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A131023 First subdiagonal 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 A131023 #18 Sep 08 2022 08:45:30
%S A131023 1,2,3,4,5,7,14,37,101,256,593,1267,2534,4825,8921,16384,30581,58975,
%T A131023 117950,242461,504605,1048576,2156201,4371451,8742902,17308657,
%U A131023 34085873,67108864,132623405,263652487,527304974,1059392917,2133134741
%N A131023 First subdiagonal 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 A131023 Also first differences of main diagonal A129339.
%H A131023 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,6).
%F A131023 a(1) = 1, a(2) = 2, a(3) = 3, a(4) = 4; for n > 4, a(n) = 5*a(n-1) - 9*a(n-2) + 6*a(n-3).
%F A131023 G.f.: x*(1-3*x+2*x^2+x^3)/((1-2*x)*(1-3*x+3*x^2)).
%F A131023 a(n) = A057681(n-1) + 2^(n-2), a(1) = 1. - _Bruno Berselli_, Feb 17 2011
%e A131023 For first seven rows of T see A131022 or A129339.
%o A131023 (Magma) m:=34; 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[n+1, n]: n in [1..m-1] ];
%o A131023 (PARI) {m=33; v=concat([1, 2, 3, 4],vector(m-4)); for(n=5, m, v[n]=5*v[n-1]-9*v[n-2]+6*v[n-3]); v}
%Y A131023 Cf. A131022 (T read by rows), A129339 (main diagonal of T), A131024 (row sums of T), A131025 (antidiagonal sums of T). First through sixth column of T are in A088911, A131026, A131027, A131028, A131029, A131030 resp.
%K A131023 nonn,easy
%O A131023 1,2
%A A131023 _Klaus Brockhaus_, following a suggestion of _Paul Curtz_, Jun 10 2007