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 A175676 a(n) = binomial(n,3) mod n. 6

%I

%S 0,0,1,0,0,2,0,0,3,0,0,4,0,0,5,0,0,6,0,0,7,0,0,8,0,0,9,0,0,10,0,0,11,

%T 0,0,12,0,0,13,0,0,14,0,0,15,0,0,16,0,0,17,0,0,18,0,0,19,0,0,20,0,0,

%U 21,0,0,22,0,0,23,0,0,24,0,0,25,0,0,26,0,0,27,0,0,28,0,0,29,0,0,30,0,0,31,0

%N a(n) = binomial(n,3) mod n.

%C Number of partitions of n+3 into 3 parts that are in arithmetic progression. - _Wesley Ivan Hurt_, Dec 07 2020

%H Altug Alkan, <a href="/A175676/b175676.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).

%F a(n) = n/3 if n==0 (mod 3) else a(n) = 0.

%F G.f.: x^3 / ( (x-1)^2*(1+x+x^2)^2 ). - _R. J. Mathar_, Mar 11 2011

%F a(n) = A008620(n-1)*A079978(n). - _Bruno Berselli_, Jun 22 2012

%F a(n) = (n + 2*n*cos((2*n*Pi)/3))/9. - _Kritsada Moomuang_, Apr 02 2018

%t Table[Mod[Binomial[n,3],n],{n,150}]

%o (PARI) a(n)=if(n%3, 0, n/3); \\ _Charles R Greathouse IV_, Sep 02 2015 [Corrected by _Altug Alkan_, Apr 02 2018]

%o (PARI) a(n)=!(n%3)*(1-n)\-3; \\ _Altug Alkan_, Apr 02 2018

%o (GAP) List([1..100],n->Binomial(n,3) mod n); # _Muniru A Asiru_, Apr 05 2018

%Y Cf. A007290.

%K nonn,easy

%O 1,6

%A _Zak Seidov_, Aug 07 2010

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Last modified June 14 15:19 EDT 2021. Contains 345025 sequences. (Running on oeis4.)