%I #15 May 08 2024 14:27:18

%S 1,1,1,1,5,1,1,9,28,1,1,13,109,153,1,1,17,244,1617,3751,1,1,21,433,

%T 5929,52501,175760,1,1,25,676,14625,258751,3261249,6835648,1,1,29,973,

%U 29241,810001,19763200,148756357,1051779953,1,1,33,1324,51313,1968751,73559825,1086478912,23937893793,364668913756,1

%N Square array A(m,n) = Resultant(X^m+n,(X+1)^m+n), read by (falling) antidiagonals, m >= 1, n >= 0.

%C This polynomial resultant gives the period for solutions to the equations A255852 - A255869. For example, A010034(n) = A255859(17) + A(17,9)*(n-1). In general, there may be more than one starting solutions (cf. A118119).

%e The square array starts at its upper left as follows:

%e [ 1 1 1 1 1 1 1 ... ]

%e [ 1 5 9 13 17 21 25 ... ]

%e [ 1 28 109 244 433 676 973 ... ]

%e [ 1 153 1617 5929 14625 29241 51313 ... ]

%e [ 1 3751 52501 258751 810001 1968751 4072501 ... ]

%e [ 1 175760 3261249 19763200 73559825 207499536 488999665 ... ]

%e [ : : : : : : : ·. ]

%e [ : : : : : : : ·.]

%o (PARI) A255831(m,n)=polresultant('x^m+n,('x+1)^m+n)

%o (Python)

%o from sympy import resultant

%o from sympy.abc import x

%o def A255831_T(m,n): return resultant(x**m+n,(x+1)**m+n) # _Chai Wah Wu_, May 08 2024

%K nonn,tabl

%O 0,5

%A _M. F. Hasler_, Mar 17 2015

%E Edited by _Max Alekseyev_, Aug 07 2015