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A054845
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Number of ways of representing n as the sum of one or more consecutive primes.
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15
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0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 2, 1, 1, 0, 0, 0, 2, 1, 0, 1, 0, 1, 1, 1, 2, 0, 0, 0, 0, 2, 1, 0, 1, 0, 3, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 2, 0, 0, 1, 0, 1, 2, 2, 1, 0, 0, 0, 0, 0, 2, 1, 0, 0, 2, 2, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 3, 1, 0, 0, 0, 1, 1, 2, 0, 0, 0, 0, 1, 0, 2, 1, 0, 2, 2
(list;
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history;
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OFFSET
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0,6
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COMMENTS
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Moser shows that the average order of a(n) is log 2, that is, Sum_{i=1..n} a(i) ~ n log 2. This shows that a(n) = 0 infinitely often (and with positive density); Moser asks if a(n) = 1 infinitely often, if a(n) = k is solvable for all k, whether these have positive density, and whether the sequence is bounded. - Charles R Greathouse IV, Mar 21 2011
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REFERENCES
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R. K. Guy, Unsolved Problems In Number Theory, C2.
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LINKS
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Carlos Rivera, Problem 9, The Prime Puzzles and Problems Connection.
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FORMULA
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G.f.: Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^prime(k). - Ilya Gutkovskiy, Apr 18 2019
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EXAMPLE
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a(5)=2 because of 2+3 and 5. a(17)=2 because of 2+3+5+7 and 17.
41 = 41 = 11+13+17 = 2+3+5+7+11+13, so a(41)=3.
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MAPLE
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local a, mipri, npr, ps ;
a := 0 ;
for mipri from 1 do
for npr from 1 do
ps := add(ithprime(i), i=mipri..mipri+npr-1) ;
if ps = n then
a := a+1 ;
elif ps >n then
break;
end if;
end do:
if ithprime(mipri) > n then
break ;
end if;
end do:
a ;
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MATHEMATICA
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f[n_] := Block[{p = Prime@ Range@ PrimePi@ n}, len = Length@ p; Count[(Flatten[#, 1] &)[Table[ p[[i ;; j]], {i, len}, {j, i, len}]], q_ /; Total@ q == n]]; f[0] = 0; Array[f, 102, 0](* Jean-François Alcover, Feb 16 2011 *) (* fixed by Robert G. Wilson v *)
nn=100; p=Prime[Range[PrimePi[nn]]]; t=Table[0, {nn}]; Do[s=0; j=i; While[s=s+p[[j]]; s<=nn, t[[s]]++; j++], {i, Length[p]}]; Join[{0}, t]
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PROG
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(PARI){/* program gives values of a(n) for n=0..precprime(nn)-1 */
nn=2000; t=vector(nn+1); forprime(x=2, nn, s=x;
forprime(y=x+1, nn, if(s<=nn, t[s+1]++, break()); s=s+y)); t[1..precprime(nn)]} \\ Zak Seidov, Feb 17 2011, corrected by Michael S. Branicky, Feb 28 2022
(Magma) S:=[0]; for n in [1..104] do count:=0; for q in PrimesUpTo(n) do p:=q; s:=p; while s lt n do p:=NextPrime(p); s+:=p; end while; if s eq n then count+:=1; end if; end for; Append(~S, count); end for; S; // Klaus Brockhaus, Feb 17 2011
(Perl) use ntheory ":all"; my $n=10000; my @W=(0)x($n+1); forprimes { my $s=$_; do { $W[$s]++; $s += ($_=next_prime($_)); } while $s <= $n; } $n; print "$_ $W[$_]\n" for 0..$#W; # Dana Jacobsen, Aug 22 2018
(Python)
from sympy import primerange
def aupton(nn): # modification of PARI by Zak Seidov
alst = [0 for n in range(nn+1)]
for x in primerange(2, nn+1):
s = x
alst[s] += 1
for y in primerange(x+1, nn+1):
s += y
if s <= nn:
alst[s] += 1
else:
break
return alst
(Python) # alternate version for going to large n
import heapq
from sympy import prime
from itertools import islice
def agen(): # generator of terms
p = v = prime(1); h = [(p, 1, 1)]; nextcount = 2; oldv = ways = 0
while True:
(v, s, l) = heapq.heappop(h)
if v == oldv: ways += 1
else:
yield ways
for n in range(oldv+1, v): yield 0
ways = 1
if v >= p:
p += prime(nextcount)
heapq.heappush(h, (p, 1, nextcount))
nextcount += 1
oldv = v
v -= prime(s); s += 1; l += 1; v += prime(l)
heapq.heappush(h, (v, s, l))
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CROSSREFS
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KEYWORD
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nice,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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