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A111979
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Column 0 of the matrix logarithm (A111978) of triangle A111975, which shifts columns left and up under matrix square; these terms are the result of multiplying the element in row n by n!.
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3
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0, 1, 0, 16, 0, 1536, 0, -319488, 0, 36007575552, 0, -53682434054553600, 0, 1790644857560674043166720, 0, -1280831660558056667387645027942400, 0, 18961467116136182692294341450867551502336000, 0
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OFFSET
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0,4
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COMMENTS
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Let q=2; the g.f. of column k of A111975^m (matrix power m) is: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} A(q^j*x).
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LINKS
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FORMULA
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E.g.f. A(x): x-x^2 = Sum_{j>=1}(1-2^j*x)/j!*Prod_{i=0..j-1}A(2^i*x). E.g.f. A(x): x+x^2 = Sum_{j>=1}(1-4^j*x^2)/j!*Prod_{i=0..j-1}A(2^i*x).
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EXAMPLE
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E.g.f. A(x) = x + 16/3!*x^3 + 1536/5!*x^5 - 319488/7!*x^7
+ 36007575552/9!*x^9 - 53682434054553600/11!*x^11 +...
where A(x) satisfies:
x*(1-x) = (1-2*x)*A(x) + (1-2^2*x)*A(x)*A(2*x)/2!
+ (1-2^3*x)*A(x)*A(2*x)*A(2^2*x)/3! +...
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PROG
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(PARI) {a(n, q=2)=local(A=Mat(1), B); if(n<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!*B[n+1, 1]))}
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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