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A191837
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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.
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2
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48, 108, 204, 324, 624, 630, 1050, 1320, 1590, 2100, 2400, 2730, 3570, 3960, 4830, 5460, 5880, 6930, 7770, 9240, 9450, 11970, 12810, 13020, 14910, 14910, 17430, 18480, 20160, 21630, 23100, 24150, 28770, 28770, 31290, 32760, 32760, 36960, 36960, 39270, 39270, 50190, 51870, 51870
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OFFSET
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2,1
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COMMENTS
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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.
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.
The integers in this sequence are all congruent to 0 mod 6.
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.
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
All a(n) are congruent to 0 mod 6=2*3.
All a(n) >= a(7)=630 are congruent to 0 mod 30=2*3*5.
All a(n) >= a(16)=4830 are congruent to 0 mod 210=2*3*5*7.
All a(n) >= a(279)=3513510 are congruent to 0 mod 2310=2*3*5*7*11.
All a(n) >= a(1440)=137507370 are congruent to 0 mod 30030=2*3*5*7*11*13. (End)
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LINKS
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EXAMPLE
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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.
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.
a(2) = 48 = 5+7+17+19 = 7+11+13+17
a(3) = 108 = 5+7+11+19+29+37
a(9) = 1320 = 13+17+19+23+29+31+37+41+43+61+71+83+89+97+103+107+149+307
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MATHEMATICA
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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*)
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PROG
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(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))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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a(2) to a(5) verified; a(6) to a(10) added by S Kolman, Jul 03 2011
a(11) to a(13) added by S Kolman, Jul 04 2011
a(14) to a(14) added by S Kolman, Jul 05 2011
Corrected a(8)-a(14) and extended to a(2500). - J. Stauduhar, Jul 12 2011
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STATUS
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approved
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