In this connexion, it may be that Conway's relation twixt the triangular and pentagonal numbers is not as well known as it deserves to be. Divide the triangular numbers by 3 (when you can exactly): 0 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 ... 0 - 1 2 - 5 7 - 12 15 - 22 26 - 35 40 - 51 ... - - + + - - + + - - + ... and you get the pentagonal numbers in pairs, one of positive rank and the other negative. Then append signs according as the pair have the same (+) or opposite (-) parity. Then Euler's pentagonal number theorem is easy to remember: p(n-0)-p(n-1)-p(n-2)+p(n-5)+p(n-7)-p(n-12)-p(n-15)++-- =0^n where p(n) is the partition function, the left side terminates before the argument becomes negative, and 0^n = 1 if n = 0 and = 0 if n > 0. E.g., p(0)=1, p(7)=p(7-1)+p(7-2)-p(7-5)-p(7-7)+0^7=11+7-2-1+0=15. This, of course, is how Major Percy MacMahon did his calculations up to p(200), against which Hardy & Ramanujan were able to check their famous formula (too large to go in this margin, but see the BoN again). There are pictures for the relations (see The BoN) a a a a a a a a a a a a a a a b a a b a a b a a a b b c c c b c c b b a a b b a a a b b '''' c c c b b c c c b b c c c b b b c c c c b b c c c c b b b c c c c c b b b where the pentagonal numbers are well known (?) to be made of three triangles, one of which differs in rank by one from the other two, more or less according as the pentagon is of pos or neg rank: 0+0+1=1 1+1+0=2 1+1+3=5 3+3+1=7 3+3+6=12 6+6+3=15 6+6+10=22 ... + - + - + - + x y y y z x x x x y y z x x y y z z x x x y z y x x z x x y z z x x x y z z z x x y y x z x x z z y y x z z x x x z z z x x x x z z z z [deform these carefully into regular pentagons!] Best to all for 2006, R. On Tue, 27 Dec 2005, Andrew Plewe wrote:
Point taken concerning the algebra -- I'll work it out. I'd like to come up with a generalized equation similar to that for A005449, or (n(3n+1))/2, which will give all values for the desired sequence with f(x), x = 0 as the "starting" point. Incidentally, I find these sequences interesting because I ran into them while doing the following simple transformation (add the first two terms and redistribute over the remaining terms, then sum the result):
1 + 1 + 2+3 = 3+4 = 7 1 + 2 + 3+4+5 = 4+5+6 = 15 1 + 3 + 4+5+6+7 = 5+6+7+8 = 26
which can be described by 7 + 8(x) + (3 * tri(x-1), starting with x = 0.
And:
2 + 1 + 2+3+4 = 3+4+5 = 12 2 + 2 + 3+4+5+6 = 4+5+6+7 = 22 2 + 3 + 4+5+6+7+8 = 5+6+7+8+9 = 35
which can be described by 12 + 10(x) + (3 * tri(x-1)), starting with x = 0.
Anyway, thanks for you help!
-Andrew Plewe-
-----Original Message----- From: Marc LeBrun [mailto:mlb@fxpt.com] Sent: Thursday, December 22, 2005 7:03 PM To: seqfan Subject: Re: Help extending equation -- sequence related to pentagonal numbers
I'm not 100% sure what you are asking, but maybe this will help:
You have a bunch of statements of the form
"Equation to produce sequence from Kth term: u + v(x) + (3 * tri(x-1))"
So let's restate this using a general parameterized formula we'll define: f[u,v](x) = u + v x + 3 tri(x-1).
(You usually make this statement for K=4--except the first time, when you use K=3 for some reason--but no matter...)
You give a few "starting values" and then you're basically saying that everything after that is f(1), f(2), f(3) and so on.
But, at least for all those I checked, your formula applied to x<1, that is f(0), f(-1), f(-2) and so on, seems to give the "starting values" as well.
So I'll leap to the conclusion that what you want is the formula, a(N) defined so that it that "starts" at the first term (ie N=1), rather than f which "starts" at the Kth term.
This is easy: because a(K)=f(1), a(K+1)=f(2), and so on, it's simply an offset, so a(N)=f(N-K+1).
So now all we have to do is substitute x=N-K-1 to get the formula for a(N):
a(N) = f(N-K+1) = u + v (N-K-1) + 3 tri(N-K-2)
I'll leave simplifying the algebra to you.
By the way, I remark that these a(N) are just quadratics in N, and so their properties are well understood (eg 3-term recurrence, rational 2nd-degree generating functions, etc), so you probably should only trouble to submit them if they have some other especially interesting property (such as being pentagonal or whatever).
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At 05:14 PM 12/22/2005, Andrew Plewe wrote:
My new sequence follows from extending a group of partial equations for other sequences related to pentagonal/triangle numbers in the OEIS. To demonstrate the pattern, I've listed the sequences in the OEIS with their corresponding partial generating equations. I'd like to find a.) an equation that fully describes my new sequence, and b.) a general method for deriving equations for other similar sequences. "tri", here, represents the function for generating triangle numbers ((n * (n + 1))/2:
A005449 = 0, 2, 7, 15, 26, 40, etc. Equation to produce sequence from third term: 7 + 8(x) + (3 * tri(x-1))
A000326 = 0, 1, 5, 12, 22, 35, etc. Equation to produce sequence from fourth term: 12 + 10(x) + (3 * tri(x-1))
A045943 = 0, 3, 9, 18, 30, 45, etc. Equation to produce sequence from fourth term: 18 + 12(x) + (3 * tri(x-1))
A095794 = 1, 6, 14, 25, 39, 56, etc. Equation to produce sequence from fourth term: 25 + 14(x) + (3 * tri(x-1))
And my new sequence:
A000001 = 3, 10, 20, 33, 49, 68, etc. Equation to produce sequence from fourth term: 33 + 16(x) + (3 * tri(x-1))
As you can see there is a pattern to these generating equations. Looking at the equations already in the database for the sequences listed above, I was unable to determine a general method for generating equations which would fully describe sequences derived in this manner. I'm not sure if it's o.k. to submit equations that partially describe a sequence to the OEIS, so I'd like to find one which fully describes my sequence before submitting it. Any help is appreciated. Thanks!