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A111845
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Triangle P, read by rows, that satisfies [P^4](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(4*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(k,k)=1 and P(k+1,1)=P(k+1,0) for k>=0.
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5
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1, 1, 1, 4, 4, 1, 40, 40, 16, 1, 1040, 1040, 544, 64, 1, 78240, 78240, 48960, 8320, 256, 1, 18504256, 18504256, 13110400, 2878720, 131584, 1024, 1, 14463224448, 14463224448, 11192599808, 2982187520, 180270080, 2099200, 4096, 1
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OFFSET
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0,4
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COMMENTS
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Column 0 and column 1 are equal for n>0.
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LINKS
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FORMULA
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Let q=4; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x = Sum_{n>=1} -(-1)^n/n!*Product_{j=0..n-1} L(q^j*x); L(x) equals the g.f. of column 0 of the matrix log of P (A111849).
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EXAMPLE
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Let q=4; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x = L(x) - L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! -+ ...
and L(x) = x + 4/2!*x^2 + 56/3!*x^3 + 1728/4!*x^4 +...(A111849).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(4*x) + m^3/3!*L(x)*L(4*x)*L(4^2*x) +
m^4/4!*L(x)*L(4*x)*L(4^2*x)*L(4^3*x) + ...
Triangle P begins:
1;
1,1;
4,4,1;
40,40,16,1;
1040,1040,544,64,1;
78240,78240,48960,8320,256,1;
18504256,18504256,13110400,2878720,131584,1024,1; ...
where P^4 shifts columns left and up one place:
1;
4,1;
40,16,1;
1040,544,64,1;
78240,48960,8320,256,1; ...
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PROG
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(PARI) P(n, k, q=4)=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]=(A^q)[i-1, 1], B[i, j]=(A^q)[i-1, j-1])); )); A=B); return(A[n+1, k+1]))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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