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