Fibbinary sequence
Gottfried Helms
Annette.Warlich at t-online.de
Mon Nov 6 13:05:20 CET 2006
A courious result by chance:
A003714 Fibbinary numbers:
if n = F_i1+F_i2+...+F_ik is the Zeckendorf representation of n
(i.e. write n in Fibonacci number system)
then
a(n) = 2^{i1-2}+2^{i2-2}+...+2^{ik-2}.
0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85,
128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149,
160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260
I was checking some variants of the Pascal-matrix and got the
following
Let Px =
1 . . . . . . .
1 -1 . . . . . .
3/2 -3 1 . . . . .
4 -9 6 -1 . . . .
15 -35 30 -10 1 . . .
137/2 -345/2 165 -75 15 -1 . .
1491/4 -2009/2 1050 -560 315/2 -21 1 .
2374 -6720 7588 -4515 1540 -294 28 -1
then the index r of the rows, r=1..inf(??) , which contain only integers, is A003714
1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37,
40, 41, 42, 64,
(although I checked this only up to N=64)
Explanation: -----------------------------------------------------------
The matrix Px is constructed as a variant of Pj
Pj:
1 . . . . . . .
1 -1 . . . . . .
1 -2 1 . . . . .
1 -3 3 -1 . . . .
1 -4 6 -4 1 . . .
1 -5 10 -10 5 -1 . .
1 -6 15 -20 15 -6 1 .
1 -7 21 -35 35 -21 7 -1
which is the column-signed version of the Pascalmatrix P
Pj = P*J
where J is diagonal unit with alternating signs (diag(1,-1,1,-1,...))
The matrix S2 of the stirling-numbers of 2'nd kind
1 . . . . . . .
1 1 . . . . . .
1 3 1 . . . . .
1 7 6 1 . . . .
1 15 25 10 1 . . .
1 31 90 65 15 1 . .
1 63 301 350 140 21 1 .
1 127 966 1701 1050 266 28 1
and S1 of the stirling numbers of first kind (where also S1 = S2^-1)
1 . . . . . . .
-1 1 . . . . . .
2 -3 1 . . . . .
-6 11 -6 1 . . . .
24 -50 35 -10 1 . . .
-120 274 -225 85 -15 1 . .
720 -1764 1624 -735 175 -21 1 .
-5040 13068 -13132 6769 -1960 322 -28 1
with the eigenvalues Z(1) = diag(1,1/2,1/3,1/4) such that
G = S2 * Z(1) * S2^-1
G:
1 . . . . . . .
1/2 1/2 . . . . . .
1/6 1/2 1/3 . . . . .
0 1/4 1/2 1/4 . . . .
-1/30 0 1/3 1/2 1/5 . . .
0 -1/12 0 5/12 1/2 1/6 . .
1/42 0 -1/6 0 1/2 1/2 1/7 .
0 1/12 0 -7/24 0 7/12 1/2 1/8
form an eigensystem of G.
G ist interesting as it contains the bernoulli-
numbers, and coefficients of the integrals of the
bernoulli-functions.
Also G is Eigenmatrix of the signed Pascal-matrix Pj,
with the eigenvalues J = diag(1,-1,1,-1,...)
Pj = G * J * G^-1
and Pj is used in divergent Euler-summation, Hasse's
zeta-summation and elsewhere.
Now I tried a variant of P by varying G, where the eigenvalues
are not Z(1), but are V(1/2) = diag(1,1/2,1/4,1/8,....) for some
analyses about summing zeta series, since Z(1) and V(1/2) are
closely related in this subject and complementary/mutually supplementary
in certain manners.
Now with the G-variant
Gx = S2 * V(1/2) * S2^-1
Gx:
1 . . . . . . .
1/2 1/2 . . . . . .
0 3/4 1/4 . . . . .
-1/4 3/8 3/4 1/8 . . . .
0 -5/8 15/16 5/8 1/16 . . .
1/2 -15/16 -15/32 45/32 15/32 1/32 . .
0 7/4 -105/32 35/64 105/64 21/64 1/64 .
-17/8 147/32 7/32 -735/128 35/16 105/64 7/32 1/128
I come to
Px = Gx * J * Gx^-1
= ( S2 * V(1/2) * S2^-1 ) * J * (S2 * V(2) * S2^-1)
(while Pj is
Pj = G * J * G ^-1
= ( S2 * Z(-1) * S2^-1 ) * J * (S2 * Z(1) * S2^-1)
)
and as shown above:
Px:
1 . . . . . . .
1 -1 . . . . . .
3/2 -3 1 . . . . .
4 -9 6 -1 . . . .
15 -35 30 -10 1 . . .
137/2 -345/2 165 -75 15 -1 . .
1491/4 -2009/2 1050 -560 315/2 -21 1 .
2374 -6720 7588 -4515 1540 -294 28 -1
and the rows containing only integers, are uniquely
indicated by A003714, up to r=64, where the rownumber r
starts with 1.
Don't have an idea, how this coincidence could be interesting -
I thought I'd just sentd the observation..
Regards -
Gottfried Helms
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