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%I #64 Feb 28 2022 10:52:41
%S 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,
%T 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,
%U 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
%N Number of ways of representing n as the sum of one or more consecutive primes.
%C 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
%D R. K. Guy, Unsolved Problems In Number Theory, C2.
%H T. D. Noe, <a href="/A054845/b054845.txt">Table of n, a(n) for n = 0..10000</a>
%H Leo Moser, <a href="https://doi.org/10.4153/CMB-1963-013-1">Notes on number theory. III. On the sum of consecutive primes</a>, Canad. Math. Bull. 6 (1963), pp. 159-161.
%H Carlos Rivera, <a href="http://www.primepuzzles.net/problems/prob_009.htm">Problem 9</a>, The Prime Puzzles and Problems Connection.
%F G.f.: Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^prime(k). - _Ilya Gutkovskiy_, Apr 18 2019
%e a(5)=2 because of 2+3 and 5. a(17)=2 because of 2+3+5+7 and 17.
%e 41 = 41 = 11+13+17 = 2+3+5+7+11+13, so a(41)=3.
%p A054845 := proc(n)
%p local a,mipri,npr,ps ;
%p a := 0 ;
%p for mipri from 1 do
%p for npr from 1 do
%p ps := add(ithprime(i),i=mipri..mipri+npr-1) ;
%p if ps = n then
%p a := a+1 ;
%p elif ps >n then
%p break;
%p end if;
%p end do:
%p if ithprime(mipri) > n then
%p break ;
%p end if;
%p end do:
%p a ;
%p end proc: # _R. J. Mathar_, Nov 27 2018
%t 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_ *)
%t 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]
%o (PARI){/* program gives values of a(n) for n=0..precprime(nn)-1 */
%o nn=2000;t=vector(nn+1);forprime(x=2,nn,s=x;
%o 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
%o (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
%o (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
%o (Python)
%o from sympy import primerange
%o def aupton(nn): # modification of PARI by Zak Seidov
%o alst = [0 for n in range(nn+1)]
%o for x in primerange(2, nn+1):
%o s = x
%o alst[s] += 1
%o for y in primerange(x+1, nn+1):
%o s += y
%o if s <= nn:
%o alst[s] += 1
%o else:
%o break
%o return alst
%o print(aupton(101)) # _Michael S. Branicky_, Feb 17 2022
%o (Python) # alternate version for going to large n
%o import heapq
%o from sympy import prime
%o from itertools import islice
%o def agen(): # generator of terms
%o p = v = prime(1); h = [(p, 1, 1)]; nextcount = 2; oldv = ways = 0
%o while True:
%o (v, s, l) = heapq.heappop(h)
%o if v == oldv: ways += 1
%o else:
%o yield ways
%o for n in range(oldv+1, v): yield 0
%o ways = 1
%o if v >= p:
%o p += prime(nextcount)
%o heapq.heappush(h, (p, 1, nextcount))
%o nextcount += 1
%o oldv = v
%o v -= prime(s); s += 1; l += 1; v += prime(l)
%o heapq.heappush(h, (v, s, l))
%o print(list(islice(agen(), 102))) # _Michael S. Branicky_, Feb 17 2022
%Y Cf. A000586, A054859.
%K nice,nonn
%O 0,6
%A _Jud McCranie_, May 25 2000
%E Edited by _N. J. A. Sloane_, Oct 27 2008 at the suggestion of Jake M. Foster
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