A273866 - OEIS

%I #62 Oct 06 2019 02:37:54

%S 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,

%T 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,

%U 57,66,57,36,16,5,1

%N 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).

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

%H Giedrius Alkauskas, <a href="http://arxiv.org/abs/0801.0805">One curious proof of Fermat's little theorem</a>, arXiv:0801.0805 [math.NT], 2008.

%H Giedrius Alkauskas, <a href="https://www.jstor.org/stable/40391097">A curious proof of Fermat's little theorem</a>, Amer. Math. Monthly 116(4) (2009), 362-364.

%H H. Gingold, H. W. Gould, and Michael E. Mays, <a href="https://www.researchgate.net/publication/268023169_Power_product_expansions">Power Product Expansions</a>, Utilitas Mathematica 34 (1988), 143-161.

%H H. Gingold and A. Knopfmacher, <a href="http://dx.doi.org/10.4153/CJM-1995-062-9">Analytic properties of power product expansions</a>, Canad. J. Math. 47 (1995), 1219-1239.

%H W. Lang, <a href="/A157162/a157162.txt">Recurrences for the general problem</a>.

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

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

%F 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]

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

%F 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

%e a{1}(h) = h,

%e a{2}(h) = h,

%e a{3}(h) = h^2 + h,

%e a{4}(h) = h^3 + h^2 + h,

%e a{5}(h) = h^4 + 2*h^3 + 2*h^2 + h,

%e a{6}(h) = h^5 + 2*h^4 + 2*h^3 + 2*h^2 + h,

%e a{7}(h) = h^6 + 3*h^5 + 5*h^4 + 5*h^3 + 3*h^2 + h,

%e a{8}(h) = h^7 + 3*h^6 + 6*h^5 + 7*h^4 + 6*h^3 + 3*h^2 + h,

%e a{9}(h) = h^8 + 4*h^7 + 9*h^6 + 13*h^5 + 13*h^4 + 9*h^3 + 4*h^2 + h

%e ...

%e and the corresponding a(k,m) table is:

%e   1,

%e   1,

%e   1,  1,

%e   1,  1,  1,

%e   1,  2,  2,  1,

%e   1,  2,  2,  2,  1,

%e   1,  3,  5,  5,  3,  1,

%e   1,  3,  6,  7,  6,  3,  1,

%e   1,  4,  9, 13, 13,  9,  4,  1,

%e   ...

%e 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

%p with(ListTools), with(numtheory), with(combinat);

%p L := product(1-a[k]*x^k, k = 1 .. 600);

%p S := Flatten([seq(-h, i = 1 .. 100)]);

%p Sabs := Flatten([seq(i, i = 1 .. 100)]);

%p seq(assign(a[i] = solve(coeff(L, x^i) = `if`(is(i in Sabs), S[Search(i, Sabs)], 0), a[i])), i = 1 .. 20);

%p map(coeffs, [seq(simplify(a[i]), i = 1 .. 20)]);

%t strictrees[n_Integer?Positive]:=Prepend[Join@@Function[ptn,Tuples[strictrees/@ptn]]/@Select[IntegerPartitions[n],And[Length[#]>1,UnsameQ@@#]&],n];

%t 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 *)

%t (* second program *)

%t A[m_, n_] :=

%t   A[m, n] =

%t    Which[m == 1, -h, m > n >= 1, 0, True,

%t     A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];

%t a[n_] := Expand[-A[n, n]];

%t a /@ Range[1, 25] (* _Petros Hadjicostas_, Oct 04 2019, courtesy of _Jean-François Alcover_ *)

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

%K nonn,tabf

%O 1,9

%A _Gevorg Hmayakyan_, Jun 01 2016