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A185958 Accumulation array of the array max{n,k}, by antidiagonals. 2

%I #16 Sep 04 2020 15:40:10

%S 1,3,3,6,7,6,10,13,13,10,15,21,22,21,15,21,31,34,34,31,21,28,43,49,50,

%T 49,43,28,36,57,67,70,70,67,57,36,45,73,88,94,95,94,88,73,45,55,91,

%U 112,122,125,125,122,112,91,55,66,111,139,154,160,161,160,154,139,111,66,78,133,169,190,200,203,203,200,190,169,133,78,91,157,202,230,245,251,252,251,245,230,202,157,91,105,183,238,274,295,305,308,308,305,295,274,238,183,105

%N Accumulation array of the array max{n,k}, by antidiagonals.

%C A member of the accumulation chain

%C ... < A185917 < A051125 < A185958 < ...,

%C where A051125, written as a rectangular array M, is given by M(n,k)=max{n,k}. See A144112 for the definition of accumulation array.

%C row 1: A000217

%C row 2: A002061

%C diag (1,7,...): A002412

%C diag (3,13,..): A016061

%C antidiagonal sums: A070893

%F From _Yu-Sheng Chang_, Jun 05 2020: (Start)

%F O.g.f.: F(z,v) = -(v^2*z^3+v*z^3-3*v*z^2+1)/((v*z^2-v*z-z+1)^2*(v*z^2-1)*(z-1)*(v*z-1)).

%F T(n,k) = [v^k] 1/2*n^2*(v^(n+2)+1)/(1-v)^2+1/2*n*(3*v^(n+3)-7*v^(n+2)+7*v-3)/(-1+v)^3-1/2*v*((1-v^(1/2))^4*(-1)^n+(1+v^(1/2))^4)*v^(1/2*n)/(1-v)^4+(6*v^2+6*v^(n+2)+v^(n+4)-3*v^(n+3)-3*v+1)/(1-v)^4.

%F (End)

%e Northwest corner:

%e 1....3....6....10....15

%e 3....7....13...21....31

%e 6....13...22...34....49

%e 10...21...34...50....70

%p A := proc(n,k) option remember; ## n >= 0 and k = 0 .. n

%p if k < 0 or k > n then

%p 0

%p elif n = 0 then

%p 1

%p else

%p A(n-1,k) + A(n-1,k-1) - A(n-2,k-1) + max(n-k+1,k+1)

%p end if

%p end proc: # _Yu-Sheng Chang_, Jun 05 2020

%Y Cf. A144112, A051125.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Feb 07 2011

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Last modified April 19 15:11 EDT 2024. Contains 371794 sequences. (Running on oeis4.)