I need to compute the coefficients of the expansion of (1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)...(1+x^n)?

Answer 1

I don't know whether there is an easy way to do this. Sheer hard slog suggests that the answer may be
1 + x + x^2 +2(x^3 + x^4 + x^5) + 3(x^6 + x^7 + x^8) + 4(x^9 ... with every three powers having the same multiplier in front.
hope this helps.

Edit. No it's not that simple. Will come back to it.

Further edit. After alot more slog I now think thatit will be
1 + x + x^2 + 2x^3 + 2x^4 + 3x^5 + 4x^5 + 5x^6 + 6x^7 . . . with the general term from x^5 onwards being (n - 1)x^n.

No even this breaks down with x^7. Sorry that's it.

Answer 2

This expansion occurs in the theory of partitions - the coeffiecients of the first few terms are listed under Sloanes's sequence number A000009, and can be viewed here: (and see below)

http://www.research.att.com/~njas/sequences/A000009

Euler enjoyed playing around with this and similar expansions. Here are the first few coefficients of x⁰, x¹, x², x³, x⁴, . . . . as per the link given above:

1, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 27, 32, 38, 46, 54, 64, 76, 89, 104, 122, 142, 165, 192, 222, 256, 296, 340, 390, 448, 512, 585, 668, 760, 864, 982, 1113, 1260, 1426, 1610, 1816, 2048, 2304, 2590, 2910, 3264, 3658, 4097, 4582, 5120, 5718, . . . .

Answer 3

basically, this boils down into a problem of how many ways can you make a number (sum - ie adding) using only numbers less than itself, including zero.

so fo x^3 it is x^3 * 1, x^2 * x - ie 2 ways

for x^6 it is x^6 * 1, x^5 * x, x^4 * x^2 but not x^3 * x^3 - cannot repeat. However dont forget x^3 * x^2 * x or x^4 * x^2 * 1

thus sums for 6 are (6,0) (5,1) (4,2) (3,2,1)
so the coefficient of x^6 is 4.

for power 7 we have (7,0) (6,1) (5,2) (4,3) (4,2,1) so 5 ways.

note that (5,2,0) is the same as (5,2)

I hope this will throw some light, but unfortunately not an nth term formula.

Answer 4

It is an interesting question.

Consider the following
Let x<=1

1/(1-x)=1+x+x^2+x^3+x^4+x^5+x^6+x^7+........

=(1+x)+(x^2+x^3)+(x^4+x^5)+(x^6+x^7)+.........

=(1+x)+x^2(1+x)+x^4(1+x)+x^6(1+x)+.........
=(1+x)(1+x^2+x^4+x^6........)
=(1+x) { (1+x^2)+x^4(1+x^2)+...........)
=(1+x)(1+x^2)(1+x^4)

Similarly
1/(1-x^3)=1+x^3+x^6+x^9+.....= (1+x^3)(1+x^6)+...........

Now x^5 is left out

1/(1-x^5)=1+x^5+x^10+x^15+..........=((1+x^5)(1+x^10)+........


From the above it could be easily deducted that
(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5).......

=1/(1-x). 1/(1-x3)./(1-x^5).(1-x^7)+
= (1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10+x^11+...)......
(1+x^3+ +x^6+ +x^9+ ) .
( 1 +x^5+ +x^10 +)
(1 +x^7+ ......................)
( 1 +x^9 + .................)
----------------------------------------------------------------------------------
= 1+x+x^2+2x^3+..................

Thus the Co-efficients of expansion can be found out

Answer 5

Add the exponents then work out the rest