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A345441
Numerators of irregular triangle of fractions arising from a problem of projecting into Hilbert space.
1
0, 1, 0, 1, 1, 1, 0, 3, 1, 1, 3, 1, 1, 0, 7, 5, 1, 1, 7, 5, 7, 1, 0, 15, 17, 3, 1, 1, 1, 15, 17, 29, 9, 3, 0, 31, 49, 23, 1, 3, 1, 1, 31, 49, 95, 47, 7, 3, 1, 0, 63, 129, 9, 39, 75, 7, 1, 1, 63, 129, 273, 189, 153, 17, 9, 1, 0, 127, 321, 201, 75, 171, 103, 13, 1, 1
OFFSET
1,8
LINKS
H. H. Bauschke and R. M. Corless, Analyzing a Projection Method with Maple, MapleTech Journal, 4:1 (1997), 2-7.
EXAMPLE
Triangle begins:
[0],
[1/2],
[0,1/4],
[1/4,1/8],
[0,3/16,1/8],
[1/8,3/32,1/16,1/16],
[0,7/64,5/32,1/16],
[1/16,7/128,5/64,7/64,1/32],
[0,15/256,17/128,3/32,1/32,1/64],
[1/32,15/512,17/256,29/256,9/128,3/128],
[0,31/1024,49/512,23/256,1/16,3/64,1/128],
...
PROG
(PARI) htabl(nn) = {my(A0 = matrix(nn, nn), A1 = matrix(nn, nn), B1 = matrix(nn, nn)); A0[2, 1] = 1/2; B1[2, 1] = 1/2; A0[3, 2] = 1/4; A1[3, 1] = 1/2; B1[3, 1] = 1/4; for (n=4, nn, for (m=1, n, if (m%2, if (m==1, A0[n, 1] = A0[n-2, 1]/2; A1[n, 1] = A1[n-2, 1]/2; B1[n, 1] = B1[n-1, 1]/2, A0[n, m] = (A0[n-2, m]+A0[n-2, m-1])/2; A1[n, m] = (A1[n-2, m]+A1[n-2, m-1])/2; B1[n, m] = (B1[n-1, m]+B1[n-1, m-1])/2), A0[n, m] = (A0[n-1, m-1]+A0[n-1, m])/2; A1[n, m] = (A1[n-1, m-1]+A1[n-1, m])/2; B1[n, m] = (B1[n-2, m]+B1[n-2, m-1])/2; ); ); ); A0; }
len(v, n) = my(k=n); forstep(i=n, 1, -1, if (!v[i], k--, break)); if (k==0, 1, k);
tabfn(nn) = my(m = htabl(nn), v = vector(nn, n, apply(numerator, Vec(m[n, ], len(m[n, ], n))))); for (n=1, #v, for (k=1, #v[n], print1(v[n][k], ", "))); \\ Michel Marcus, Mar 20 2023
CROSSREFS
Cf. A000337 (third column), A345442 (denominators).
Sequence in context: A218332 A322506 A103496 * A267863 A262681 A076498
KEYWORD
nonn,tabf,frac
AUTHOR
N. J. A. Sloane, Jun 27 2021
EXTENSIONS
More terms from Michel Marcus, Mar 20 2023
STATUS
approved