A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)*x^k) = 1 - h*x/(1-x).

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 3, 5, 5, 3, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 9, 13, 13, 9, 4, 1, 1, 4, 10, 17, 20, 17, 10, 4, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 5, 16, 36, 57, 66, 57, 36, 16, 5, 1

Offset

1,9

Comments

The a(k,m) form a table where each row has k-1 elements starting from 2 and the a(1,1) = 1.

Links

Table of n, a(n) for n=1..67.

Giedrius Alkauskas, One curious proof of Fermat's little theorem, arXiv:0801.0805 [math.NT], 2008.

Giedrius Alkauskas, A curious proof of Fermat's little theorem, Amer. Math. Monthly 116(4) (2009), 362-364.

H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.

H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 1219-1239.

W. Lang, Recurrences for the general problem.

Formula

a(k,m) = a(k, k-m).

For prime p: Sum_{m = 1..p-1} a(p, m) = (2^p - 2)/p.

a{k}(h) satisfies Sum_{d|k} (1/d)*(a{k/d}(h))^d = ((h+1)^k - 1)/k. [Corrected by Petros Hadjicostas, Oct 04 2019]

For prime p: a{p}(h) = ((h+1)^p - h^p - 1)/p.

See A273873 for the definition of strict tree. Then a(n,m) = Sum_t (-1)^{v(t)-1} where the sum is over all strict trees of weight n with m leaves, and v(t) is the number of nodes in t (including the leaves, which are positive integers). See example 2 and the first Mathematica program. - Gus Wiseman, Nov 14 2016

Example

a{1}(h) = h,
a{2}(h) = h,
a{3}(h) = h^2 + h,
a{4}(h) = h^3 + h^2 + h,
a{5}(h) = h^4 + 2*h^3 + 2*h^2 + h,
a{6}(h) = h^5 + 2*h^4 + 2*h^3 + 2*h^2 + h,
a{7}(h) = h^6 + 3*h^5 + 5*h^4 + 5*h^3 + 3*h^2 + h,
a{8}(h) = h^7 + 3*h^6 + 6*h^5 + 7*h^4 + 6*h^3 + 3*h^2 + h,
a{9}(h) = h^8 + 4*h^7 + 9*h^6 + 13*h^5 + 13*h^4 + 9*h^3 + 4*h^2 + h
...
and the corresponding a(k,m) table is:
  1,
  1,
  1,  1,
  1,  1,  1,
  1,  2,  2,  1,
  1,  2,  2,  2,  1,
  1,  3,  5,  5,  3,  1,
  1,  3,  6,  7,  6,  3,  1,
  1,  4,  9, 13, 13,  9,  4,  1,
  ...
a(7,3) = 5 because there are six strict trees contributing positive one {{5,1},1}, {{4,2},1}, {{4,1},2}, {{3,2},2}, {4,{2,1}}, {{3,1},3} and there is one strict tree contributing negative one {4,2,1}. - Gus Wiseman, Nov 14 2016

Maple

with(ListTools), with(numtheory), with(combinat);
L := product(1-a[k]*x^k, k = 1 .. 600);
S := Flatten([seq(-h, i = 1 .. 100)]);
Sabs := Flatten([seq(i, i = 1 .. 100)]);
seq(assign(a[i] = solve(coeff(L, x^i) = `if`(is(i in Sabs), S[Search(i, Sabs)], 0), a[i])), i = 1 .. 20);
map(coeffs, [seq(simplify(a[i]), i = 1 .. 20)]);

Mathematica

strictrees[n_Integer?Positive]:=Prepend[Join@@Function[ptn, Tuples[strictrees/@ptn]]/@Select[IntegerPartitions[n], And[Length[#]>1, UnsameQ@@#]&], n];
Table[Sum[(-1)^(Count[tree, _, {0, Infinity}]-1), {tree, Select[strictrees[n], Length[Flatten[{#}]]===m&]}], {n, 1, 9}, {m, 1, n-1/.(0->1)}] (* Gus Wiseman, Nov 14 2016 *)
(* second program *)
A[m_, n_] :=
  A[m, n] =
   Which[m == 1, -h, m > n >= 1, 0, True,
    A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
a[n_] := Expand[-A[n, n]];
a /@ Range[1, 25] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

Crossrefs

Cf. A196545, A220418, A220420, A273866, A273873, A279785, A290261, A290262, A290971, A295632.

Sequence in context: A113971 A109338 A291191 * A071202 A207379 A220163

Adjacent sequences: A273863 A273864 A273865 * A273867 A273868 A273869

Keyword

nonn,tabf

Author

Gevorg Hmayakyan, Jun 01 2016

Status

approved