A213325 Number of ways to write n = q + sum_{k=1}^m(-1)^{m-k}p_k, where p_k is the k-th prime, and q is a practical number with q-4 and q+4 also practical.

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

Offset

1,13

Comments

Conjecture: a(n)>0 for all n>8.

The author has verified this for n up to 5*10^6.

Links

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

Zhi-Wei Sun, More conjectures on alternating sums of consecutive primes, a message to Number Theory List, March 2, 2013.

Example

a(11)=1 since 11=8+(7-5+3-2) with 4, 8, 12 all 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)
q[n_]:=q[n]=pr[n-4]==True&&pr[n]==True&&pr[n+4]==True
s[0_]:=0
s[n_]:=s[n]=Prime[n]-s[n-1]
a[n_]:=a[n]=Sum[If[n-s[m]>0&&q[n-s[m]], 1, 0], {m, 1, n}]
Table[a[n], {n, 1, 100}]

Crossrefs

Cf. A008347, A005153, A208246, A213202.

Sequence in context: A303401 A327491 A099910 * A043555 A242753 A232443

Adjacent sequences: A213322 A213323 A213324 * A213326 A213327 A213328

Keyword

nonn

Author

Zhi-Wei Sun, Mar 03 2013

Status

approved