A327449 Number of ways the first n primes can be partitioned into three sets with equal sums.
0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 423, 0, 2624, 0, 13474, 0, 0, 0, 611736, 0, 4169165, 0, 30926812, 0, 214975174, 0, 1590432628, 0, 11431365932, 0, 83946004461, 0, 0, 0, 4615654888831, 0, 35144700468737, 0, 271133285220726, 0, 2103716957561013, 0, 0, 0, 0, 0, 990170108748552983, 0, 7855344215856348141
Offset: 1
Keywords
Examples
One of the three solutions for n = 10: 3 + 17 + 23 = 2 + 5 + 7 + 29 = 11 + 13 + 19.
References
- Keith F. Lynch, Posting to Math Fun Mailing List, Sep 17 2019.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..102
Crossrefs
Programs
-
Maple
s:= proc(n) option remember; `if`(n<2, 0, ithprime(n)+s(n-1)) end: b:= proc(n, x, y) option remember; `if`(n=1, 1, (p-> (l-> add(`if`(p>l[i], 0, b(n-1, sort(subsop(i=l[i]-p, l)) [1..2][])), i=1..3))([x, y, s(n)-x-y]))(ithprime(n))) end: a:= n-> `if`(irem(2+s(n), 3, 'q')=0, b(n, q-2, q)/2, 0): seq(a(n), n=1..40); # Alois P. Heinz, Sep 19 2019
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Mathematica
s[n_] := s[n] = If[n < 2, 0, Prime[n] + s[n - 1]]; b[n_, x_, y_] := b[n, x, y] = If[n == 1, 1, Function[p, Function[l, Sum[If[ p > l[[i]], 0, b[n - 1, Sequence @@ Sort[ReplacePart[l, i -> l[[i]] - p]][[1;; 2]]]], {i, 1, 3}]][{x, y, s[n] - x - y}]][Prime[n]]]; a[n_] := If[Mod[2+s[n], 3]==0, q = Quotient[2+s[n], 3]; b[n, q-2, q]/2, 0]; Array[a, 40] (* Jean-François Alcover, Apr 09 2020, after Alois P. Heinz *)
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PARI
EqSumThreeParts(v)={ my(n=#v, vs=vector(n), m=vecsum(v)/3, brk=0); for(i=1, n-1, vs[i+1]=vs[i]+v[i]; if(vs[i]<=min(1000,m), brk=i)); my(q=Vecrev(prod(i=1, brk, 1+x^v[i]+y^v[i]))); my(recurse(k,s,p)=if(k==brk, if(s<#q, polcoef(p*q[s+1],m,y)), if(s<=vs[k], self()(k-1, s, p*(1 + y^v[k]))) + if(s>=v[k], self()(k-1, s-v[k], p)) )); if(frac(m), 0, recurse(n-1, m, 1 + O(y*y^m))/2); } a(n)={EqSumThreeParts(primes(n))} \\ Andrew Howroyd, Sep 19 2019
Formula
Extensions
Corrected and a(30)-a(52) added by Andrew Howroyd, Sep 19 2019
a(53) and beyond from Alois P. Heinz, Sep 19 2019
Comments