login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

a(1)=1, a(2)=3; thereafter, a(n) = least positive integer > a(n-1) and not equal to a(i)+a(j)+a(k) for 1<=i<=j<=k<=n-1.
1

%I #31 Apr 15 2023 04:19:11

%S 1,3,4,13,14,23,24,33,34,43,44,53,54,63,64,73,74,83,84,93,94,103,104,

%T 113,114,123,124,133,134,143,144,153,154,163,164,173,174,183,184,193,

%U 194,203,204,213,214,223,224,233,234,243,244,253,254,263,264,273,274,283,284

%N a(1)=1, a(2)=3; thereafter, a(n) = least positive integer > a(n-1) and not equal to a(i)+a(j)+a(k) for 1<=i<=j<=k<=n-1.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F a(n) = max{1, 5*n-9+2*(-1)^n}.

%F From _Colin Barker_, Jul 22 2012: (Start)

%F a(n) = a(n-1) + a(n-2) - a(n-3) for n>4.

%F G.f.: x*(1+2*x+7*x^3)/((1-x)^2*(1+x)). (End)

%F Conjecture: Except for the first term, a(n)=10*n-a(n-1)-23 (with a(2)=3). - _Vincenzo Librandi_, Dec 07 2010 [This is easily proved. - _N. J. A. Sloane_, Aug 07 2017]

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - sqrt(1-2/sqrt(5))*Pi/(10*phi) + log(phi)/(2*sqrt(5)) - log(2)/5, where phi is the golden ratio (A001622). - _Amiram Eldar_, Apr 15 2023

%t Join[{1},LinearRecurrence[{1,1,-1},{3,4,13},60]] (* _Harvey P. Dale_, Aug 19 2014 *)

%Y Cf. A001622.

%K nonn,easy

%O 1,2

%A Bela Bajnok (bbajnok(AT)gettysburg.edu), Aug 10 2005

%E Definition corrected by Bela Bajnok (bbajnok(AT)gettysburg.edu) and _N. J. A. Sloane_, Aug 07 2017