%I #18 Jan 04 2017 02:45:21
%S 1,1,1,1,2,1,2,1,1,2,3,4,1,2,1,1,1,2,3,1,1,2,1,2,1,2,1,1,2,1,2,1,2,3,
%T 6,1,2,1,1,2,1,1,1,2,3,1,2,3,1,2,1,1,1,1,2,3,4,1,1,2,3,1,1,2,3,4,1,2,
%U 3,1,1,2,1,2,3,1,2,1,1,1
%N Least positive integer m such that n - p(m) = x*(3x-1)/2 + y*(3y+1)/2 for some nonnegative integers x and y, or 0 if no such m exists, where p(.) is the partition function given by A000041.
%C The conjecture in A280455 asserts that a(n) > 0 for all n > 0.
%H Zhi-Wei Sun, <a href="/A280444/b280444.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/160p.pdf">Problems on combinatorial properties of primes</a>, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28--Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
%e a(12) = 4 since 12 - p(4) = 12 - 5 = 7 = 0*(3*0-1)/2 + 2*(3*2+1)/2.
%e a(35) = 6 since 35 - p(6) = 35 - 11 = 24 = 4*(3*4-1)/2 + 1*(3*1+1)/2.
%e a(4327) = 15 since 4327 - p(15) = 4327 - 176 = 4151 = 16*(3*16-1)/2 + 50*(3*50+1)/2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t p[n_]:=p[n]=PartitionsP[n];
%t Pen[n_]:=Pen[n]=SQ[24n+1]&&Mod[Sqrt[24n+1],6]==1;
%t Do[m=1;Label[bb];If[p[m]>n,Goto[cc]];Do[If[Pen[n-p[m]-x(3x-1)/2],Print[n," ",m];Goto[aa]],{x,0,(Sqrt[24(n-p[m])+1]+1)/6}];m=m+1;Goto[bb];Label[cc];Print[n," ",0];Label[aa];Continue,{n,1,80}]
%Y Cf. A000041, A000326, A005449, A280455.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Jan 03 2017
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