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
Keywords
Examples
a(11)=1 since 11=8+(7-5+3-2) with 4, 8, 12 all practical.
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.
Programs
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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}]
Comments