|
%I #67 Nov 29 2013 21:19:42
%S 48,108,204,324,624,630,1050,1320,1590,2100,2400,2730,3570,3960,4830,
%T 5460,5880,6930,7770,9240,9450,11970,12810,13020,14910,14910,17430,
%U 18480,20160,21630,23100,24150,28770,28770,31290,32760,32760,36960,36960,39270,39270,50190,51870,51870
%N Least even number m which can be written as sum of 2n primes p(1) < ... < p(2n) < m/2 such that m-p(i) is also prime for i=1,...,2n.
%C Original definition: In the Goldbach partitions of 2n, find the first 2n with four prime elements to sum to it; find the first 2n with six elements summing to 2n; and so for 2k elements.
%C Whenever there is more than one decomposition of m as sum of primes, it must be odd+odd=even. Then, only an even number of (odd prime) summands can yield m. Moreover, we restrict these summands to be the lesser one of the decompositions p+q=m, therefore we need more than 2 such summands to yield m, and a(1) is undefined.
%C The integers in this sequence are all congruent to 0 mod 6.
%C There can be more than one composition of m. E.g., for m=48, 48=5+7+17+19 and 48=7+11+13+17.
%C Conjecture: For all a(n), a(n)-1 can be found in A014092 (numbers not the sum of two primes), and a(n)+1 can be found in A007921. (numbers not the difference of two primes). - _J. Stauduhar_, Aug 28 2012
%C From _J. Stauduhar_, Aug 22 2011: (Start)
%C All a(n) are congruent to 0 mod 6=2*3.
%C All a(n) >= a(7)=630 are congruent to 0 mod 30=2*3*5.
%C All a(n) >= a(16)=4830 are congruent to 0 mod 210=2*3*5*7.
%C All a(n) >= a(279)=3513510 are congruent to 0 mod 2310=2*3*5*7*11.
%C All a(n) >= a(1440)=137507370 are congruent to 0 mod 30030=2*3*5*7*11*13. (End)
%H J. Stauduhar, <a href="/A191837/b191837.txt">Table of n, a(n) for n = 2..2500</a>
%H J. Stauduhar, <a href="/A191837/a191837.c.txt">C program to generate sequence A191837</a>
%e For 48, we have 48=5+43=7+41=17+31=19+29 (ignoring 11+37), and use 5+7+17+19 to give the first even number having four such primes summing to itself.
%e Similarly, 108 is the least even number with six prime elements summing to itself: 5+103=7+101=11+97=19+89=29+79=37+71 and taking 5+7+11+19+29+37=108.
%e a(2) = 48 = 5+7+17+19 = 7+11+13+17
%e a(3) = 108 = 5+7+11+19+29+37
%e a(9) = 1320 = 13+17+19+23+29+31+37+41+43+61+71+83+89+97+103+107+149+307
%t nCk[a_, b_]:=Block[{ndx=ns= a, i=rs=b, ct=t=0}, If[(d[[1]]-1)==(ns-rs), For[ct=1, ct<=rs, ct++, t+=s[[d[[ct]]]]]; If[t==m, Print[sm/2, " ", t]; sm+=2; m-=6; Return[False], Return[False]]]; While[d[[i]]==ndx && i>1, --i; --ndx]; d[[i]]+=1; i++; While[i<=rs, d[[i]]=d[[i-1]]+1; ++i;]; For[ct=1, ct<=rs, ct++, t+=s[[d[[ct]]]]; If[t>m, Break[]]]; If[t==m, Print[sm/2, " ", t]; sm+=2; m-=6; Return[False]]; Return[True]]; For[sm=4; m=6, sm<=60, m+=6, s={}; sum=smndct=pct=0; For[p=5, p<m/2, p+=2, If[PrimeQ[p] && PrimeQ[m-p], If[sum+p>m, Break[]]; If[smndct++<sm-1, sum+=p]; AppendTo[s, p]; pct++]]; If[pct >= sm, d=Range[sm]; While[nCk[Length[s], sm]]]]; (* _J. Stauduhar_, Sep 07 2012*)
%o (PARI) a(n)=forstep(m=2,1e9,2,L=[]; forprime(p=1,m\2-1,isprime(m-p)|next;L=concat(L,p)); #L<2*n&next; sum(i=#L-2*n+1,#L,L[i])<m&next;forvec(v=vector(2*n,i,[1,#L]), sum(i=1,2*n,L[v[i]])==m & return(m),2))
%K nonn
%O 2,1
%A _J. M. Bergot_, Jun 17 2011
%E a(4)-a(5) from _M. F. Hasler_, Jun 21 2011
%E a(2) to a(5) verified; a(6) to a(10) added by _S Kolman_, Jul 03 2011
%E a(11) to a(13) added by _S Kolman_, Jul 04 2011
%E a(14) to a(14) added by _S Kolman_, Jul 05 2011
%E Confirmed a(7). a(6) corrected by _J. Stauduhar_, Jul 08 2011
%E Corrected a(8)-a(14) and extended to a(2500). - _J. Stauduhar_, Jul 12 2011
%E Edited by _J. Stauduhar_, Aug 28 2012
|
