A185150 -id:A185150 - cp's OEIS frontend

A191004 Number of ways to write n = p+q+(n mod 2)q, where p is an odd prime and q<=n/2 is a prime such that JacobiSymbol[q,n]=1 if n is odd, and JacobiSymbol[(q+1)/2,n+1]=1 if n is even.

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 3, 3, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 4, 2, 2, 2, 2, 2, 2, 1, 1, 2, 4, 3, 5, 4, 1, 4, 1, 2, 3, 2, 2, 2, 3, 1, 4, 1, 2, 4, 2, 2, 3, 1, 2, 4, 5, 3, 3, 1, 4, 3, 2, 3, 5, 3, 4, 8, 2, 2, 7, 4, 4, 5, 2, 2, 6, 3, 3, 4, 4, 2, 4, 2, 1, 4, 4

Offset: 1

Zhi-Wei Sun, Dec 30 2012

nonn

Conjecture: a(n)>0 for all n>5.
We have verified this for n up to 10^9. It is stronger than Goldbach's conjecture and Lemoine's conjecture.
Zhi-Wei Sun also conjectured the following refinement: Any odd number 2n+1>64 not among 105, 247, 255, 1105 can be written as p+2q, where p and q are primes, and JacobiSymbol[q,p']=1 for any prime divisor p' of 2n+1; also, any even number 2n>8 not among 32 and 152 can be written as p+q, where p and q<=n/2 are primes, and JacobiSymbol[(q+1)/2,p']=1 for any prime divisor p' of 2n+1.

			a(19)=1 since 19=5+2*7 with JacobiSymbol[7,19]=1.
a(32)=1 since 32=29+3 with JacobiSymbol[(3+1)/2,32+1]=1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..20000
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

Cf. A002375, A046927, A185150.

a[n_]:=a[n]=Sum[If[(Mod[n,2]==1&&PrimeQ[n-2Prime[k]]==True&&JacobiSymbol[Prime[k],n]==1)||(Mod[n,2]==0&&n-Prime[k]>2&&PrimeQ[n-Prime[k]]==True&&JacobiSymbol[(Prime[k]+1)/2,n+1]==1),1,0],{k,1,PrimePi[n/2]}]
Do[Print[n," ",a[n]],{n,1,200}]

A190898 Least odd prime p>n^2 with (n/p) = 1, where ( / ) is the Legendre symbol.

Original entry on oeis.org

3, 7, 11, 17, 29, 43, 53, 71, 83, 107, 127, 157, 173, 199, 229, 257, 293, 337, 379, 401, 457, 499, 541, 577, 631, 683, 733, 787, 857, 911, 967, 1031, 1091, 1163, 1229, 1297, 1373, 1447, 1553, 1601, 1697, 1787, 1867, 1973, 2029, 2129, 2213, 2339, 2411, 2503, 2617, 2707, 2819, 2927, 3041, 3137, 3251, 3457, 3491, 3607

Offset: 1

Zhi-Wei Sun, Dec 29 2012

Conjecture: a(n)<(n+1)^2 for all n>0. (See also A185150.)

This conjecture implies that a(1),a(2),a(3),... are pairwise distinct.

			a(2)=7 since 7 is the first prime p>2^2 with (2/p) = 1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

Cf. A180150, A014085, A185636.

Do[Do[If[n^2+k>2&&PrimeQ[n^2+k]==True&&JacobiSymbol[n,n^2+k]==1,Print[n," ",n^2+k];Goto[aa]],{k,1,2n}];
Label[aa];Continue,{n,1,100}]
js[n_]:=Module[{p=NextPrime[n^2]},While[JacobiSymbol[n,p]!=1,p= NextPrime[ p]];p]; Join[{3},Array[js,60,2]] (* Harvey P. Dale, Jan 29 2023 *)

A192056 a(n) = |{0

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 2, 2, 1, 3, 1, 2, 2, 3, 3, 2, 1, 3, 4, 2, 6, 2, 1, 8, 3, 3, 6, 2, 1, 3, 3, 1, 5, 7, 5, 4, 4, 3, 3, 6, 3, 3, 6, 3, 5, 3, 7, 5, 7, 6, 4, 5, 1, 8, 8, 2, 4, 6, 1, 5, 2, 4, 9, 8, 3, 6, 7, 3, 5, 5, 5, 3, 3, 5, 9, 4, 13, 6, 5, 9, 7, 7, 3, 10, 9, 8, 9, 7, 4, 7, 13, 5, 7, 10, 4, 4, 11, 4, 5

Offset: 1

Zhi-Wei Sun, Dec 30 2012

nonn

Conjecture: a(n)>0 for all n>2. Moreover, if n>2 is not among 12, 18, 105, 522, then there is 0

k^2+(n-k)^2-3(n-1 mod 2) is prime and also JacobiSymbol[k,p]=1 for any prime divisor p of n+3(n-1 mod 2).

This conjecture has been verified for n up to 2*10^8. It is stronger than Ming-Zhi Zhang's conjecture that any odd integer n>1 can be written as x+y (x,y>0) with x^2+y^2 prime (see A036468).

			a(33)=1 since 4^2+29^2=857 is prime and JacobiSymbol[4,33]=1.
a(24)=1 since 10^2+14^2-3=293 is prime and JacobiSymbol[10,24+3]=1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

Cf. A036468, A191004, A185150, A220554.

a[n_]:=a[n]=Sum[If[PrimeQ[k^2+(n-k)^2-3Mod[n-1,2]]==True&&JacobiSymbol[k,n+3Mod[n-1,2]]==1,1,0],{k,1,(n-1)/2}]
Do[Print[n," ",a[n]],{n,1,100}]

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cp's OEIS Frontend

A191004 Number of ways to write n = p+q+(n mod 2)q, where p is an odd prime and q<=n/2 is a prime such that JacobiSymbol[q,n]=1 if n is odd, and JacobiSymbol[(q+1)/2,n+1]=1 if n is even.

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A190898 Least odd prime p>n^2 with (n/p) = 1, where ( / ) is the Legendre symbol.

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A192056 a(n) = |{0

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