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Search: id:A111978
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| A111978 |
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Matrix log of triangle A111975, which shifts columns left and up under matrix square; these terms are the result of multiplying each element in row n and column k by (n-k)!. |
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+0
3
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| 0, 1, 0, 0, 2, 0, 16, 0, 4, 0, 0, 32, 0, 8, 0, 1536, 0, 64, 0, 16, 0, 0, 3072, 0, 128, 0, 32, 0, -319488, 0, 6144, 0, 256, 0, 64, 0, 0, -638976, 0, 12288, 0, 512, 0, 128, 0, 36007575552, 0, -1277952, 0, 24576, 0, 1024, 0, 256, 0, 0, 72015151104, 0, -2555904, 0, 49152, 0, 2048, 0, 512, 0
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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Column k equals 2^k multiplied by column 0 (A111979) when ignoring zeros above the diagonal.
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FORMULA
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T(n, k) = 2^k*T(n-k, 0) = 2^k*A111979(n-k) for n>=k>=0.
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EXAMPLE
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Matrix log of A111975, with factorial denominators, begins:
0;
1/1!, 0;
0/2!, 2/1!, 0;
16/3!, 0/2!, 4/1!, 0;
0/4!, 32/3!, 0/2!, 8/1!, 0;
1536/5!, 0/4!, 64/3!, 0/2!, 16/1!, 0;
0/6!, 3072/5!, 0/4!, 128/3!, 0/2!, 32/1!, 0;
-319488/7!, 0/6!, 6144/5!, 0/4!, 256/3!, 0/2!, 64/1!, 0; ...
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PROGRAM
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(PARI) {T(n, k, q=2)=local(A=Mat(1), B); if(n<k|k<0, 0, for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i, B[i, j]=1, if(j==1, B[i, j]=if(i>2, (A^q)[i-1, 2], 1), B[i, j]=(A^q)[i-1, j-1])); )); A=B); B=sum(i=1, #A, -(A^0-A)^i/i); return((n-k)!*B[n+1, k+1]))}
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CROSSREFS
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Cf. A111975 (triangle), A111979 (column 0).
Sequence in context: A088504 A074031 A086261 this_sequence A025600 A009006 A057375
Adjacent sequences: A111975 A111976 A111977 this_sequence A111979 A111980 A111981
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KEYWORD
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frac,sign,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Aug 25 2005
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