(x^4) - 4*F(x^5) + 4*F(x^10) + 4*F(x^20), where F(x) = x/(1 - x)^2.

%F O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = Sum_{d divides 20} (phi(d)/d) * log(1/(1 - x^d)) = log(1/(1 - x)) + (1/2)*log(1/(1 - x^2)) + (2/4)*log(1/(1 - x^4)) + (4/5)*log(1/(1 - x^5)) + (4/10)*log(1/(1 - x^10)) + (8/20)*log(1/(1 - x^20)), where phi(n) denotes the Euler totient function A000010. (End)

%F From _Amiram Eldar_, Nov 25 2022: (Start)

%F Multiplicative with a(2^e) = 2^max(0, e-2), a(5^e) = 5^max(0,e-1), and a(p^e) = p^e otherwise.

%F Dirichlet g.f.: zeta(s-1)*(1 - 1/2^s - 1/4^s - 4/5^s + 4/10^s + 4/20^s).

%F Sum_{k=1..n} a(k) ~ (231/800) * n^2. (End)

%p seq(numer(n/(n+20)),n=0..80); # _Muniru A Asiru_, Feb 19 2019

%t f[n_]:=Numerator[n/(n+20)];Array[f,100,0] (* _Vladimir Joseph Stephan Orlovsky_, Feb 17 2011 *)

%o (Sage) [lcm(20,n)/20 for n in range(0, 80)] # _Zerinvary Lajos_, Jun 12 2009

%o (Magma) [Numerator(n/(n+20)): n in [0..100]]; // _Vincenzo Librandi_, Mar 06 2018

%o (PARI) a(n) = numerator(n/(n+20)); \\ _Michel Marcus_, Mar 07 2018

%o (GAP) List([0..80],n->NumeratorRat(n/(n+20))); # _Muniru A Asiru_, Feb 19 2019

%Y Cf. A000010, A010879.

%Y Cf. Sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106620 (k = 13 thru 19).

%K nonn,easy,frac,mult

%O 0,4

%A _N. J. A. Sloane_, May 15 2005

%E Keyword:mult added by _Andrew Howroyd_, Jul 25 2018