A226743 Number of alternating sums of consecutive primes with result 2n-1.

1, 1, 2, 2, 3, 1, 2, 3, 2, 3, 5, 4, 4, 4, 5, 4, 5, 5, 4, 6, 5, 4, 9, 5, 5, 7, 6, 6, 10, 7, 9, 5, 11, 6, 6, 9, 8, 8, 9, 9, 9, 12, 8, 8, 10, 7, 9, 9, 12, 11, 8, 11, 12, 6, 10, 6, 8, 14, 10, 12, 13, 10, 11, 5, 11, 9, 11, 16, 11, 11, 14, 10, 10, 13, 10, 17, 12, 11, 18, 13, 13, 11, 18, 11, 13, 12, 14, 16, 17, 14, 10, 15, 11, 12

Offset: 1

Author

Ralf Stephan, Sep 01 2013

Keywords

• nonn

Comments

Since A008347 has no duplicate values, a(n) must be finite. This is not true for even results of the sum.

Sums of a single term are not included. - Robert Israel, Feb 06 2025

Examples

			n=5: 11-7+5=2*5-1, 13-11+7=2*5-1, 19-17+13-11+7-5+3=2*5-1, so a(5)=3.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A084143.

Programs

Maple

N:= 100: M:= 2*N-1: # for a(1)..a(N)
p:= 1: b:= 0: B:= NULL:
for i from 1 do
  p:= nextprime(p);
  b:= b + (-1)^i*p;
  B:= B,b;
  if b > M then nB:= i; break fi;
od:
V:= Vector(M):
for j from 2 to nB by 2 while B[j] <= M do V[B[j]]:= 1 od:
for i from 1 to nB do
  for j from i+3 to nB by 2 do
    r:= abs(B[j]-B[i]);
    if r <= M then V[r]:= V[r]+1 else break fi;
od od:
seq(V[i],i=1..M,2); # Robert Israel, Feb 06 2025

PARI

vb=vecsmall(500);for(k=2,1000,forstep(l=k-1,1,-1,t=sum(i=l,k,prime(i)*(-1)^(k-i));if(t<500,vb[t]=vb[t]+1)))