Sequences based on algorithms
Andrew Plewe
aplewe at sbcglobal.net
Tue Nov 13 17:39:26 CET 2007
say, the "nextprime" function in Mathematica, or in some other context. In
scheme in other versions may be different.
-Andrew Plewe-
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Date: Tue, 13 Nov 2007 07:51:30 -0800 (PST)
From: <mathoflove-seqfan at yahoo.com>
Reply-To: mathoflove-seqfan at yahoo.com
Subject: papers on sequences
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Hello seqfans,
I started writing papers about sequences. Two of them are almost ready
- I am waiting for the help with my English.
What are the standard places to publish such papers?
Are there arXivs?
What are the standard journals?
Best, Tanya
Andrew, I think those sequences should certainly be submitted to
the OEIS.
I agree about the ambiguity of "next prime" - maybe include
a comment saying which definition is being used?
Best regards
Neil
I have submitted the sequence, with notes about "nextprime" and "sqrtint"
(thanks to Ray Chandler for suggesting sequences that match the output of
those functions, which I noted in the comments). I expanded the sequence to
cover all integers. Here are all of the values below 1200:
4, 6, 9, 10, 14, 22, 25, 35, 49, 55, 65, 77, 85, 91, 119, 121, 143, 169,
187, 209, 221, 247, 253, 289, 299, 319, 323, 361, 377, 391, 407, 437, 493,
527, 529, 551, 589, 629, 667, 697, 703, 713, 841, 851, 899, 943, 961, 989,
1073, 1081, 1147, 1189
I wonder; are 4, 10, 14, and 22 the only even integers in the sequence? I
believe that is true; it appears that, if n = ab, and b is larger than a,
then a > b/x, where x is a lower bound that slowly increase from 6 to about
1 as n increases. Therefore the number of times that a particular prime can
appear as a factor in the sequence is finite.
-----Original Message-----
Andrew, I think those sequences should certainly be submitted to
the OEIS.
I agree about the ambiguity of "next prime" - maybe include
a comment saying which definition is being used?
Best regards
Neil
----__JWM__J0c15.6f97S.527bM
Seqfans,=20
=20
Given the functional equation:=20
(1) r*A(x) =3D c + b*x + A(x)^n=20
=20
the following expressions for A(x) are equivalent:=20
=20
(3) A(x) =3D (c+bx)/r * G( (c+bx)^(n-1)/r^n )=20
=20=20
where G(x) satisfies: G(x) =3D 1 + x*G(x)^n
and G(x)^n =3D (1/x)*Series_Reversion( x/(1+x)^n ) ;
=20
(4) A(x) =3D 1 + Series_Reversion[ (1 + r*x - (1+x)^n )/b ] .
=20
See examples below.
=20=20
But do these formulas (3) and (4) help any in obtaining=20
a formula for the n-th term a(n)?=20
=20
If there are no further insights from seqfans regarding this,=20
I will submit the formulas and say no more on the subject.=20
Regards,=20
=20
--------------------------------------------------------
formula for A(x):
A120588 : 3*A(x) =3D 2 + x + A(x)^2 : [1,1,1]:
A120590 : 4*A(x) =3D 3 + x + A(x)^3 : [1,1,3]:
A120592 : 5*A(x) =3D 4 + 4*x + A(x)^3 : [1,2,6]:=20=20
A120593 : 5*A(x) =3D 4 + x + A(x)^4 : [1,1,6]:=20=20
A120594 : 8*A(x) =3D 7 + 8*x + A(x)^4 : [1,2,6]:=20=20
A120595 : 13*A(x) =3D 12 + 27*x + A(x)^4 : [1,3,6]:
A120596 : 6*A(x) =3D 5 + x + A(x)^5 : [1,1,10]:
A120597 : 9*A(x) =3D 8 + 8*x + A(x)^5 : [1,2,10]:
A120598 : 30*A(x) =3D 29 + 125*x + A(x)^5 : [1,5,10]:=20
A120599 : 13*A(x) =3D 12 + 32*x + A(x)^5 : [1,4,20]:
A120600 : 7*A(x) =3D 6 + x + A(x)^6 : [1,1,15]:
A120601 : 15*A(x) =3D 14 + 27*x + A(x)^6 : [1,3,15]:=20
A120602 : 31*A(x) =3D 30 + 125*x + A(x)^6 : [1,5,15]:=20
A120603 : 16*A(x) =3D 15 + 27*x + A(x)^7 : [1,3,21]:=20=20
A120604 : 24*A(x) =3D 23 + 64*x + A(x)^8 : [1,4,28]:=20=20
A120605 : 25*A(x) =3D 24 + 64*x + A(x)^9 : [1,4,36]:=20=20
A120606 : 36*A(x) =3D 35 + 81*x + A(x)^9 : [1,3,12]:=20=20
A120607 : 37*A(x) =3D 36 + 81*x + A(x)^10 : [1,3,15]:=20=20
-------------------------------------------------------
END.
_____________________________________________________________
It pays to Discover.=20
Apply now! 0% into APR on balance transfers.
