OFFSET
1,2
COMMENTS
Each n-th row polynomial, P(n,z), has a trivial zero at z = 0; for odd rows, P(2n+1,z) also has zeros at z = -2n, z = -(2n+1), for n > 0.
FORMULA
The n-th row polynomials, P(n, z), satisfy 1 + Sum_{n>=1} P(n, z) x^n = (Sum_{k>=1} x^(k-1)/k)^z.
EXAMPLE
P(1,z) = z/2,
P(2,z) = (5z + 3z^2)/24,
P(3,z) = (6z + 5z^2 + z^3)/48,
P(4,z) = (502z + 485z^2 + 150z^3 + 15z^4)/5760,
P(5,z) = (760z + 802z^2 + 305z^3 + 50z^4 +3z^5)/11520,
P(6,z) = (152696z + 171150z^2 + 73801z^3 + 15435z^4 + 1575z^5
+ 63z^6)/2903040,
P(7,z) = (252336z + 295748z^2 + 139020z^3 + 33817z^4 + 4515z^5
+ 315z^6 + 9z^7)/5806080,
P(8,z) = (51360816z + 62333204z^2 + 31231500z^3 + 8437975z^4
+ 1334760z^5 + 124110z^6 + 6300z^7 + 135z^8)/1393459200.
MAPLE
nmax:=8; A053657 := proc(n) local P, p, q, s, r; P := select(isprime, [$2..n]); r:=1; for p in P do s := 0; q := p-1; do if q > (n-1) then break fi; s := s + iquo(n-1, q); q := q*p; od; r := r * p^s; od; r end: f(z) := convert(series((-ln(1-x)/x)^z, x, nmax+2), polynom): for n from 1 to nmax do f(n) := A053657(n+1)*coeff(f(z), x, n) od: for n from 1 to nmax do for m from 1 to n do a(n, m) := coeff(f(n), z, m) od: od: seq(seq(a(n, m), m=1..n), n=1..nmax); # Johannes W. Meijer, Jun 08 2009, revised Nov 25 2012
MATHEMATICA
rows = 9; A053657[n_] := Product[p^Sum[Floor[(n-1)/((p-1) p^k)], {k, 0, n}], {p, Prime[Range[n]]}]; (Rest[CoefficientList[#, z]]& /@ Rest @ CoefficientList[(-Log[1-x]/x)^z + O[x]^(rows+1), x]) * Array[A053657, rows, 2] // Flatten (* Jean-François Alcover, Nov 22 2016 *)
PROG
(PARI) {T(n, k)=local(X=x+x^2*O(x^n)); D=1; for(j=0, n, D=lcm(D, denominator( polcoeff(polcoeff((-log(1-X)/x)^z+z*O(z^j), j, z), n, x)))); return(D*polcoeff(polcoeff((-log(1-X)/x)^z+z*O(z^k), k, z), n, x))}
CROSSREFS
KEYWORD
AUTHOR
Paul D. Hanna, Sep 15 2002; revised Jun 27 2005
STATUS
approved