A290262 -id:A290262 - cp's OEIS frontend

A290261 Write 1 - x/(1-x) as an inverse power product 1/(1 + a(1)x) 1/(1 + a(2)x^2) 1/(1 + a(3)x^3) 1/(1 + a(4)x^4) ...

1, 2, 2, 6, 6, 10, 18, 54, 54, 114, 186, 334, 630, 1314, 2106, 5910, 7710, 15642, 27594, 57798, 97902, 207762, 364722, 712990, 1340622, 2778930, 4918482, 10437702, 18512790, 37500858, 69273666, 154021590, 258155910, 535004610, 981288906

Offset: 1

Gus Wiseman, Jul 24 2017

nonn

The initial terms are given on page 1234 of Gingold, Knopfmacher, 1995.
From Petros Hadjicostas, Oct 04 2019: (Start)
In Section 3 of Gingold and Knopfmacher (1995), it is proved that, if f(z) = Product_{n >= 1} (1 + g(n))*z^n = 1/(Product_{n >= 1} (1 - h(n))*z^n), then g(2*n - 1) = h(2*n - 1) and Sum_{d|n} (1/d)*h(n/d)^d = -Sum_{d|n} (1/d)*(-g(n/d))^d. The same results were proved more than ten years later by Alkauskas (2008, 2009). [If we let a(n) = -g(n), then Alkauskas works with f(z) = Product_{n >= 1} (1 - a(n))*z^n; i.e., a(2*n - 1) = -h(2*n - 1) etc.]
The PPE of 1 - x/(1-x) is given in A220418, which is also studied in Gingold and Knopfmacher (1995) starting at p. 1223.
(End)

Seiichi Manyama, Table of n, a(n) for n = 1..3333
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.

Cf. A220418, A273866, A289501, A290262.

nn=20;Solve[Table[Expand[SeriesCoefficient[Product[1/(1+a[k]x^k),{k,n}],{x,0,n}]]==-1,{n,nn}],Table[a[n],{n,nn}]][[1,All,2]]
(* Second program: *)
A[m_, n_] := A[m, n] = Which[m == 1, 2^(n-1), m > n >= 1, 0, True, A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1] ];
a[n_] := A[n, n];
a /@ Range[1, 55] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

a(n) = Sum_t (-1)^v(t) where the sum is over all enriched p-trees of weight n (see A289501 for definition) and v(t) is the number of nodes (branchings and leaves) in t.
From Petros Hadjicostas, Oct 04 2019: (Start)
a(n) satisfies Sum_{d|n} (1/d)*(-a(n/d))^d = -(2^n - 1)/n. Thus, a(n) = Sum_{d|n, d>1} (1/d)*(-a(n/d))^d + (2^n - 1)/n.
a(2*n - 1) = A220418(2*n - 1) for n >= 1 because A220418 gives the PPE of 1 - x/(1-x).
Define (A(m,n): n,m >= 1) by A(m=1,n) = 2^(n-1) for n >= 1, A(m,n) = 0 for m > n >= 1 (upper triangular), and A(m,n) = A(m-1,n) - A(m-1,m-1) * A(m,n-m+1) for n >= m >= 2. Then a(n) = A(n,n). [Theorem 3 in Gingold et al. (1988).]
(End)

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

Original entry on oeis.org

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

Gevorg Hmayakyan, Jun 01 2016

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

			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

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.

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

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)]);

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 *)

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

A295632 Write 1/Product_{n > 1}(1 - 1/n^s) in the form Product_{n > 1}(1 + a(n)/n^s).

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1

Offset: 2

Gus Wiseman, Nov 24 2017

sign

First negative entry is a(1024) = -4.

