Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #8 Mar 15 2021 01:55:51
%S 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 2380,201376,67945521,1,7,28,165,3060,237336,74974368,89356415775,1,8,
%U 36,220,3876,278256,82598880,94525795200,396861704798625,1,9,45,286,4845,324632,90858768,99949406400,409663695276000,6098989894499557055
%N Square array, read by antidiagonals, where T(n,k) = binomial(2^k + n-1, k).
%C 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 G. C. Greubel, <a href="/A136555/b136555.txt">Antidiagonal rows n = 0..50, flattened</a>
%F G.f. for row n: Sum_{i>=0} (1 + 2^i*x)^(n-1) * log(1 + 2^i*x)^i / i!.
%F From _G. C. Greubel_, Mar 14 2021: (Start)
%F For the square array:
%F T(n, n) = A060690(n).
%F T(n+1, n) = A132683(n), T(n+2, n) = A132684(n).
%F T(2*n+1, n) = A132685(n), T(2*n, n) = A132686(n).
%F T(3*n+2, n) = A132689(n), T(3*n+1, n) = A132688(n), T(3*n, n) = A132687(n).
%F For the number triangle:
%F t(n, k) = T(n-k, k) = binomial(2^k + n - k -1, k).
%F Sum_{k=0..n} t(n,k) = Sum_{k=0..n} T(n-k, k) = A136557(n). (End)
%e Square array begins:
%e 1, 1, 3, 35, 1365, 169911, 67945521, 89356415775, ... A136556;
%e 1, 2, 6, 56, 1820, 201376, 74974368, 94525795200, ... A014070;
%e 1, 3, 10, 84, 2380, 237336, 82598880, 99949406400, ... A136505;
%e 1, 4, 15, 120, 3060, 278256, 90858768, 105637584000, ... A136506;
%e 1, 5, 21, 165, 3876, 324632, 99795696, 111600996000, ... ;
%e 1, 6, 28, 220, 4845, 376992, 109453344, 117850651776, ... ;
%e 1, 7, 36, 286, 5985, 435897, 119877472, 124397910208, ... ;
%e 1, 8, 45, 364, 7315, 501942, 131115985, 131254487936, ... ;
%e ...
%e Form column vector R_{n} out of row n of this array;
%e then row n+1 can be generated from row n by:
%e R_{n+1} = P * R_{n} for n>=0,
%e where triangular matrix P = A132625 begins:
%e 1;
%e 1, 1;
%e 2, 1, 1;
%e 14, 4, 1, 1;
%e 336, 60, 8, 1, 1;
%e 25836, 2960, 248, 16, 1, 1;
%e 6251504, 454072, 24800, 1008, 32, 1, 1; ...
%e where row n+1 of P = row n of P^(2^n) with appended '1' for n>=0.
%p 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 Table[Binomial[2^k +n-k-1, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Mar 14 2021 *)
%o (PARI) T(n,k)=binomial(2^k+n-1,k)
%o (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 (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 (Magma) [Binomial(2^k +n-k-1, k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Mar 14 2021
%Y Rows: A014070, A136505, A136506, A136556.
%Y Diagonals: A060690, A132683, A132684.
%Y Cf. A136557 (antidiagonal sums).
%Y Cf. A132625.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Jan 07 2008