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

[Ron Hardin]

(k+1)^(n+1)-k^(n+1) is the number of _first_ differences of length n+1 arrays of numbers in 0..k (already in OEIS transposed as A047969)

Going to second differences, take a 0..k array of length n+2
Form an array of second differences
How many of the second difference arrays are there?

The column recurrences come out highly systematic (x-k)*(x-k-1) for n>2k+1

Is this enough to suggest a formula?


/tmp/din
T(n,k)=Number of second differences of arrays of length n+2 of numbers in 0..k

Table starts
....5......9......13.......17........21.........25........29........33.......37
...15.....49.....103......177.......271........385.......519.......673......847
...31....199.....625.....1429......2731.......4651......7309.....10825....15319
...63....665....3151.....9705.....23351......47953.....88215....149681...238735
..127...2059...14053....58141....176851.....439927....951049...1854553..3342151
..255...6305...58975...320481...1225631....3693505...9399615..21108545.43067071
..511..19171..242461..1688101...8006491...29066311..86929081.224817481.........
.1023..58025..989527..8717049..50556551..219071473.766106895...................
.2047.175099.4017157.44633821.313882531.1609259287.............................

Empirical for column k:
k=1: a(n)=3*a(n-1)-2*a(n-2) for n>3
k=2: a(n)=5*a(n-1)-6*a(n-2) for n>5
k=3: a(n)=7*a(n-1)-12*a(n-2) for n>7
k=4: a(n)=9*a(n-1)-20*a(n-2) for n>9
k=5: a(n)=11*a(n-1)-30*a(n-2) for n>11
k=6: a(n)=13*a(n-1)-42*a(n-2) for n>13
k=7: a(n)=15*a(n-1)-56*a(n-2) for n>15
Empirical for row n:
n=1: a(n) = 4*n + 1
n=2: a(n) = 10*n^2 + 4*n + 1
n=3: a(n) = 20*n^3 + 9*n^2 + 1*n + 1
n=4: a(n) = 35*n^4 + 14*n^3 - 17*n^2 + 30*n + 1
n=5: a(n) = 56*n^5 + 14*n^4 - 108*n^3 + 289*n^2 - 125*n + 1
n=6: a(n) = 84*n^6 - 402*n^4 + 1656*n^3 - 1860*n^2 + 776*n + 1
n=7: a(n) = 120*n^7 - 42*n^6 - 1158*n^5 + 6945*n^4 - 13980*n^3 + 13512*n^2 - 4887*n + 1


 
rhhardin at mindspring.com
rhhardin at att.net (either)