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A131024 Row sums of triangular array T: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j-1,k) for 2 <= k <= j.

%I A131024 #11 Jan 29 2025 08:03:54
%S A131024 1,3,7,11,16,22,36,73,175,431,1024,2290,4824,9649,18571,34955,65536,
%T A131024 124510,242460,484921,989527,2038103,4194304,8565754,17308656,
%U A131024 34617313,68703187,135812051,268435456,532087942,1059392916,2118785833,4251920575,8546887871
%N A131024 Row sums of triangular array T: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j-1,k) for 2 <= k <= j.
%C A131024 Sum of n-th row equals (n+1)-th term of main diagonal minus (n+1)-th term of first column. A088911 has offset 0, so a(n) = A129339(n+1) - A088911(n).
%H A131024 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (6,-14,14,0,-14,15,-6).
%F A131024 G.f.: x*(1-3*x+3*x^2-3*x^3+6*x^4-4*x^5+x^6)/((1-x)*(1+x)*(1-2*x)*(1-x+x^2)*(1-3*x+3*x^2)).
%e A131024 For first seven rows of T see A131022 or A129339.
%o A131024 (PARI) lista(m) = my(M=matrix(m, m)); for(j=1, m, M[j, 1]=if((j-1)%6<3, 1, 0)); for(k=2, m, for(j=k, m, M[j, k]=M[j-1, k-1]+M[j, k-1])); for(j=1, m, print1(sum(k=1, j, M[j, k]), ", "))
%o A131024 (Magma) m:=32; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do if (j-1) mod 6 lt 3 then M[j, 1]:=1; end if; end for; for k:=2 to m do for j:=k to m do M[j, k]:=M[j-1, k-1]+M[j, k-1]; end for; end for; [ &+[ M[j, k]: k in [1..j] ]: j in [1..m] ];
%Y A131024 Cf. A131022 (T read by rows), A129339 (main diagonal of T), A131023 (first subdiagonal of T), A131025 (antidiagonal sums of T). First through sixth column of T are in A088911, A131026, A131027, A131028, A131029, A131030 resp.
%K A131024 nonn,easy
%O A131024 1,2
%A A131024 _Klaus Brockhaus_, following a suggestion of _Paul Curtz_, Jun 10 2007