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Variant of triangle A008301, read by rows of 2*n+1 terms, such that the first column is the secant numbers (A000364).
8

%I #34 Oct 14 2022 06:59:27

%S 1,1,3,1,5,15,21,15,5,61,183,285,327,285,183,61,1385,4155,6681,8475,

%T 9129,8475,6681,4155,1385,50521,151563,247065,325947,378105,396363,

%U 378105,325947,247065,151563,50521,2702765,8108295,13311741,17908935

%N Variant of triangle A008301, read by rows of 2*n+1 terms, such that the first column is the secant numbers (A000364).

%C Foata and Han refer to this as the triangle of Poupard numbers h_n(k). - _N. J. A. Sloane_, Feb 17 2014

%C Central terms (A125054) equal the binomial transform of the tangent numbers (A000182).

%H Reinhard Zumkeller, <a href="/A125053/b125053.txt">Rows n = 0..100 of triangle, flattened</a>

%H Dominique Foata and Guo-Niu Han, <a href="http://www-irma.u-strasbg.fr/~foata/paper/pub123Seidel.pdf">Seidel Triangle Sequences and Bi-Entringer Numbers</a>, Nov 20 2013;

%H Foata, Dominique; Han, Guo-Niu; Strehl, Volker <a href="https://doi.org/10.1016/j.laa.2016.09.016">The Entringer-Poupard matrix sequence</a>. Linear Algebra Appl. 512, 71-96 (2017).

%F Sum_{k=0..2n} C(2n,k)*T(n,k) = 4^n * A000182(n), where A000182 are the tangent numbers.

%F Sum_{k=0..2n} (-1)^n*C(2n,k)*T(n,k) = (-4)^n.

%e If we write the triangle like this:

%e ......................... ...1;

%e ................... ...1, ...3, ...1;

%e ............. ...5, ..15, ..21, ..15, ...5;

%e ....... ..61, .183, .285, .327, .285, .183, ..61;

%e . 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385;

%e then the first nonzero term is the sum of the previous row:

%e 1385 = 61 + 183 + 285 + 327 + 285 + 183 + 61,

%e the next term is 3 times the first:

%e 4155 = 3*1385,

%e and the remaining terms in each row are obtained by the rule illustrated by:

%e 6681 = 2*4155 - 1385 - 4*61;

%e 8475 = 2*6681 - 4155 - 4*183;

%e 9129 = 2*8475 - 6681 - 4*285;

%e 8475 = 2*9129 - 8475 - 4*327;

%e 6681 = 2*8475 - 9129 - 4*285;

%e 4155 = 2*6681 - 8475 - 4*183;

%e 1385 = 2*4155 - 6681 - 4*61.

%e An alternate recurrence is illustrated by:

%e 4155 = 1385 + 2*(61 + 183 + 285 + 327 + 285 + 183 + 61);

%e 6681 = 4155 + 2*(183 + 285 + 327 + 285 + 183);

%e 8475 = 6681 + 2*(285 + 327 + 285);

%e 9129 = 8475 + 2*(327);

%e and then for k>n, T(n,k) = T(n,2*n-k).

%p T := proc(n, k) option remember; local j;

%p if n = 1 then 1

%p elif k = 1 then add(T(n-1, j), j=1..2*n-3)

%p elif k = 2 then 3*T(n, 1)

%p elif k > n then T(n, 2*n-k)

%p else 2*T(n, k-1) - T(n, k-2) - 4*T(n-1, k-2)

%p fi end:

%p seq(print(seq(T(n,k), k=1..2*n-1)), n=1..5); # _Peter Luschny_, May 11 2014

%t t[n_, k_] := t[n, k] = If[2*n < k || k < 0, 0, If[n == 0 && k == 0, 1, If[k == 0, Sum[t[n-1, j], {j, 0, 2*n-2}], If[k <= n, t[n, k-1] + 2*Sum[t[n-1, j], {j, k-1, 2*n-1-k}], t[n, 2*n-k]]]]]; Table[t[n, k], {n, 0, 6}, {k, 0, 2*n}] // Flatten (* _Jean-François Alcover_, Dec 06 2012, translated from Pari *)

%o (PARI) T(n,k)=if(2*n<k || k<0,0,if(n==0 && k==0,1,if(k==0,sum(j=0,2*n-2,T(n-1,j)), if(k<=n,T(n,k-1)+2*sum(j=k-1,2*n-1-k,T(n-1,j)),T(n,2*n-k)))))

%o (Haskell)

%o a125053 n k = a125053_tabf !! n !! k

%o a125053_row n = a125053_tabf !! n

%o a125053_tabf = iterate f [1] where

%o f zs = zs' ++ reverse (init zs') where

%o zs' = (sum zs) : g (map (* 2) zs) (sum zs)

%o g [x] y = [x + y]

%o g xs y = y' : g (tail $ init xs) y' where y' = sum xs + y

%o -- _Reinhard Zumkeller_, Mar 17 2012

%Y Cf. A008301, A000364 (secant numbers, which are the row sums), A125054 (central terms), A125055, A000182, A008282.

%Y Cf. A210111 (left half).

%K nonn,tabf,nice

%O 0,3

%A _Paul D. Hanna_, Nov 21 2006, Dec 20 2006