A055878
Least positive sequence with Hankel transform {1,1,1,1,1,...}.
Original entry on oeis.org
1, 1, 2, 1, 6, 1, 43, 1, 7794, 1, 387355897233, 1, 58111569344245781528237519282283636, 1
Offset: 1
- M M Mangontarum, O I Cauntongan, A P Macodi-Ringia, The Noncentral Version of the Whitney Numbers: A Comprehensive Study, International Journal of Mathematics and Mathematical Sciences, Volume 2016, Article ID 6206207, 16 pages; http://dx.doi.org/10.1155/2016/6206207. See reference 21.
Next term is too large to include.
A376277
The least increasing sequence starting with 1, such that the determinants of the Hankel matrices H1 = [a(0), a(1), ..., a(n); ...; a(n), a(n+1), ..., a(2*n)] and H2 = [a(1), a(2), ..., a(n+1); ...; a(n+1), a(n+2), ..., a(2*n+1)] are > 0.
Original entry on oeis.org
1, 2, 5, 13, 35, 98, 287, 883, 2858, 9708, 34411, 126337, 476767, 1836851, 7185420, 28420613, 113317776, 454468077, 1830556209, 7397188271, 29965426959, 121620119888, 494365414071, 2011965781648, 8196475452837, 33419092543257, 136353532725534, 556669705441210
Offset: 0
Cf.
A000108 (We obtain the Catalan numbers if we use "least positive sequence" in the definition instead of "least increasing").
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hankelok(s) = {my(m1=floor((#s+1)/2)); my(m2=floor(#s/2)); my(h1=matrix(m1,m1,x,y,s[x+y-1])); my(h2=matrix(m2,m2,x,y,s[x+y])); return((matdet(h1) > 0) && (matdet(h2) > 0))}
a(max_n) = {my(s=[1,2],k=3); while(#s < max_n, while(hankelok(concat(s,[k]))==0,k=k+1); s=concat(s,[k])); return(s)}
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my(N=30, x='x+O('x^N)); Vec(1/(1-2*x/(1-(1/2)*x/(1-(1/2)*x/(1-2*x/(1-((1-sqrt(1-4*x))/(2*x))*x))))))
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a(n) = if(n<3, [1, 2, 5][n+1], sum(k=1, floor((n+1)/2), (binomial(n-k+1, k)+binomial(n-k, k-1)-binomial(n-k-3, k-4))*(-1)^(k+1)*a(n-k)))
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