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A126155 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. 6
1, 1, 5, 1, 7, 35, 55, 35, 7, 139, 695, 1195, 1415, 1195, 695, 139, 5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, 5473, 357721, 1788605, 3175705, 4343885, 5126905, 5403005, 5126905, 4343885, 3175705, 1788605, 357721 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,3
LINKS
FORMULA
Sum_{k=0,2n} (-1)^k*C(2n,k)*T(n,k) = (-8)^n.
EXAMPLE
The triangle begins:
1;
1, 5, 1;
7, 35, 55, 35, 7;
139, 695, 1195, 1415, 1195, 695, 139;
5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, 5473;
357721, 1788605, 3175705, 4343885, 5126905, 5403005, 5126905, 4343885, 3175705, 1788605, 357721; ...
If we write the triangle like this:
.......................... ....1;
................... ....1, ....5, ....1;
............ ....7, ...35, ...55, ...35, ....7;
..... ..139, ..695, .1195, .1415, .1195, ..695, ..139;
.5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, .5473;
then the first term in each row is the sum of the previous row:
5473 = 139 + 695 + 1195 + 1415 + 1195 + 695 + 139
the next term is 5 times the first:
27365 = 5*5473,
and the remaining terms in each row are obtained by the rule
illustrated by:
48145 = 2*27365 - 5473 - 8*139;
63365 = 2*48145 - 27365 - 8*695;
69025 = 2*63365 - 48145 - 8*1195;
63365 = 2*69025 - 63365 - 8*1415;
48145 = 2*63365 - 69025 - 8*1195;
27365 = 2*48145 - 63365 - 8*695;
5473 = 2*27365 - 48145 - 8*139.
An alternate recurrence is illustrated by:
27365 = 5473 + 4*(139 + 695 + 1195 + 1415 + 1195 + 695 + 139);
48145 = 27365 + 4*(695 + 1195 + 1415 + 1195 + 695);
63365 = 48145 + 4*(1195 + 1415 + 1195);
69025 = 63365 + 4*(1415);
and then for k>n, T(n,k) = T(n,2n-k).
MAPLE
T := proc(n, k) option remember; local j;
if n = 1 then 1
elif k = 1 then add(T(n-1, j), j=1..2*n-3)
elif k = 2 then 5*T(n, 1)
elif k > n then T(n, 2*n-k)
else 2*T(n, k-1)-T(n, k-2)-8*T(n-1, k-2)
fi end:
seq(print(seq(T(n, k), k=1..2*n-1)), n=1..6); # Peter Luschny, May 12 2014
MATHEMATICA
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]];
Table[T[n, k], {n, 1, 6}, {k, 1, 2n-1}] (* Jean-François Alcover, Jun 15 2019, from Maple *)
PROG
(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))))))}
for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print(""))
(PARI) /* Alternate Recurrence: */
{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)))))}
for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print(""))
(SageMath)
from functools import cache
@cache
def R(n, k):
return (1 if n == 1 else sum(R(n-1, j) for j in range(1, 2*n-2))
if k == 1 else 5*R(n, 1) if k == 2 else R(n, 2*n-k)
if k > n else 2*R(n, k-1) - R(n, k-2) - 8*R(n-1, k-2))
def A126155(n, k): return R(n+1, k+1)
for n in range(5): print([A126155(n, k) for k in range(2*n+1)])
# Peter Luschny, Dec 14 2023
CROSSREFS
Cf. A126156 (column 0); diagonals: A126157, A126158; A126159; variants: A008301 (p=1), A125053 (p=2), A126150 (p=3).
Sequence in context: A211849 A363419 A222182 * A021197 A286872 A363514
KEYWORD
nonn,tabl
AUTHOR
Paul D. Hanna, Dec 20 2006
STATUS
approved

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Last modified March 29 07:27 EDT 2024. Contains 371265 sequences. (Running on oeis4.)