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Search: id:A117269
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| A117269 |
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Triangle T, read by rows, that satisfies matrix equation: T - (T-I)^2 = C, where C is Pascal's triangle. |
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+0
4
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| 1, 1, 1, 3, 2, 1, 19, 9, 3, 1, 207, 76, 18, 4, 1, 3211, 1035, 190, 30, 5, 1, 64383, 19266, 3105, 380, 45, 6, 1, 1581259, 450681, 67431, 7245, 665, 63, 7, 1, 45948927, 12650072, 1802724, 179816, 14490, 1064, 84, 8, 1, 1541641771, 413540343, 56925324, 5408172
(list; table; graph; listen)
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OFFSET
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0,4
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COMMENT
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E.g.f. of column 0 is F(x) = (3-sqrt(5-4*exp(x)))/2 since F(x) satisfies the characteristic equation: F - (F-1)^2 = exp(x). The matrix log of T is the integer triangle A117270.
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FORMULA
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T(n,k) = A052886(n-k)*C(n,k) for n>k, with T(n,n) = 1.
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EXAMPLE
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Triangle T begins:
1;
1,1;
3,2,1;
19,9,3,1;
207,76,18,4,1;
3211,1035,190,30,5,1;
64383,19266,3105,380,45,6,1;
1581259,450681,67431,7245,665,63,7,1; ...
where (T-I)^2 =
0;
0,0;
2,0,0;
18,6,0,0;
206,72,12,0,0;
3210,1030,180,20,0,0;
64382,19260,3090,360,30,0,0; ...
and T - (T-I)^2 = Pascal's triangle.
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PROGRAM
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(PARI) {T(n, k)=local(C=matrix(n+1, n+1, r, c, if(r>=c, binomial(r-1, c-1))), M=C); for(i=1, n+1, M=(M-M^0)^2+C); return(M[n+1, k+1])}
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CROSSREFS
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Cf. A117270 (log), A117271, A052886.
Sequence in context: A106208 A129377 A136733 this_sequence A107862 A117265 A107727
Adjacent sequences: A117266 A117267 A117268 this_sequence A117270 A117271 A117272
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Mar 05 2006
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