[seqfan] Re: Arrays that are second differences of a 0..k array

Richard J. Mathar mathar at mpia-hd.mpg.de
Wed Jul 3 10:25:29 CEST 2013


Quoting http://list.seqfan.eu/pipermail/seqfan/2013-July/011374.html :

rh> (=A047969 transposed)
rh> T(n,k)=(k+1)^(n+1)-k^(n+1)
rh> 
rh> Table starts
rh> ....3.....5......7.......9.......11........13........15.........17.........19
rh> ....7....19.....37......61.......91.......127.......169........217........271
rh> ...15....65....175.....369......671......1105......1695.......2465.......3439
rh> ...31...211....781....2101.....4651......9031.....15961......26281......40951
rh> ...63...665...3367...11529....31031.....70993....144495.....269297.....468559
rh> ..127..2059..14197...61741...201811....543607...1273609....2685817....5217031
rh> ..255..6305..58975..325089..1288991...4085185..11012415...26269505...56953279
rh> ..511.19171.242461.1690981..8124571..30275911..93864121..253202761..612579511
rh> .1023.58025.989527.8717049.50700551.222009073.791266575.2413042577.6513215599
rh> 
rh> Empirical for column k:
rh> k=1: a(n)=3*a(n-1)-2*a(n-2)
rh> k=2: a(n)=5*a(n-1)-6*a(n-2)
rh> k=3: a(n)=7*a(n-1)-12*a(n-2)
rh> k=4: a(n)=9*a(n-1)-20*a(n-2)
rh> k=5: a(n)=11*a(n-1)-30*a(n-2)
rh> k=6: a(n)=13*a(n-1)-42*a(n-2)
rh> k=7: a(n)=15*a(n-1)-56*a(n-2)

This is T(n,k) = (2*k+1)*T(n-1,k)-k*(k+1)*T(n-2,k) and proven easily
by a Maple program:

> T := (k+1)^(n+1)-k^(n+1) ;
                           (n + 1)    (n + 1)
               T := (k + 1)        - k

> T-(2*k+1)*subs(n=n-1,T)+k*(k+1)*subs(n=n-2,T) ;
      (n + 1)    (n + 1)                     n    n                      (n - 1)    (n - 1)
(k +1)        - k        - (2 k + 1) ((k + 1)  - k ) + k (k + 1) ((k + 1)        - k       )

> expand(%) ;
                                    2        n            n
                             n     k  (k + 1)    k (k + 1)
                     -(k + 1)  k + ----------- + ----------
                                      k + 1        k + 1

> factor(%) ;


                                                                 0


rh> Empirical for row n:
rh> n=1: a(n) = 2*n + 1
rh> n=2: a(n) = 3*n^2 + 3*n + 1
rh> n=3: a(n) = 4*n^3 + 6*n^2 + 4*n + 1
rh> n=4: a(n) = 5*n^4 + 10*n^3 + 10*n^2 + 5*n + 1
rh> n=5: a(n) = 6*n^5 + 15*n^4 + 20*n^3 + 15*n^2 + 6*n + 1
rh> n=6: a(n) = 7*n^6 + 21*n^5 + 35*n^4 + 35*n^3 + 21*n^2 + 7*n + 1
rh> n=7: a(n) = 8*n^7 + 28*n^6 + 56*n^5 + 70*n^4 + 56*n^3 + 28*n^2 + 8*n + 1

These are simply the usual recurrences with binomial coefficients for
the polynomials in k, as expected.

-- 
Richard J. Mathar



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