A338088 Smallest prime numbers which can be represented as x^2 + h*y^2 with x > 0 for every h in the first n idoneal numbers.

2, 17, 73, 73, 241, 241, 1009, 1009, 1009, 1009, 1009, 2521, 2521, 2521, 2521, 2521, 8089, 8089, 8089, 8089, 8089, 8089, 19009, 19009, 19009, 19009, 19009, 19009, 53881, 53881, 53881, 53881, 53881, 53881, 53881, 605641, 605641, 605641, 605641, 605641, 605641

Offset: 1

Author

Marco Frigerio, Oct 09 2020

Keywords

• nonn
• fini
• full

Comments

The sequence lists prime numbers, in nondecreasing order, such that each of them can be written, in a unique way, in the form x^2 + h*y^2, where x, y are natural numbers, while h takes an increasing number of values of the sequence A000926 (idoneal numbers). See examples.

Examples

			a(1) = 2 because, for A000926(1) = 1, 2 = 1^2+A000926(1)*1^2.
a(2) = 17 because, considered the first two idoneal numbers, A000926(1) = 1 and A000926(2) = 2, 17 = 1^2+A000926(1)*4^2 = 3^2+A000926(2)*2^2.
The prime 1009 is present in the sequence 5 times because:
a(7) = 15^2+1*28^2 = 19^2+2*18^2 = 31^2+3*4^2 = 15^2+4*14^2 = 17^2+5*12^2 = 25^2+6*12^2 = 1^2+7*12^2, with idoneal numbers up to A000926(7), and also:
a(8) = 19^2+8*9^2,
a(9) = 28^2+9*5^2,
a(10) = 3^2+10*10^2,
a(11) = 31^2+12*2^2,
with idoneal numbers from  A000926(8) to A000926(11).
1083289 is the last term of the sequence since for every idoneal number h there are x, y such that 1083289 = x^2 + h*y^2 and this is the least prime for which this is possible.
		

Links

• Marco Frigerio, Table of n, a(n) for n = 1..65

Crossrefs

Cf. A000926.

Programs

PARI

isok(p,u)={for(i=1, #u, my(s=qfbsolve(Qfb(1,0,u[i]),p)); if(s==0 || s[1]==0, return(0))); 1}
idoneal()={select(m->!#select(k->k<>2, quadclassunit(-4*m).cyc), [1..1848])}
seq()={my(u=idoneal(), v=[1], L=List()); forprime(p=2, oo, if(isok(p,v), listput(L,p); my(k=#v); while(k<#u && isok(p,[u[k+1]]), listput(L,p); k++); if(k==#u, return(Vec(L))); v=u[1..k+1]))}  \\ Andrew Howroyd, Nov 05 2020