A185150 -id:A185150

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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

+10
2

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,14

COMMENTS

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.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..20000

Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

EXAMPLE

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.

MATHEMATICA

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}]

CROSSREFS

Cf. A002375, A046927, A185150.

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 30 2012

STATUS

approved

A190898

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

+10
1

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

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.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

EXAMPLE

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

MATHEMATICA

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 *)

CROSSREFS

Cf. A180150, A014085, A185636.

KEYWORD

nonn,nice

AUTHOR

Zhi-Wei Sun, Dec 29 2012

STATUS

approved

A192056

a(n) = |{0<k<n/2: k^2+(n-k)^2-3(n-1 mod 2) is prime and JacobiSymbol[k,n+3(n-1 mod 2)]=1}|

+10
1

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,6

COMMENTS

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

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).

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

EXAMPLE

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.

MATHEMATICA

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}]

CROSSREFS

Cf. A036468, A191004, A185150, A220554.

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 30 2012

STATUS

approved

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