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A174960
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Smallest prime p such that p + n*(n+1)/2 is prime, or 0 if no such prime exists.
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3
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2, 2, 2, 5, 3, 2, 2, 3, 5, 2, 0, 5, 5, 0, 2, 7, 3, 0, 2, 3, 13, 2, 0, 5, 7, 0, 2, 5, 3, 0, 2, 3, 13, 2, 0, 11, 7, 0, 2, 7, 3, 2, 0, 7, 7, 0, 0, 23, 5, 0, 2, 41, 3, 2, 2, 3, 5, 0, 0, 7, 17, 0, 0, 11, 3, 0, 2, 3, 5, 2, 0, 23, 5, 0, 2, 7, 13, 0, 2, 3, 11, 2, 0, 5, 11, 0, 0, 5, 3, 2, 0, 31, 5, 2, 0, 7, 7, 0, 0
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OFFSET
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0,1
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COMMENTS
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If n(n+1)/2 is odd, m+n(n+1)/2 can be prime only for m = 2, since otherwise m+n(n+1)/2 is divisible by 2. Hence a(n) = 0 if n(n+1)/2 is odd and 2+n(n+1)/2 is not prime.
For n > 0 also smallest m such that all eigenvalues of the n X n matrix M_m,n are prime, where M_m,n(j,k) = j for j <> k, M_m,n(j,k) = m+j for j = k.
The eigenvalues of M_m,n are m+n(n+1)/2, and m with the multiplicity n-1; cf. reference for proof. Thus all eigenvalues can be prime only if m is prime.
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REFERENCES
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J.-M. Monier, Algebre et geometrie, exercices corriges. Dunod, 1997, p. 78.
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LINKS
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EXAMPLE
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(in Maple notation)
For n = 1 and m = 2, eigenvals(matrix(1,1, [[3]])) = {3}, so a(1) = 2.
For n = 2 and m = 2, eigenvals(matrix(2,2, [[3,1],[2,4]]) = {2,5} so a(2) = 2.
For n = 3 and m = 2, eigenvals(matrix(3,3, [[3,1,1],[2,4,2],[3,3,5]])) = {2,2,8} and 8 is not prime; for m = 3, eigenvals(matrix(3,3, [[4,1,1],[2,5,2],[3,3,6]])) = {3,3,9} and 9 is not prime; for m = 5, eigenvals(matrix(3,3, [[6,1,1],[2,7,2],[3,3,8]])) = {5,5,11} and 11 is prime, so a(3) = 5;
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MAPLE
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with(numtheory):for n from 1 to 200 do:nn:=1:for k from 2 to 1000 do: x:=k + n*(n+1)/2:if (type(x, prime)=true)and(type(k, prime)=true)and nn=1 then print(k):nn:=2:else fi:od:od:
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MATHEMATICA
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a[n_] := (p = 2; q = n*(n+1)/2; While[p > 0, If[ PrimeQ[p+q], Break[], p = If[ OddQ[q], 0, NextPrime[p]]]]; p); Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 03 2011 *)
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PROG
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(Magma) SmallestP:=function(n) for p in PrimesUpTo(1000) do if IsPrime(p + n*(n+1) div 2) then return p; end if; end for; return 0; end function; [SmallestP(n): n in [0..100]]; // Klaus Brockhaus, Apr 10 2010
(Magma) SmallestQ:=function(n) for m in PrimesUpTo(1000) do E:=Eigenvalues(Matrix([&cat[ [j ne k select j else m+j]: k in [1..n]]: j in [1..n] ])); if forall(t){x: x in E | IsPrime(x[1])} then return m; end if; end for; return 0; end function; [2] cat [SmallestQ(n): n in [1..100]]; // Klaus Brockhaus, Apr 10 2010
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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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STATUS
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approved
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