This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A126155 #26 Feb 03 2025 13:45:30 %S A126155 1,1,5,1,7,35,55,35,7,139,695,1195,1415,1195,695,139,5473,27365,48145, %T A126155 63365,69025,63365,48145,27365,5473,357721,1788605,3175705,4343885, %U A126155 5126905,5403005,5126905,4343885,3175705,1788605,357721 %N A126155 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 5 times first term. %F A126155 Sum_{k=0..2n} (-1)^k*C(2n,k)*T(n,k) = (-8)^n. %e A126155 The triangle begins: %e A126155 1; %e A126155 1, 5, 1; %e A126155 7, 35, 55, 35, 7; %e A126155 139, 695, 1195, 1415, 1195, 695, 139; %e A126155 5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, 5473; %e A126155 357721, 1788605, 3175705, 4343885, 5126905, 5403005, 5126905, 4343885, 3175705, 1788605, 357721; ... %e A126155 If we write the triangle like this: %e A126155 .......................... ....1; %e A126155 ................... ....1, ....5, ....1; %e A126155 ............ ....7, ...35, ...55, ...35, ....7; %e A126155 ..... ..139, ..695, .1195, .1415, .1195, ..695, ..139; %e A126155 .5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, .5473; %e A126155 then the first term in each row is the sum of the previous row: %e A126155 5473 = 139 + 695 + 1195 + 1415 + 1195 + 695 + 139 %e A126155 the next term is 5 times the first: %e A126155 27365 = 5*5473, %e A126155 and the remaining terms in each row are obtained by the rule illustrated by: %e A126155 48145 = 2*27365 - 5473 - 8*139; %e A126155 63365 = 2*48145 - 27365 - 8*695; %e A126155 69025 = 2*63365 - 48145 - 8*1195; %e A126155 63365 = 2*69025 - 63365 - 8*1415; %e A126155 48145 = 2*63365 - 69025 - 8*1195; %e A126155 27365 = 2*48145 - 63365 - 8*695; %e A126155 5473 = 2*27365 - 48145 - 8*139. %e A126155 An alternate recurrence is illustrated by: %e A126155 27365 = 5473 + 4*(139 + 695 + 1195 + 1415 + 1195 + 695 + 139); %e A126155 48145 = 27365 + 4*(695 + 1195 + 1415 + 1195 + 695); %e A126155 63365 = 48145 + 4*(1195 + 1415 + 1195); %e A126155 69025 = 63365 + 4*(1415); %e A126155 and then for k>n, T(n,k) = T(n,2n-k). %p A126155 T := proc(n,k) option remember; local j; %p A126155 if n = 1 then 1 %p A126155 elif k = 1 then add(T(n-1, j), j=1..2*n-3) %p A126155 elif k = 2 then 5*T(n, 1) %p A126155 elif k > n then T(n, 2*n-k) %p A126155 else 2*T(n, k-1)-T(n, k-2)-8*T(n-1, k-2) %p A126155 fi end: %p A126155 seq(print(seq(T(n,k), k=1..2*n-1)), n=1..6); # _Peter Luschny_, May 12 2014 %t A126155 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 A126155 Table[T[n, k], {n, 1, 6}, {k, 1, 2n-1}] (* _Jean-François Alcover_, Jun 15 2019, from Maple *) %o A126155 (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 A126155 for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print("")) %o A126155 (PARI) /* Alternate Recurrence: */ %o A126155 {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 A126155 for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print("")) %o A126155 (SageMath) %o A126155 from functools import cache %o A126155 @cache %o A126155 def R(n, k): %o A126155 return (1 if n == 1 else sum(R(n-1, j) for j in range(1, 2*n-2)) %o A126155 if k == 1 else 5*R(n, 1) if k == 2 else R(n, 2*n-k) %o A126155 if k > n else 2*R(n, k-1) - R(n, k-2) - 8*R(n-1, k-2)) %o A126155 def A126155(n, k): return R(n+1, k+1) %o A126155 for n in range(5): print([A126155(n, k) for k in range(2*n+1)]) %o A126155 # _Peter Luschny_, Dec 14 2023 %Y A126155 Cf. A126156 (column 0); diagonals: A126157, A126158; A126159; variants: A008301 (p=1), A125053 (p=2), A126150 (p=3). %K A126155 nonn,tabl %O A126155 0,3 %A A126155 _Paul D. Hanna_, Dec 20 2006