A120291 Numerator of determinant of n X n matrix with elements M[i,j] = (1+Prime[i])/Prime[i] if i=j and 1 otherwise.

3, 1, 11, 3, 29, 1, 59, 1, 101, 1, 1, 3, 239, 47, 1, 191, 21, 251, 569, 64, 1, 12, 25, 482, 1061, 1, 1, 98, 1481, 797, 1721, 926, 3, 8, 3, 1214, 1, 458, 1, 1544, 99, 1724, 1213, 1916, 1, 2, 1, 3, 4889, 853, 5351, 1, 49, 3041, 2113, 3301, 6871, 3571, 2473, 10, 2661

Offset: 1

Author

Alexander Adamchuk, Jul 08 2006, Aug 19 2006

Keywords

• frac
• nonn

Comments

Many a(n), such as 3,11,29,59,101,239,569,1061,1481,1721,4889.., are primes of form p(1)+...+p(k)+1 where p(i) =i-th prime A053845. It appeares that all primes of this form are presented in a(n) in their natural order.

Indices n such that a(n) = 1 are {2,6,8,10,11,15,21,26,27,37,39,45,47,52,75,84,87,88,91,94,...} = A121744[n] Numbers n such that (1 + Sum[Prime[k],{k,1,n}]) = (1 + A007504[n]) divides primorial number p(n)# = Product[Prime[k],{k,1,n}] = A002110[n].

Crossrefs

Cf. A024528, A053845.

Cf. A121744, A007504, A002110.

Programs

Mathematica

Numerator[Table[Det[DiagonalMatrix[Table[1/Prime[i],{i,1,n}]]+1],{n,1,70}]]
Table[Numerator[(1+Sum[Prime[k],{k,1,n}])/Product[Prime[k],{k,1,n}]],{n,1,100}]

Formula

a(n) = numerator[Det[DiagonalMatrix[Table[1/Prime[i],{i,1,n}]]+1]].

a(n) = Numerator[ (1 + Sum[ Prime[k], {k,1,n} ]) / Product[ Prime[k], {k,1,n} ] ]. a(n) = Numerator[ (1 + A007504[n]) / A002110[n] ].