A185636 -id:A185636 - cp's OEIS frontend

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

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 *)

A224030 a(n) = |{0

Original entry on oeis.org

0, 1, 0, 0, 2, 2, 1, 2, 2, 2, 2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 3, 1, 2, 1, 1, 2, 4, 3, 4, 2, 1, 2, 2, 2, 1, 1, 2, 3, 2, 2, 4, 3, 3, 5, 4, 3, 3, 1, 4, 3, 2, 2, 2, 3, 2, 1, 3, 3, 4, 3, 7, 2, 5, 2, 3, 5, 5, 5, 4, 3, 2, 3, 2, 3, 5, 2, 2, 4, 5, 4, 4, 2, 4, 9, 4, 6, 7, 5, 3, 3, 4, 3, 3, 9, 5, 3, 3, 3, 5

Offset: 1

Zhi-Wei Sun, Apr 15 2013

nonn

Conjecture: a(n)>0 for all n>4.
This has been verified for n up to 10^8.
We also conjecture that for any integer n>1 there is an integer 0

			a(7) = 1 since 2*7+5 = 19 and 2*7^3+5^3 = 811 are both prime.
a(57) = 1 since 2*57+23 = 137 and 2*57^3+23^3 = 382553 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
D. R. Heath-Brown, Primes represented by x^3 + 2y^3. Acta Mathematica 186 (2001), pp. 1-84.
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588 [math.NT], 2012-2017.

Cf. A185636, A204065, A220413.

a[n_]:=a[n]=Sum[If[PrimeQ[2n+k]==True&&PrimeQ[2n^3+k^3]==True,1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A232109 Least prime p < n + 5 with n + (p-1)*(p-3)/8 prime, or 0 if such a prime p does not exist.

Original entry on oeis.org

5, 3, 3, 5, 3, 5, 3, 7, 11, 5, 3, 5, 3, 7, 17, 5, 3, 5, 3, 7, 11, 5, 3, 23, 17, 7, 11, 5, 3, 5, 3, 13, 11, 7, 19, 5, 3, 7, 17, 5, 3, 5, 3, 7, 17, 5, 3, 23, 11, 7, 11, 5, 3, 23, 17, 7, 11, 5, 3, 5, 3, 31, 11, 7, 19, 5, 3, 7, 11, 5, 3, 5, 3, 13, 17, 7, 19, 5, 3, 7, 17, 5, 3, 23, 17, 7, 11, 5, 3, 29, 11, 13, 11, 7, 19, 5, 3, 7, 11, 5

Offset: 1

Zhi-Wei Sun, Nov 18 2013

nonn

Conjecture: a(n) > 0 for all n > 0. Moreover, for any integer n > 1 there exists a prime p < 2*sqrt(n)*log(7n) such that n + (p-1)*(p-3)/8 is prime.

This implies that any integer n > 1 can be written as (p-1)/2 + q with q a positive integer, and p and (p^2-1)/8 + q both prime.

			a(1) = 5 since neither 1 + (2-1)*(2-3)/8 = 7/8 nor 1 + (3-1)*(3-3)/8 = 1  is prime, but 1 + (5-1)*(5-3)/8 = 2 is prime.

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

Cf. A000040, A185636, A204065, A227898, A228425, A231201, A231557.

Do[Do[If[PrimeQ[n+(Prime[k]-1)(Prime[k]-3)/8],Goto[aa]],{k,1,PrimePi[n+4]}];
Print[n," ",0];Label[aa];Continue,{n,1,100}]

A239238 a(n) = |{0 <= k < n: q(n+k*(k+1)/2) + 1 is prime}|, where q(.) is the strict partition function given by A000009.

Original entry on oeis.org

1, 2, 3, 2, 3, 1, 4, 5, 2, 4, 5, 4, 4, 4, 2, 4, 3, 6, 3, 1, 3, 5, 5, 5, 2, 9, 8, 7, 5, 3, 3, 4, 3, 7, 4, 8, 6, 2, 6, 6, 5, 2, 5, 5, 3, 3, 4, 4, 7, 7, 8, 5, 5, 4, 8, 6, 3, 4, 3, 5, 11, 2, 2, 4, 6, 6, 5, 5, 4, 4, 5, 6, 6, 8, 4, 9, 4, 6, 4, 3

Offset: 1

Zhi-Wei Sun, Mar 13 2014

nonn

We note that a(n) > 0 for n up to 3580 with the only exception n = 1831. Also, for n = 722, there is no number k among 0, ..., n with q(n+k*(k+1)/2) - 1 prime.

			a(6) = 1 since q(6+0*1/2) + 1 = q(6) + 1 = 5 is prime.
a(20) = 1 since q(20+8*9/2) + 1 = q(56) + 1 = 7109 is prime.
a(104) = 1 since q(104+15*16/2) + 1 = q(224) + 1 = 1997357057 is prime.
a(219) = 1 since q(219+65*66/2) + 1 = q(2364) + 1 = 111369933847869807268722580000364711 is prime.
a(1417) > 0 since q(1417+1347*1348/2) + 1 = q(909295) + 1 is prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..600

Cf. A000009, A000040, A000217, A185636, A239232.

q[n_]:=PartitionsQ[n]
a[n_]:=Sum[If[PrimeQ[q[n+k(k+1)/2]+1],1,0],{k,0,n-1}]
Table[a[n],{n,1,80}]

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

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A224030 a(n) = |{0

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A232109 Least prime p < n + 5 with n + (p-1)*(p-3)/8 prime, or 0 if such a prime p does not exist.

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A239238 a(n) = |{0 <= k < n: q(n+k*(k+1)/2) + 1 is prime}|, where q(.) is the strict partition function given by A000009.

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