A126155 - OEIS

%I #20 Dec 14 2023 10:10:44

%S 1,1,5,1,7,35,55,35,7,139,695,1195,1415,1195,695,139,5473,27365,48145,

%T 63365,69025,63365,48145,27365,5473,357721,1788605,3175705,4343885,

%U 5126905,5403005,5126905,4343885,3175705,1788605,357721

%N Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301.

%F Sum_{k=0,2n} (-1)^k*C(2n,k)*T(n,k) = (-8)^n.

%e The triangle begins:

%e 1;

%e 1, 5, 1;

%e 7, 35, 55, 35, 7;

%e 139, 695, 1195, 1415, 1195, 695, 139;

%e 5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, 5473;

%e 357721, 1788605, 3175705, 4343885, 5126905, 5403005, 5126905, 4343885, 3175705, 1788605, 357721; ...

%e If we write the triangle like this:

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

%e ................... ....1, ....5, ....1;

%e ............ ....7, ...35, ...55, ...35, ....7;

%e ..... ..139, ..695, .1195, .1415, .1195, ..695, ..139;

%e .5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, .5473;

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

%e 5473 = 139 + 695 + 1195 + 1415 + 1195 + 695 + 139

%e the next term is 5 times the first:

%e 27365 = 5*5473,

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

%e illustrated by:

%e 48145 = 2*27365 - 5473 - 8*139;

%e 63365 = 2*48145 - 27365 - 8*695;

%e 69025 = 2*63365 - 48145 - 8*1195;

%e 63365 = 2*69025 - 63365 - 8*1415;

%e 48145 = 2*63365 - 69025 - 8*1195;

%e 27365 = 2*48145 - 63365 - 8*695;

%e 5473 = 2*27365 - 48145 - 8*139.

%e An alternate recurrence is illustrated by:

%e 27365 = 5473 + 4*(139 + 695 + 1195 + 1415 + 1195 + 695 + 139);

%e 48145 = 27365 + 4*(695 + 1195 + 1415 + 1195 + 695);

%e 63365 = 48145 + 4*(1195 + 1415 + 1195);

%e 69025 = 63365 + 4*(1415);

%e and then for k>n, T(n,k) = T(n,2n-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 5*T(n, 1)

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

%p else 2*T(n, k-1)-T(n, k-2)-8*T(n-1, k-2)

%p   fi end:

%p seq(print(seq(T(n,k), k=1..2*n-1)), n=1..6); # _Peter Luschny_, May 12 2014

%t T[n_, k_] := T[n, k] = Which[n==1, 1, k==1, Sum[T[n-1, j], {j, 1, 2n-3}], k==2, 5 T[n, 1], k>n, T[n, 2n-k], True, 2 T[n, k-1] - T[n, k-2] - 8 T[n-1, k-2]];

%t Table[T[n, k], {n, 1, 6}, {k, 1, 2n-1}] (* _Jean-François Alcover_, Jun 15 2019, from Maple *)

%o (PARI) {T(n,k) = local(p=4);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==1,(p+1)*T(n,0),if(k<=n,2*T(n,k-1)-T(n,k-2)-2*p*T(n-1,k-2),T(n,2*n-k))))))}

%o for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print(""))

%o (PARI) /* Alternate Recurrence: */

%o {T(n,k) = local(p=4);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)+p*sum(j=k-1,2*n-1-k,T(n-1,j)),T(n,2*n-k)))))}

%o for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print(""))

%o (SageMath)

%o from functools import cache

%o @cache

%o def R(n, k):

%o       return (1 if n == 1 else sum(R(n-1, j) for j in range(1, 2*n-2))

%o                 if k == 1 else 5*R(n, 1) if k == 2 else R(n, 2*n-k)

%o                 if k > n else 2*R(n, k-1) - R(n, k-2) - 8*R(n-1, k-2))

%o def A126155(n, k): return R(n+1, k+1)

%o for n in range(5): print([A126155(n, k) for k in range(2*n+1)])

%o # _Peter Luschny_, Dec 14 2023

%Y Cf. A126156 (column 0); diagonals: A126157, A126158; A126159; variants: A008301 (p=1), A125053 (p=2), A126150 (p=3).

%K nonn,tabl

%O 0,3

%A _Paul D. Hanna_, Dec 20 2006