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%I #13 Oct 31 2015 14:01:42
%S 1,2,2,2,2,1,1,3,3,4,3,4,3,4,2,2,4,4,4,5,3,1,4,4,2,5,2,4,4,4,2,2,3,5,
%T 6,2,4,5,5,4,4,4,3,9,5,4,2,5,6,7,6,7,6,3,3,9,6,6,8,5,3,5,5,4,8,7,6,5,
%U 5,3,3,5,6,8,6,6,4,8,2,6,5,5,8,8,2,5,7,4,9,7,5,5,6,5,4,4,5,6,7,6
%N Number of ordered ways to write n as x^2 + 2*y^2 + phi(z^2) (x >= 0, y >= 0 and z > 0) such that y or z has the form p-1 with p prime, where phi(.) is Euler's totient function.
%C Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 6, 7, 22, 3447.
%C This is similar to the conjecture in A262311, and we have verified it for n up to 10^6.
%H Zhi-Wei Sun, <a href="/A263992/b263992.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;6f4ac126.1510">Some mysterious representations of integers</a>, a message to Number Theory Mailing List, Oct. 25, 2015.
%e a(1) = 1 since 1 = 0^2 + 2*0^2 + phi(1^2) with 1 + 1 = 2 prime.
%e a(6) = 1 since 6 = 2^2 + 2*0^2 + phi(2^2) with 2 + 1 = 3 prime.
%e a(7) = 1 since 7 = 2^2 + 2*1^2 + phi(1^2) with 1 + 1 = 2 prime.
%e a(22) = 1 since 22 = 0^2 + 2*1^2 + phi(5^2) with 1 + 1 = 2 prime.
%e a(3447) = 1 since 3447 = 42^2 + 2*29^2 + phi(1^2) with 1 + 1 = 2 prime.
%t phi[n_]:=phi[n]=EulerPhi[n]
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
%t Do[r=0;Do[If[n-z*phi[z]<0,Goto[aa]];Do[If[SQ[n-z*phi[z]-2y^2]&&(PrimeQ[y+1]||PrimeQ[z+1]),r=r+1],{y,0,Sqrt[(n-z*phi[z])/2]}];Label[aa];Continue,{z,1,n}];Print[n," ",r];Continue,{n,1,100}]
%Y Cf. A000010, A000290, A006093, A262311, A263998.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Oct 31 2015
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