login
A340880
Irregular table: n-th row polynomial given by the formal power series expansion of Sum_{k >= 0} (1 + q)^(n*k + k*(k+1)/2)* Product_{j = 1..k} (1 - (1 + q)^j), n >= 1.
3
1, 1, 2, 1, 5, 8, 3, 1, 9, 33, 60, 53, 23, 4, 1, 14, 87, 310, 685, 965, 893, 546, 214, 49, 5, 1, 20, 185, 1040, 3930, 10469, 20229, 29037, 31480, 25975, 16278, 7643, 2611, 614, 89, 6, 1, 27, 345, 2760, 15415, 63535, 199724, 490019, 955902, 1504600, 1931685, 2037007
OFFSET
1,3
COMMENTS
The polynomial identity Product_{k = 1..N} (1 - q^k) = 1 - Sum_{k = 0..N-1} q^(k+1)*Product_{i = 1..k} (1 - q^i), N = 1,2,..., is easily established by induction on N. This identity is the starting point in Andrew's exposition of Euler's proof of the pentagonal number theorem.
Replacing q with 1 + q and letting N tend to infinity leads to the formal power series identity S(1,q) := Sum_{k >= 0} (1 + q)^k*Product_{j = 1..k} (1 - (1 + q)^j) = 1/(1 + q).
More generally, for n = 1,2,..., it can be shown that the series S(n,q) := Sum_{k >= 0} (1 + q)^(n*k)*Product_{j = 1..k} (1 - (1 + q)^j) is a rational function of q of the form R(n,q)/(1 + q)^(n*(n+1)/2), where R(n,q) is a polynomial of degree n*(n-1)/2 in q. The proof makes use of the following formal series identity, valid for n >= 1:
t^n*Sum_{k >= 0} x^(n*k+n*(n+1)/2)*(Product_{i = 1..k} 1 - t*x^i) + (-1)^(n+1)*(1-x)*(1-x^2)*...*(1-x^(n-1))*( Product_{i >= 0} 1 - t*x^i ) = Sum_{k = 0..n-1} t^k*x^(k*(k+1)/2)*Product_{i = k+1..n-1} (x^i - 1).
The first few polynomials are R(1,q) = 1, R(2,q) = 1 + 2*q and R(3,q) = 1 + 5*q + 8*q^2 + 3*q^3. The present table lists the coefficients of the polynomials R(n,q) for n >= 1.
Conjecture: For fixed integers q and N, the sequence ( R(n,q) mod N )n>=1 is eventually periodic.
LINKS
George E. Andrews, Euler's Pentagonal Number Theorem, Mathematics Magazine, Vol. 56, No. 5 (Nov., 1983), pp. 279-284.
FORMULA
n-th row polynomial: R(n,q) = Sum_{k = 0..n-1} x^(k*(k+1)/2)* ( Product_{j = k+1..n-1} x^j - 1 ), where x = 1 + q.
In the q-adic topology, lim_{n -> oo} R(2*n+1, q - 1) = -lim_{n -> oo} R(2*n, q - 1) = Product_{n >= 1} (1 - q^n)^2.
EXAMPLE
The table begins
n\k 0 1 2 3 4 5 6 7 8 9 10
----------------------------------------------------------------------
1 1,
2 1, 2,
3 1, 5, 8, 3,
4 1, 9, 33, 60, 53, 23, 4,
5 1, 14, 87, 310, 685, 965, 893, 546, 214, 49, 5,
6 1, 20, 185, 1040, 3930, 10469, 20229, 29037, 31480, 25975, ...
MAPLE
row := n -> add( (1+q)^((1/2)*k*(k+1))*mul((1+q)^j-1, j = k+1..n-1), k = 0..n-1 ):
seq(seq(coeff(row(n), q, k), k = 0..(1/2)*n*(n-1)), n = 1..10);
CROSSREFS
Row sums A340881. Cf. A340882.
Sequence in context: A082635 A166623 A348729 * A094510 A023677 A108599
KEYWORD
nonn,tabf,easy
AUTHOR
Peter Bala, Feb 16 2021
STATUS
approved