A142961  Wolfdieter Lang (after loss, rewritten Aug 23 2019)
The irregular triangle a(k, p) begins:


k\p      0    1      2     3      4      5      6     7      8     9 ...
------------------------------------------------------------------------
0:       1
1:       1
2:       1    3
3:       3    5
4:      -2    1     30    35 
5:     -10    5     70    63
6:       8   -2    -75    35    315    231
7:      56  -14   -245   105    693    429
8:    -272   36   2268  -525  -5880   2310  12012  6435
9:   -2448  324   9660 -2037 -16632   6006  25740  1215
10:   3968 -304 -31260  3840  73395 -14091 -90090 30030 109395 46189
...

The row polynomials are P(k, x=n) = sum(a(k, p)*n^p, p = 0..r(k)-1), with the row length sequence  r(0) = 1 = r(1) and r(k) = 2*(floor(k/2)) = A001790(k), k = 2,  3, ... .

The original k-family of convolutions of {n^k*binomial(2*n, n)}_{n>=0}  and  {binomial(2*n, n)}_{n>=0}, namely  Sigma(k, n) := sum(p^k*binomial(2*p, p)*binomial(2*(n-p), n-p), p=0..n), for k >= 0 and n >= 0, begins:


k\n  0 1    2       3        4          5           6             7              8               9               10 ...
-----------------------------------------------------------------------------------------------------------------------
0:   1 4   16      64      256       1024        4096         16384          65536          262144          1048576 
1:   0 2   16      96      512       2560       12288         57344         262144         1179648          5242880
2:   0 2   28     240     1664      10240       58368        315392        1638400         8257536         40632320    
3:   0 2   52     648     5888      44800      304128       1906688       11272192        63700992        347340800    
4:   0 2  100    1824    21776     205120     1659648      12078080       81289216       515211264       3113615360   
5:   0 2  196    5256    82592     964000     9301248      78600704      602374144      4282564608      28688384000 
6:   0 2  388   15360   318224    4606240    53024448     520470272     4542853120     36233773056     269081477120   
7:   0 2  772   45288  1239008   22259200   305846208    3488041088    34680881152    310370476032    2555413299200   
8:   0 2 1540  134304  4859216  108430720  1779097728   23580677120   267128525056   2682698425344   24490848604160
9:   0 2 3076  399816 19155872  531311200 10414355328  160465620224  2071470367744  23347797447168  236357135667200
10:  0 2 6148 1193280 75799184 2614947040 61257073728 1097514397952 16147851927040 204291592229376 2293521422597120
...

The structure of scaled Sigma(k, n) in terms of n is:
k = 0:  Sigma(0, n)/(4^n/1) = n^0, [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] =  A000012(n). 
k = 1:  Sigma(1, n)/(4^n/2) = n^1, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...] = A001477(n).
k = 2:  Sigma(2, n)/(4^n/8) = n^1*(1 + 3*n), [0, 4, 14, 30, 52, 80, 114, 154, 200, 252, 310, ...] = 2*A005449(n). 
k = 3:  Sigma(3, n)/(4^n/16) = n^2*(3 + 5*n),  [0, 8, 52, 162, 368, 700, 1188, 1862, 2752, 3888, 5300, ...] = 2*A142962(n), for n >= 1, and 0 for n = 0.
k = 4:  Sigma(4, n)/(4^n/128) = n^1*(-2 + n + 30*n^2 + 35*n^3), [0, 64, 800, 3648, 10888, 25640, 51864, 94360, 158768, 251568, 380080, ...].
k = 5:  Sigma(5, n)/(4^n/256) = n^2*(-10 + 5*n + 70*n^2 +  63*n^3), [0, 128, 3136, 21024, 82592, 241000, 581328, 1228136, 2353024, 4182192, 7004000]. 
k = 6:  Sigma(6, n)/(4^n/1024) = n^1*(8 - 2*n - 75*n^2 + 35*n^3 + 315*n^4 + 231*n^5), [0, 512, 24832, 245760, 1272896, 4606240, 13256112, 32529392, 70982080, 141538176, 262774880]. 
...

Here the numbers A046161(k) = denominator(binomial(2*k, k)/4^k), for k >= 0  enter. This is the sequence [1, 2, 8, 16, 128, 256, 1024, 2048, 32768, 65536, 262144, ... ].

(Sigma(k, n)/(4^n/A046161(k))) equals c(k, n)*P(k, n) with the row polynomials P(k, x=n) = sum(a(k, p)*n^p, p = 0..r(k)-1), with r and the irregular triangle a(k, p), where c(0, n) = 1, c(1, n) = n, and c(k, n) = n for even k >= 2 and c(k, n) = n^2 for odd k >= 3.

The row sums of the irregular triangle a(k, p) are:
signed: [1, 1, 4, 8, 64, 128, 512, 1024, 16384, 32768, 131072, ...],
unsigned: [1, 1, 4, 8, 68, 148, 666, 1542, 29738, 75002, 402562, ...].

-----------------------------------------------------------------------------------------

In terms of the Stirling2 triangle S2(n, k) = A048993(n, k) one finds (from the o.g.f.s)

Sigma(k, n) = 4^n*sum(binomial(n, p)*(2*p -1)!!*S2(k, p)/2^p, p=0..min(n,k)).

The double factorials are givem by (2*p -1)!!= A001147(p), with (-1)!!:=1.

--------------------------------------------------  eof ----------------------------------------------------------------