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A136555 Square array, read by antidiagonals, where T(n,k) = binomial(2^k + n-1, k).

%I A136555 #8 Mar 15 2021 01:55:51
%S A136555 1,1,1,1,2,3,1,3,6,35,1,4,10,56,1365,1,5,15,84,1820,169911,1,6,21,120,
%T A136555 2380,201376,67945521,1,7,28,165,3060,237336,74974368,89356415775,1,8,
%U A136555 36,220,3876,278256,82598880,94525795200,396861704798625,1,9,45,286,4845,324632,90858768,99949406400,409663695276000,6098989894499557055
%N A136555 Square array, read by antidiagonals, where T(n,k) = binomial(2^k + n-1, k).
%C A136555 Let vector R_{n} equal row n of this array; then R_{n+1} = P * R_{n} for n>=0, where triangle P = A132625 such that row n+1 of P = row n of P^(2^n) with appended '1' for n>=0.
%H A136555 G. C. Greubel, <a href="/A136555/b136555.txt">Antidiagonal rows n = 0..50, flattened</a>
%F A136555 G.f. for row n: Sum_{i>=0} (1 + 2^i*x)^(n-1) * log(1 + 2^i*x)^i / i!.
%F A136555 From _G. C. Greubel_, Mar 14 2021: (Start)
%F A136555 For the square array:
%F A136555     T(n, n) = A060690(n).
%F A136555   T(n+1, n) = A132683(n),   T(n+2, n) = A132684(n).
%F A136555 T(2*n+1, n) = A132685(n),   T(2*n, n) = A132686(n).
%F A136555 T(3*n+2, n) = A132689(n), T(3*n+1, n) = A132688(n), T(3*n, n) = A132687(n).
%F A136555 For the number triangle:
%F A136555 t(n, k) = T(n-k, k) = binomial(2^k + n - k -1, k).
%F A136555 Sum_{k=0..n} t(n,k) = Sum_{k=0..n} T(n-k, k) = A136557(n). (End)
%e A136555 Square array begins:
%e A136555   1, 1,  3,  35, 1365, 169911,  67945521,  89356415775, ... A136556;
%e A136555   1, 2,  6,  56, 1820, 201376,  74974368,  94525795200, ... A014070;
%e A136555   1, 3, 10,  84, 2380, 237336,  82598880,  99949406400, ... A136505;
%e A136555   1, 4, 15, 120, 3060, 278256,  90858768, 105637584000, ... A136506;
%e A136555   1, 5, 21, 165, 3876, 324632,  99795696, 111600996000, ... ;
%e A136555   1, 6, 28, 220, 4845, 376992, 109453344, 117850651776, ... ;
%e A136555   1, 7, 36, 286, 5985, 435897, 119877472, 124397910208, ... ;
%e A136555   1, 8, 45, 364, 7315, 501942, 131115985, 131254487936, ... ;
%e A136555   ...
%e A136555 Form column vector R_{n} out of row n of this array;
%e A136555 then row n+1 can be generated from row n by:
%e A136555 R_{n+1} = P * R_{n} for n>=0,
%e A136555 where triangular matrix P = A132625 begins:
%e A136555         1;
%e A136555         1,      1;
%e A136555         2,      1,     1;
%e A136555        14,      4,     1,    1;
%e A136555       336,     60,     8,    1,  1;
%e A136555     25836,   2960,   248,   16,  1, 1;
%e A136555   6251504, 454072, 24800, 1008, 32, 1, 1; ...
%e A136555 where row n+1 of P = row n of P^(2^n) with appended '1' for n>=0.
%p A136555 A136555:= (n,k) -> binomial(2^k +n-k-1, k); seq(seq(A136555(n,k), k=0..n), n=0..12); # _G. C. Greubel_, Mar 14 2021
%t A136555 Table[Binomial[2^k +n-k-1, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Mar 14 2021 *)
%o A136555 (PARI) T(n,k)=binomial(2^k+n-1,k)
%o A136555 (PARI) /* Coefficient of x^k in g.f. of row n: */ T(n,k)=polcoeff(sum(i=0,k,(1+2^i*x+x*O(x^k))^(n-1)*log((1+2^i*x)+x*O(x^k))^i/i!),k)
%o A136555 (Sage) flatten([[binomial(2^k +n-k-1, k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Mar 14 2021
%o A136555 (Magma) [Binomial(2^k +n-k-1, k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Mar 14 2021
%Y A136555 Rows: A014070, A136505, A136506, A136556.
%Y A136555 Diagonals: A060690, A132683, A132684.
%Y A136555 Cf. A136557 (antidiagonal sums).
%Y A136555 Cf. A132625.
%K A136555 nonn,tabl
%O A136555 0,5
%A A136555 _Paul D. Hanna_, Jan 07 2008