|
%I #17 Nov 09 2019 01:11:38
%S 2,2,3,2,6,4,2,9,12,5,2,12,24,20,6,2,15,40,50,30,7,2,18,60,100,90,42,
%T 8,2,21,84,175,210,147,56,9,2,24,112,280,420,392,224,72,10,2,27,144,
%U 420,756,882,672,324,90,11,2,30,180,600,1260,1764,1680,1080,450,110,12,2,33
%N Triangle read by rows: T(n,k) = (k+2)*binomial(n,k) (0 <= k <= n).
%C k*binomial(n-4, k-2) counts the permutations in S_n which have zero occurrences of the pattern 213 and one occurrence of the pattern 132 and k descents.
%C Sum of row n =(n+4)*2^(n-1) (A045623). - _Emeric Deutsch_, Apr 02 2007
%C Essentially the same as A127954: obtained by dropping the first row of A127954. - _Peter Bala_, Mar 05 2013
%D D. Hök, Parvisa mönster i permutationer [Swedish], (2007).
%F G.f.: (2 - t*(2+x))/(1 - t*(1+x))^2 = 2 + (2+3*x)*t + (2+6*x+4*x^2)*t^2 + .... - _Peter Bala_, Mar 05 2013
%F Row n is the vector of polynomial coefficients of (2 + (n+2)*x)*(1+x)^(n-1). - _Peter Bala_, Mar 05 2013
%e Triangle starts:
%e 2;
%e 2, 3;
%e 2, 6, 4;
%e 2, 9, 12, 5;
%e 2, 12, 24, 20, 6;
%p T:=(n,k)->(k+2)*binomial(n,k): for n from 0 to 11 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form. - _Emeric Deutsch_, Apr 02 2007
%Y Cf. A045623, A127954.
%K nonn,tabl
%O 0,1
%A David Hoek (david.hok(AT)telia.com), Mar 23 2007
%E Edited by _Emeric Deutsch_, Apr 02 2007
|
