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Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301.
5

%I #9 Jun 14 2017 00:22:15

%S 1,1,4,1,6,24,36,24,6,96,384,636,744,636,384,96,2976,11904,20256,

%T 26304,28536,26304,20256,11904,2976,151416,605664,1042056,1407024,

%U 1650456,1736064,1650456,1407024,1042056,605664,151416,11449296,45797184

%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) = (-6)^n.

%e Triangle begins:

%e 1;

%e 1, 4, 1;

%e 6, 24, 36, 24, 6;

%e 96, 384, 636, 744, 636, 384, 96;

%e 2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, 2976;

%e 151416, 605664, 1042056, 1407024, 1650456, 1736064, 1650456, 1407024, 1042056, 605664, 151416; ...

%e If we write the triangle like this:

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

%e ................... ....1, ....4, ....1;

%e ............ ....6, ...24, ...36, ...24, ....6;

%e ..... ...96, ..384, ..636, ..744, ..636, ..384, ...96;

%e .2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, .2976;

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

%e 2976 = 96 + 384 + 636 + 744 + 636 + 384 + 96

%e the next term is 4 times the first:

%e 11904 = 4*2976,

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

%e illustrated by:

%e 20256 = 2*11904 - 2976 - 6*96;

%e 26304 = 2*20256 - 11904 - 6*384;

%e 28536 = 2*26304 - 20256 - 6*636;

%e 26304 = 2*28536 - 26304 - 6*744;

%e 20256 = 2*26304 - 28536 - 6*636;

%e 11904 = 2*20256 - 26304 - 6*384;

%e 2976 = 2*11904 - 20256 - 6*96.

%e An alternate recurrence is illustrated by:

%e 11904 = 2976 + 3*(96 + 384 + 636 + 744 + 636 + 384 + 96);

%e 20256 = 11904 + 3*(384 + 636 + 744 + 636 + 384);

%e 26304 = 20256 + 3*(636 + 744 + 636);

%e 28536 = 26304 + 3*(744);

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

%o (PARI) T(n,k)=local(p=3);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 (PARI) /* Alternate Recurrence: */ T(n,k)=local(p=3);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)))))

%Y Cf. A126151 (column 0); diagonals: A126152, A126153; A126154; variants: A008301, A125053, A126155.

%K nonn,tabl

%O 0,3

%A _Paul D. Hanna_, Dec 19 2006