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----__JWM__J0c15.6f97S.527bM
<html><P>Seqfans, <BR> OK, now there are 2 formulas for t=
he g.f. A(x) that <BR>satisfies (1) - they are formulas (3) and (4) as foll=
ows. <BR> <BR>Given the functional equation: <BR>(1) r*A(x) =3D =
c + b*x + A(x)^n <BR> <BR>the following expressions for A(x) are equiv=
R> <BR>where G(x) satisfies: G(x) =3D 1 + x*G(x)^n<BR>and G(x)^=
n =3D (1/x)*Series_Reversion( x/(1+x)^n ) ;<BR> <BR>(4) A(x) =3D 1 + S=
eries_Reversion[ (1 + r*x - (1+x)^n )/b ] .<BR> <BR>See examples below=
.<BR> <BR>But do these formulas (3) and (4) help any in obtainin=
g <BR>a formula for the n-th term a(n)? <BR> <BR>If there are no furth=
er insights from seqfans regarding this, <BR>I will submit the formulas and=
BR><FONT face=3D"Courier New">---------------------------------------------=
-----------<BR>EXAMPLES: A-numbers, g.f. equation, initial 3 terms, <BR>for=
mula for A(x):</FONT></P>
<P><FONT face=3D"Courier New">A120588 : 3*A(x) =3D 2 + x + A(x)^2 &nbs=
p; : [1,1,1]:<BR> G.f.: A(x) =3D 1 + serrever=
; : [1,1,3]:<BR> G.f.: A(x) =3D 1 + serreverse(1+4*=
x - (1+x)^3)</FONT></P>
<P><FONT face=3D"Courier New">A120592 : 5*A(x) =3D 4 + 4*x + A(x)^3 &n=
bsp; : [1,2,6]: <BR> G.f.: A(x) =3D 1 + serreverse(=
(1+5*x - (1+x)^3)/4)</FONT></P>
<P><FONT face=3D"Courier New">A120593 : 5*A(x) =3D 4 + x + A(x)^4 &nbs=
p; : [1,1,6]: <BR> G.f.: A(x) =3D 1 + s=
erreverse(1+5*x - (1+x)^4)</FONT></P>
<P><FONT face=3D"Courier New">A120594 : 8*A(x) =3D 7 + 8*x + A(x)^4 &n=
bsp; : [1,2,6]: <BR> G.f.: A(x) =3D 1 + serreverse(=
(1+8*x - (1+x)^4)/8)</FONT></P>
<P><FONT face=3D"Courier New">A120595 : 13*A(x) =3D 12 + 27*x + A(x)^4 =
; : [1,3,6]:<BR> G.f.: A(x) =3D 1 + serreverse((1+13*x - (1+x)^4)/27)=
</FONT></P>
<P><FONT face=3D"Courier New">A120596 : 6*A(x) =3D 5 + x + A(x)^5 &nbs=
p; : [1,1,10]:<BR> G.f.: A(x) =3D 1 + serreve=
rse(1+6*x - (1+x)^5)</FONT></P>
<P><FONT face=3D"Courier New">A120597 : 9*A(x) =3D 8 + 8*x + A(x)^5 &n=
bsp; : [1,2,10]:<BR> G.f.: A(x) =3D 1 + serreverse((1+9*x=
<P><FONT face=3D"Courier New">A120598 : 30*A(x) =3D 29 + 125*x + A(x)^5 : [=
1,5,10]: <BR> G.f.: A(x) =3D 1 + serreverse((1+30*x - (1+x)^5)/125)</=
FONT></P>
<P><FONT face=3D"Courier New">A120599 : 13*A(x) =3D 12 + 32*x + A(x)^5 =
; : [1,4,20]:<BR> G.f.: A(x) =3D 1 + serreverse((1+13*x - (1+x)^5)/32=
)</FONT></P>
<P><FONT face=3D"Courier New">A120600 : 7*A(x) =3D 6 + x + A(x)^6 &nbs=
p; : [1,1,15]:<BR> G.f.: A(x) =3D 1 + serreve=
rse(1+7*x - (1+x)^6)</FONT></P>
<P><FONT face=3D"Courier New">A120601 : 15*A(x) =3D 14 + 27*x + A(x)^6 =
; : [1,3,15]: <BR> G.f.: A(x) =3D 1 + serreverse((1+15*x - (1+x)^6)/2=
7)</FONT></P>
<P><FONT face=3D"Courier New">A120602 : 31*A(x) =3D 30 + 125*x + A(x)^6 : [=
1,5,15]: <BR> G.f.: A(x) =3D 1 + serreverse((1+31*x - (1+x)^6)/125)</=
FONT></P>
<P><FONT face=3D"Courier New">A120603 : 16*A(x) =3D 15 + 27*x + A(x)^7 =
; : [1,3,21]: <BR> G.f.: A(x) =3D 1 + serreverse((1+16*x - (1+x=
)^7)/27)</FONT></P>
<P><FONT face=3D"Courier New">A120604 : 24*A(x) =3D 23 + 64*x + A(x)^8 =
; : [1,4,28]: <BR> G.f.: A(x) =3D 1 + serreverse((1+24*x - (1+x=
)^8)/64)</FONT></P>
<P><FONT face=3D"Courier New">A120605 : 25*A(x) =3D 24 + 64*x + A(x)^9 =
; : [1,4,36]: <BR> G.f.: A(x) =3D 1 + serreverse((1+25*x - (1+x=
)^9)/64)</FONT></P>
<P><FONT face=3D"Courier New">A120606 : 36*A(x) =3D 35 + 81*x + A(x)^9 =
; : [1,3,12]: <BR> G.f.: A(x) =3D 1 + serreverse((1+36*x - (1+x=
)^9)/81)</FONT></P>
<P><FONT face=3D"Courier New">A120607 : 37*A(x) =3D 36 + 81*x + A(x)^10 : [=
1,3,15]: <BR> G.f.: A(x) =3D 1 + serreverse((1+37*x - (1+x)^10)=
/81)</FONT></P>
<P><FONT face=3D"Courier New">---------------------------------------------=
----------</FONT><BR>END.</P></html>
<font face=3DTimes-New-Roman size=3D2><br><br>_____________________________=
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</font>
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