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T(n, n-k, 2) = T(n, k, 2).

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Triangle read by rows: A(n,0)=A(n,n)=1 and AT(n, k, m) = (2m*(n-k) + 1)*AT(n-1, k-1, m) + (2m*k + 1)*AT(n-1, k, m) + 2m*k*(n-k)*AT(n-2, k-1, m) with T(n, 0, m) = T(n, n, m) for = 1< and m =k<n 2, read by rows.

1, 1, 1, 1, 8, 1, 1, 33, 33, 1, 1, 112, 394, 112, 1, 1, 353, 3150, 3150, 353, 1, 1, 1080, 20719, 51192, 20719, 1080, 1, 1, 3265, 122535, 620415, 620415, 122535, 3265, 1, 1, 9824, 681040, 6312360, 12805614, 6312360, 681040, 9824, 1, 1, 29505, 3643980, 57451300, 209503086, 209503086, 57451300, 3643980, 29505, 1

The row sums are 1, 2, 10, 68, 620, 7008, 94792, 1492432, 26812064, 541255744, 12129218272,....

G. C. Greubel, <a href="/A157148/b157148.txt">Rows n = 0..50 of the triangle, flattened</a>

T(n, k, m) = (m*(n-k) + 1)*T(n-1, k-1, m) + (m*k + 1)*T(n-1, k, m) + m*k*(n-k)*T(n-2, k-1, m) with T(n, 0, m) = T(n, n, m) = 1 and m = 2.

T(n, n-k) = T(n, k).

Triangle begins as:

1;

1, 1;

1, 8, 1;

1, 33, 33, 1;

1, 112, 394, 112, 1;

1, 353, 3150, 3150, 353, 1;

1, 1080, 20719, 51192, 20719, 1080, 1;

1, 3265, 122535, 620415, 620415, 122535, 3265, 1;

1, 9824, 681040, 6312360, 12805614, 6312360, 681040, 9824, 1;

1, 29505, 3643980, 57451300, 209503086, 209503086, 57451300, 3643980, 29505, 1;

if k < 0 or k> n then 0;

elif k = 0 or k = n then 1;

elif k = 0 or k = n then

1;

else (2*(n-k)+1)*procname(n-1, k-1) + (2*k+1)*procname(n-1, k) + 2*k*(n-k)*procname(n-2, k-1);

(2*(n-k)+1)*procname(n-1, k-1)

+(2*k+1)*procname(n-1, k)

+2*k*(n-k)*procname(n-2, k-1) ;

ClearT[n_, k_, m_]:= T[n, k, m]= If[k==0 || k==n, 1, (m*(n-k)+1)*T[n-1, k-1, m] + (m*k+1)*T[n-1, k, m] + m*k*(n-k)*T[A, n, -2, k, -1, m]];

A[n_, 0, m_] := 1;

ATable[T[n_, , k, 2], {n, 0, 10}, {k, 0, n_, m_}] := 1; //Flatten (* modified by _G. C. Greubel_, Jan 09 2022 *)

A[n_, k_, m_] := (m*(n - k) + 1)*A[n - 1, k - 1, m] + (m* k + 1)*A[n - 1, k, m] + m*k*(n - k)*A[n - 2, k - 1, m];

Table[A[n, k, m], {m, 0, 10}, {n, 0, 10}, {k, 0, n}];

Table[Flatten[Table[Table[A[n, k, m], {k, 0, n}], {n, 0, 10}]], {m, 0, 10}]

(Sage)

@CachedFunction

def T(n, k, m): # A157148

if (k==0 or k==n): return 1

else: return (m*(n-k) +1)*T(n-1, k-1, m) + (m*k+1)*T(n-1, k, m) + m*k*(n-k)*T(n-2, k-1, m)

flatten([[T(n, k, 2) for k in (0..n)] for n in (0..20)]) # G. C. Greubel, Jan 09 2022

Cf. A007318 (m=0), A157147 (m=1), this sequence (m=2), A157149 (m=3), A157150 (m=4), A157151 (m=5).

Cf. A157152, A157153, A157154, A157155, A157156, A157207, A157208, A157209, A157210, A157211, A157212, A157268, A157272, A157273, A157274, A157275.

Edited by G. C. Greubel, Jan 09 2022

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Triangle read by rows: A(n,0)=A general three part recursion triangle sequence: m(n,n)=2; 1 and A(n,k,m)= (m2*(n - k) + 1)*A(n - 1, k - 1, m) + (m2*k + 1)*A(n - 1, k, m) + m2*k*(n - k)*A(n - 2, k - 1, m) for 1<=k<n.

The row sums are: 1, 2, 10, 68, 620, 7008, 94792, 1492432, 26812064, 541255744, 12129218272,....

{1, 2, 10, 68, 620, 7008, 94792, 1492432, 26812064, 541255744, 12129218272,...}.

What I have done here is add a new symmetrical part

to the "zero start" Sierpinski -Pascal recursion at "down two" or n-2 in my notation:

m*k*(n - k)*A(n - 2, k - 1, m).

It uses the symmetrical k*(n-k) multiplier.

m=2;

A(n,k,m)= (m*(n - k) + 1)*A(n - 1, k - 1, m) +

(m*k + 1)*A(n - 1, k, m) +

m*k*(n - k)*A(n - 2, k - 1, m).

{1},

1;

{1, 1},;

{1, 8, 1},;

{1, 33, 33, 1},;

{1, 112, 394, 112, 1},;

{1, 353, 3150, 3150, 353, 1},;

{1, 1080, 20719, 51192, 20719, 1080, 1},;

{1, 3265, 122535, 620415, 620415, 122535, 3265, 1},;

{1, 9824, 681040, 6312360, 12805614, 6312360, 681040, 9824, 1},;

{1, 29505, 3643980, 57451300, 209503086, 209503086, 57451300, 3643980, 29505, 1},;

{1, 88552, 19035853, 485422480, 2935387954, 5249348592, 2935387954, 485422480, 19035853, 88552, 1}

A157148 := proc(n, k)

option remember;

if k < 0 or k> n then

0;

elif k = 0 or k = n then

1;

else

(2*(n-k)+1)*procname(n-1, k-1)

+(2*k+1)*procname(n-1, k)

+2*k*(n-k)*procname(n-2, k-1) ;

end if;

end proc:

seq(seq(A157148(n, k), k=0..n), n=0..10) ; # R. J. Mathar, Feb 06 2015

nonn,tabl,unedeasy

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