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A054845 Number of ways of representing n as the sum of one or more consecutive primes. 14
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; graph; refs; listen; history; text; internal format)
OFFSET

0,6

COMMENTS

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

REFERENCES

R. K. Guy, Unsolved Problems In Number Theory, C2.

LINKS

T. D. Noe, Table of n, a(n) for n = 0..10000

Leo Moser, Notes on number theory. III. On the sum of consecutive primes, Canad. Math. Bull. 6 (1963), pp. 159-161.

C. Rivera, Problem 9, Prime Puzzles.

FORMULA

G.f.: Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^prime(k). - Ilya Gutkovskiy, Apr 18 2019

EXAMPLE

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.

MAPLE

A054845 := proc(n)

    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 ;

end proc: # R. J. Mathar, Nov 27 2018

MATHEMATICA

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-Fran├žois 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]

PROG

(PARI){/* program gives nn+1 values of a(n) for n=0..nn */

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} \\ Zak Seidov, Feb 17 2011

(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

CROSSREFS

Cf. A000586, A054859.

Sequence in context: A305490 A113706 A279952 * A317991 A236853 A117163

Adjacent sequences:  A054842 A054843 A054844 * A054846 A054847 A054848

KEYWORD

nice,nonn

AUTHOR

Jud McCranie, May 25 2000

EXTENSIONS

Edited by N. J. A. Sloane, Oct 27 2008 at the suggestion of Jake M. Foster

STATUS

approved

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Last modified June 15 20:50 EDT 2019. Contains 324145 sequences. (Running on oeis4.)