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A309495
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Triangle read by rows, derived from A007318, row sums = the Bell Sequence.
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1
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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, 354, 1, 7, 27, 90, 275, 701, 1368, 1671, 1, 8, 35, 131, 460, 1411, 3579, 6986, 8536, 1, 9, 44, 182, 716, 2536, 7738, 19620, 38315, 46814, 1, 10, 54, 244, 1057, 4213, 14846, 45251, 114798, 224189, 273907
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OFFSET
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1,5
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COMMENTS
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As described in A160185, we extract eigensequences of a rotated variant of Pascal's triangle:
1;
3, 1;
3, 2, 1;
1, 1, 1, 1;
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.
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LINKS
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FORMULA
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EXAMPLE
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Row 5 of A121207 is (1, 5, 14, 31, 52). Preface with a zero and take the first difference row:
(0, 1, 5, 14, 31, 52)
(..., 1, 4, 9, 17, 21) = row 5 of the triangle.
First few rows of the triangle:
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, 354;
...
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PROG
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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}
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]))}
{ my(A=T(10)); for(n=1, #A, print(A[n])) } \\ _Andrew Howroyd_, Feb 20 2022
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CROSSREFS
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KEYWORD
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AUTHOR
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_Gary W. Adamson_, Aug 04 2019
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EXTENSIONS
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Terms a(37) and beyond from _Andrew Howroyd_, Feb 20 2022
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STATUS
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approved
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