A222580 Number of ways to write n=p_m-p_{m-1}+...+(-1)^{m-k}p_k with k<m and p_m<=3n, and p_m+1 and p_k-1 both practical, where p_j denotes the j-th prime.

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

Offset

1,4

Comments

Conjecture: All the terms are positive.

See also the comments related to A222579.

Links

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

Zhi-Wei Sun, On functions taking only prime values, arXiv:1202.6589.

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

Example

a(9)=2 since 9=11-7+5=19-17+13-11+7-5+3 with 12, 4, 20, 2 all practical.
a(806)=1 since 806=p_{358}-p_{357}+...+p_{150}-p_{149} with p_{358}=2411<=3*806=2418, and 2412 and p_{149}-1=858 are both practical.

Mathematica

f[n_]:=f[n]=FactorInteger[n]
Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
pp[k_]:=pp[k]=pr[Prime[k]+1]==True
pq[k_]:=pq[k]=pr[Prime[k]-1]==True
s[0_]:=0
s[n_]:=s[n]=Prime[n]-s[n-1]
a[n_]:=a[n]=Sum[If[pp[j]==True&&pq[i+1]==True&&s[j]-(-1)^(j-i)*s[i]==n, 1, 0], {j, PrimePi[n]+1, PrimePi[3n]}, {i, 0, j-2}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A000040, A005153, A210479, A222579.

Sequence in context: A353236 A325939 A318586 * A316978 A331023 A284345

Adjacent sequences: A222577 A222578 A222579 * A222581 A222582 A222583

Keyword

nonn

Author

Zhi-Wei Sun, Feb 25 2013

Status

approved