A238459 -id:A238459

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A239209

a(n) = |{0 < k < n: k*p(n)*(p(n)-1) + 1 is prime}|, where p(.) is the partition function (A000041).

+10
4

0, 1, 2, 2, 2, 1, 4, 2, 3, 3, 3, 2, 3, 4, 2, 4, 4, 4, 8, 3, 3, 4, 6, 5, 3, 5, 10, 4, 4, 7, 5, 4, 3, 8, 7, 6, 3, 4, 5, 4, 3, 7, 5, 5, 3, 4, 5, 11, 7, 10, 3, 10, 8, 12, 6, 4, 10, 4, 8, 5, 11, 7, 5, 14, 5, 7, 4, 10, 1, 10, 9, 12, 8, 5, 10, 1, 7, 7, 6, 5

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

OFFSET

1,3

COMMENTS

Conjecture: (i) a(n) > 0 for all n > 1. If n > 1 is not equal to 25, then k*p(n)*(p(n)+1) - 1 is prime for some 0 < k < n.

(ii) Let q(.) be the strict partition function given by A000009. Then, for any integer n > 2, there is a number k among 1, ..., n with k*q(n)^2 - 1 prime. Also, we may replace k*q(n)^2 - 1 by k*q(n)^2 + 1 or k*q(n)*(q(n)+1) + 1 or k*q(n)*(q(n)+1) - 1.

We have verified that a(n) > 0 for all n = 2..10^5.

LINKS

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

Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

EXAMPLE

a(2) = 1 since 1*p(2)*(p(2)-1) + 1 = 1*2*1 + 1 = 3 is prime.

a(6) = 1 since 3*p(6)*(p(6)-1) + 1 = 3*11*10 + 1 = 331 is prime.

a(69) = 1 since 50*p(69)*(p(69)-1) + 1 = 50*3554345*3554344 + 1 = 631668241234001 is prime.

a(76) = 1 since 24*p(76)*(p(76)-1) + 1 = 24*9289091*9289090 + 1 = 2070892855612561 is prime.

MATHEMATICA

p[n_]:=PartitionsP[n]

f[n_]:=p[n]*(p[n]-1)

a[n_]:=Sum[If[PrimeQ[k*f[n]+1], 1, 0], {k, 1, n-1}]

Table[a[n], {n, 1, 80}]

CROSSREFS

Cf. A000009, A000040, A000041, A238393, A238457, A238458, A238459, A238509, A238516, A238577, A239207, A239214.

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Mar 12 2014

STATUS

approved

A238458

Number of primes p < n with 2*P(n-p) + 1 prime, where P(.) is the partition function (A000041).

+10
3

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

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

OFFSET

1,4

COMMENTS

Conjecture: a(n) > 0 for all n > 2. Also, for each n > 3 there is a prime p < n with 2*P(n-p) - 1 prime.

We have verified the conjecture for n up to 10^5.

See also A238459 for a similar conjecture involving the strict partition function.

LINKS

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

Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

EXAMPLE

a(3) = 1 since 2 and 2*P(3-2) + 1 = 2*1 + 1 = 3 are both prime.

a(41) = 1 since 37 and 2*P(41-37) + 1 = 2*5 + 1 = 11 are both prime.

MATHEMATICA

p[n_, k_]:=PrimeQ[2*PartitionsP[n-Prime[k]]+1]

a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, PrimePi[n-1]}]

Table[a[n], {n, 1, 100}]