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%I #17 Feb 20 2022 22:53:56
%S 1,1,1,1,2,2,1,3,5,6,1,4,9,17,21,1,5,14,34,67,82,1,6,20,58,148,290,
%T 354,1,7,27,90,275,701,1368,1671,1,8,35,131,460,1411,3579,6986,8536,1,
%U 9,44,182,716,2536,7738,19620,38315,46814,1,10,54,244,1057,4213,14846,45251,114798,224189,273907
%N Triangle read by rows, derived from A007318, row sums = the Bell Sequence.
%C As described in A160185, we extract eigensequences of a rotated variant of Pascal's triangle:
%C 1;
%C 3, 1;
%C 3, 2, 1;
%C 1, 1, 1, 1;
%C Say, for these 4 columns, the eigensequence is (1, 4, 9, 15). Then preface the latter with a zero and take the first finite difference row, = (1, 3, 5, 6), fourth row of the triangle.
%H Andrew Howroyd, <a href="/A309495/b309495.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)
%F T(n,k) = A121207(n,k) - A121207(n, k-1) for k >= 2.
%e Row 5 of A121207 is (1, 5, 14, 31, 52). Preface with a zero and take the first difference row:
%e (0, 1, 5, 14, 31, 52)
%e (..., 1, 4, 9, 17, 21) = row 5 of the triangle.
%e First few rows of the triangle:
%e 1;
%e 1, 1;
%e 1, 2, 2;
%e 1, 3, 5, 6;
%e 1, 4, 9, 17, 21;
%e 1, 5, 14, 34, 67, 82;
%e 1, 6, 20, 58, 148, 290, 354;
%e ...
%o (PARI) \\ here U(n) is A121207.
%o U(n)={my(M=matrix(n,n)); for(n=1, n, M[n,1]=1; for(k=1, n-1, M[n,k+1]=sum(j=1, k, M[n-j, k-j+1]*binomial(n-2,j-1)))); M}
%o T(n)={my(A=U(n+1)); vector(n, n, my(t=A[n+1,2..n+1]); t-concat([0], t[1..n-1]))}
%o { my(A=T(10)); for(n=1, #A, print(A[n])) } \\ _Andrew Howroyd_, Feb 20 2022
%Y Row sums are A000110.
%Y Main diagonal is A032346.
%Y Cf. A007318, A121207, A160185.
%K nonn,tabl
%O 1,5
%A _Gary W. Adamson_, Aug 04 2019
%E Terms a(37) and beyond from _Andrew Howroyd_, Feb 20 2022
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