When does a polynomial f(x) generate an arithmetic sequence for consecutive values of integer x?Given polynomial f ( x ) = a 0 + a 1 x + ⋯ + a n x n , a n ≠ 0 , n ≥ 2 , k ∈ ( 0 , n ) , a k ∈ { 0 , 1 , 2 , … }, under what conditions is f ( r ) , f ( r + 1 ) , … f ( r + n ) an arithmetic sequence for integer r?When does f(x) not generate an ( n + 1 ) term arithmetic sequence? (Update: Never - based on responses to this question).Updated question: When does f(x) generate an n term arithmetic sequence? Given n values, we can fit an n-degree polynomial. The question is given the polynomial, when does it generate an n-term arithmetic sequence for n consecutive values of xNote that a k are fixed in this problem.

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When does a polynomial f(x) generate an arithmetic sequence for consecutive values of integer x?Given polynomial   f  (  x  )  =      a    0    +      a    1    x  +  ⋯  +      a    n        x    n    ,      a    n    ≠  0  ,  n  ≥  2  ,  k  ∈  (  0  ,  n  )  ,      a    k    ∈  {  0  ,  1  ,  2  ,  …  }, under what conditions is   f  (  r  )  ,  f  (  r  +  1  )  ,  …  f  (  r  +  n  ) an arithmetic sequence for integer r?When does f(x) not generate an   (  n  +  1  ) term arithmetic sequence? (Update: Never - based on responses to this question).Updated question: When does f(x) generate an n term arithmetic sequence? Given n values, we can fit an n-degree polynomial. The question is given the polynomial, when does it generate an n-term arithmetic sequence for n consecutive values of xNote that       a    k   are fixed in this problem.

Dobricap · Accepted Answer

Step 1In one direction, this obviously happens for every r when f(x) is linear. In the other direction, suppose f(x) is not linear, then we can write the   n  +  1 values   f  (  r  )  ,  f  (  r  +  1  )  ,  .  .  .  ,  f  (  r  +  n  ) as   a  +  b  x for   x  =  r  ,  r  +  1  ,  .  .  .  ,  r  +  n. Then the polynomial  g  (  x  )  =  f  (  x  )  −  (  a  +  b  x  )Step 2is a nonzero polynomial of degree n but has   n  +  1 zeroes, which is impossible by the Fundamental Theorem of Algebra. So, such an arithmetic sequence exists if and only if f(x) is linear.

blackdivcp · Accepted Answer

Explanation:If the sequence has n terms, one can fit a   n  +  1 or higher degree polinomial that agrees on that points (and has arbitrary values at other, selected points).

Given polynomial f(x)=a_0+a_1x+cdots+a_n x^n, a_n ne 0,n>=2, k in (0,n), a_k in {0,1,2,…}, under what conditions is f(r),f(r+1),…f(r+n) an arithmetic sequence for integer r?

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