A117399
Column 1 (divided by 2) of triangle A117398, which is the matrix log of A117396.
Original entry on oeis.org
1, 1, 2, 7, 34, 202, 1402, 11122, 99322, 986362, 10784122, 128720122, 1665516922, 23220588922, 347025670522, 5534133996922, 93802153836922, 1683934185324922, 31917365573484922, 636941680764780922, 13348854673487724922
Offset: 1
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m=42;
M= Table[If[k>n-1, 0, If[k==n-1, n, -1]], {n,0,m+1}, {k,0,m+1}];
T:= T= Sum[MatrixPower[M, j]/j, {j,m+1}];
Table[T[[n+1, 2]]/2, {n,2,30}] (* G. C. Greubel, Sep 06 2022 *)
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{a(n)=local(M=matrix(n+4,n+4,r,c,if(r>=c,if(r==c+1,-c,1))), L=sum(m=1,n+4,(M^0-M)^m/m));L[n+2,2]/2}
A117396
Triangle, read by rows, defined by: T(n,k) = (k+1)*T(n,k+1) - Sum_{j=1..n-k-1} T(j,0)*T(n,j+k+1) for n>k with T(n,n)=1 for n>=0.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 5, 3, 1, 1, 17, 11, 4, 1, 1, 77, 51, 19, 5, 1, 1, 437, 291, 109, 29, 6, 1, 1, 2957, 1971, 739, 197, 41, 7, 1, 1, 23117, 15411, 5779, 1541, 321, 55, 8, 1, 1, 204557, 136371, 51139, 13637, 2841, 487, 71, 9, 1, 1, 2018957, 1345971, 504739, 134597, 28041, 4807, 701, 89, 10, 1
Offset: 0
Triangle begins:
1;
1, 1;
1, 2, 1;
1, 5, 3, 1;
1, 17, 11, 4, 1;
1, 77, 51, 19, 5, 1;
1, 437, 291, 109, 29, 6, 1;
1, 2957, 1971, 739, 197, 41, 7, 1;
1, 23117, 15411, 5779, 1541, 321, 55, 8, 1;
1, 204557, 136371, 51139, 13637, 2841, 487, 71, 9, 1; ...
Matrix inverse is:
1;
-1, 1;
1, -2, 1;
1, 1, -3, 1;
1, 1, 1, -4, 1;
1, 1, 1, 1, -5, 1; ...
Matrix log is the integer triangle A117398:
0;
1, 0;
0, 2, 0;
-1, 2, 3, 0;
-3, 4, 5, 4, 0;
-9, 14, 15, 9, 5, 0;
-33, 68, 65, 34, 14, 6, 0; ...
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[k eq 0 select 1 else k*(&+[Factorial(j)/Factorial(k+1): j in [k-1..n]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 24 2021
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T[n_, k_]:= T[n, k]= If[k==0, 1, k*Sum[j!/(k+1)!, {j,k-1,n}]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 24 2021 *)
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T(n,k)=if(n
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/* Definition by Matrix Inverse: */ T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,if(r==c+1,-c,1))));(M^-1)[n+1,k+1]
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T(n,k)=if(nPaul D. Hanna, Jun 20 2006
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def A117396(n,k): return 1 if (k==0) else k*sum(factorial(j)/factorial(k+1) for j in (k-1..n))
flatten([[A117396(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 24 2021
Showing 1-2 of 2 results.
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