1,1

The polynomial is to pass through the points (k, prime(k)), k=1..n.

The constant term is always an integer because it is the same as f(0), which can be computed from the difference table of the sequence of primes. See Conway and Guy. In fact, the interpolating polynomial is integral for all integer arguments.

A plot of the first 1000 terms shows that the sequence grows exponentially and changes signs occasionally. The Mathematica lines show two ways of computing the sequence. The second, which uses the difference table, is much faster.

J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 80

T. D. Noe, Table of n, a(n) for n=1..1000

Author?, Sicurvqf

T. D. Noe, Plot of A082594

a(n) = sum{k=1, .., n} (-1)^(k+1) A007442(k)

For n=4, we fit a cubic through the 4 points (1,2),(2,3),(3,5),(4,7) to obtain a(4) = 3.

Table[Coefficient[Expand[InterpolatingPolynomial[Prime[Range[n]], x]], x, 0], {n, 50}]

Diff[lst_List] := Table[lst[[i+1]]-lst[[i]], {i, Length[lst]-1}]; n=50; dt=Table[{}, {n}]; dt[[1]]=Prime[Range[n]]; Do[dt[[i]]=Diff[dt[[i-1]]], {i, 2, n}]; Table[s=dt[[i, 1]]; Do[s=dt[[i-j, 1]]-s, {j, i-1}]; s, {i, n}]

Cf. A007442.

Sequence in context: A001371 A001037 A122086 this_sequence A051850 A077013 A086880

Adjacent sequences: A082591 A082592 A082593 this_sequence A082595 A082596 A082597

sign

Cino Hilliard (hillcino368(AT)gmail.com), May 08 2003

Edited by T. D. Noe (noe(AT)sspectra.com), May 08 2003