Alkane numbers ; Pascal's Triangle
Creighton Dement
crowdog at crowdog.de
Mon May 23 00:33:04 CEST 2005
Hi Seqfans,
The floretion used to find the Alkane numbers (with the help of force
transforms) was (1/2)('i + j' + 'ij' + e). The page
http://www.crowdog.de/Alkane.html starts out with (O-th iteration):
1jesseq[(1/2)('i + j' + 'ij' + e)]: 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1,
0, 1, 0, -1, 0, 1, 0
(breaking this down into its positive and negative components on the
next step)
Previously, I mentioned that (for reasons I admittedly don't understand)
sometimes -though not all the time- adding/subtracting the unit e to a
floretion induces a "set of (inverse) binomial transforms". Adding the
unit to the above floretion gives:
1vesseq[(1/2)('i + j' + 'ij' + e) + e]: 3, 6, 10, 16, 28, 56, 120, 256,
528, 1056, 2080, 4096, 8128, 16256, 32640, 65536
2tesseq[(1/2)('i + j' + 'ij' + e) + e]: 3, 4, 6, 12, 28, 64, 136, 272,
528, 1024, 2016, 4032, 8128, 16384, 32896, 65792
2lesseq[(1/2)('i + j' + 'ij' + e) + e]: 1, 4, 10, 20, 36, 64, 120, 240,
496, 1024, 2080, 4160, 8256, 16384, 32640, 65280
1jesseq[(1/2)('i + j' + 'ij' + e) + e]: 1, 2, 2, 0, -4, -8, -8, 0, 16,
32, 32, 0, -64, -128, -128, 0
and the inverse binomial transform of (1, 0, -1, 0, 1, 0, -1, 0, ) is
(1, -1, 0, 2, -4, 4, 0, -8, 16, -16, 0, 32) which is readily seen to be
related to (1, 2, 2, 0, -4, -8, -8, 0, 16) = 1jesseq[(1/2)('i + j' +
'ij' + e) + e] upon multiplication by 2*(-1)^n
The relationship ves = jes + les + les now applied to the above case
returns: (to be submitted shortly)
2*A038505(n+2) = A038504(n+3) + A00749(n+3) + 2*A009545(n+1)
I found no mention of this at the link provided:
http://www.theory.csc.uvic.ca/~cos/inf/trs/str/Zq/str_tr_subtr_Z2.html
However, I did find reference to sequence A038503 and the first question
I asked myself was: "Where is A038503 hiding in all this?" (see below)
http://www.research.att.com/projects/OEIS?Anum=A038505
Sum of every 4th entry of row n in Pascal's triangle, starting at "n
choose 2".
Comments: Number of strings over Z_2 of length n with trace 0 and
subtrace 1. Same as number of strings over GF(2) of length n with trace
0 and subtrace 1.
Binomial transform of (0,1,1,0,0,1,1,0,...) - Paul Barry
(pbarry(AT)wit.ie), Jul 07 2003
http://www.research.att.com/projects/OEIS?Anum=A038504
Sum of every 4th entry of row n in Pascal's triangle, starting at "n
choose 1".
Comments: Number of strings over Z_2 of length n with trace 1 and
subtrace 0. Same as number of strings over GF(2) of length n with trace
1 and subtrace 0.
http://www.research.att.com/projects/OEIS?Anum=A000749
Number of strings over Z_2 of length n with trace 1 and subtrace 1.
Comments: Same as number of strings over GF(2) of length n with trace 1
and subtrace 1. Also expansion of bracket function.
Without the two initial zeros, binomial transform of A007877. - Henry
Bottomley (se16(AT)btinternet.com), Jun 04 2001
http://www.research.att.com/projects/OEIS?Anum=A009545
Expansion of sin(x)*exp(x).
To close - notice what happens when we apply a force tranforms
iteratively to jesseq:
1st iteration: 1, 0, 0, -2, -6, -10, -10, -2, 14, 30, 30, -2, -66,
-130, -130, -2
2nd iteration: 1, 0, -2, -2, -4, -6, -4, 6, 24, 42, 44, 14, -48, -110,
-108, 22
3rd iteration: 1, 0, -2, -4, -2, -2, 0, 8, 22, 34, 28, -12, -86, -162,
-176, -64
...
6th iteration: 1, 0, -2, -4, -4, 0, 8, 18, 8, -2, -12, -32, -64, -84,
-40, 118
...
(n -> infty) iteration: 1, 0, -2, -4, -4, 0, 8, 16, 16, 0, -32, -64,
-64, 0, 128
...
Thus, we have gone from "Expansion of sin(x)*exp(x)" (0th iteration)
to "Expansion of cos(x)/exp(x)" (infinitely iterated)!
I then remembered my submission
http://www.research.att.com/projects/OEIS?Anum=A100216
Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).
A100215(n) = ((-1)^n)*A009116(n+3) + A100216(n) +
A038503(n+1)
And, thus, A038503 has decided to "turn up at the party" after all.
However, it is apparently too late- sadly, the others (A038504, A038505)
have already gone home.
Sincerely,
Creighton
-- Leise fliehen meine Lieder durch die Nacht zu dir... (Georgette Dee)
http://dict.leo.org/?lang=en&lp=ende&search=
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