Cf. A001055, A045778, A050376, A220418, A220420, A273866, A273873, A289501, A290261, A290262, A290971, A290973, A295279, A295635, A295636.

nn=100;
facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
Solve[Table[Length[facs[n]]==Sum[Times@@a/@f,{f,Select[facs[n],UnsameQ@@#&]}],{n,2,nn}],Table[a[n],{n,2,nn}]][[1,All,2]]

A295635 Write 2 - Zeta(s) in the form 1/Product_{n > 1}(1 + a(n)/n^s).

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 4, 1, 2, 2, 6, 1, 4, 1, 4, 2, 2, 1, 8, 2, 2, 2, 4, 1, 6, 1, 6, 2, 2, 2, 12, 1, 2, 2, 8, 1, 6, 1, 4, 4, 2, 1, 16, 2, 4, 2, 4, 1, 8, 2, 8, 2, 2, 1, 16, 1, 2, 4, 10, 2, 6, 1, 4, 2, 6, 1, 24, 1, 2, 4, 4, 2, 6, 1, 16, 6, 2, 1, 16, 2, 2

Offset: 2

Gus Wiseman, Nov 24 2017

nonn

Cf. A001055, A045778, A050376, A220418, A220420, A273866, A273873, A289501, A290261, A290262, A290971, A290973, A295279, A295632, A295636.

nn=100;
facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
-Solve[Table[-1==Sum[Times@@a/@f,{f,facs[n]}],{n,2,nn}],Table[a[n],{n,2,nn}]][[1,All,2]]

A295636 Write 2 - Zeta(s) in the form Product_{n > 1}(1 - a(n)/n^s).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 4, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 1, 8, 1, 2, 2, 4, 1, 6, 1, 6, 2, 2, 2, 8, 1, 2, 2, 8, 1, 6, 1, 4, 4, 2, 1, 16, 1, 4, 2, 4, 1, 8, 2, 8, 2, 2, 1, 16, 1, 2, 4, 8, 2, 6, 1, 4, 2, 6, 1, 24, 1, 2, 4, 4, 2, 6, 1, 16, 3, 2, 1, 16, 2, 2, 2

Offset: 2

Gus Wiseman, Nov 24 2017

nonn

Cf. A001055, A045778, A050376, A220418, A220420, A273866, A273873, A289501, A290261, A290262, A290971, A290973, A295279, A295632, A295635.

nn=100;
facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
-Solve[Table[-1==Sum[Times@@a/@f,{f,Select[facs[n],UnsameQ@@#&]}],{n,2,nn}],Table[a[n],{n,2,nn}]][[1,All,2]]

a(n) = Sum_t (-1)^(v(t)-1) where the sum is over all strict tree-factorizations of n (see A295279 for definition) and v(t) is the number of nodes (branchings and leaves) in t.

A290320 Write 1 - t * x/(1-x) as an inverse power product 1/(1+c(1)x) * 1/(1+c(2)x^2) * 1/(1+c(3)x^3) * ... The sequence is a regular triangle where T(n,k) is the coefficient of t^k in c(n), 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 0, 1, 2, 2, 1, 1, 2, 2, 1, 0, 1, 3, 4, 2, 0, 0, 1, 3, 5, 5, 3, 1, 0, 1, 4, 9, 13, 13, 9, 4, 1, 1, 4, 9, 13, 13, 9, 4, 1, 0, 1, 5, 14, 25, 30, 24, 12, 3, 0, 0, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 0, 1, 6, 21, 48, 75, 81, 60, 30, 10, 2, 0, 0

Offset: 1

Gus Wiseman, Jul 27 2017

An irregular triangle with only the nonzero coefficients is given by A290262.

			Triangle begins:
  1;
  1,  1;
  1,  1,  0;
  1,  2,  2,  1;
  1,  2,  2,  1,  0;
  1,  3,  4,  2,  0,  0;
  1,  3,  5,  5,  3,  1,  0;
  1,  4,  9, 13, 13,  9,  4,  1;
  1,  4,  9, 13, 13,  9,  4,  1,  0;
  1,  5, 14, 25, 30, 24, 12,  3,  0,  0;
  1,  5, 15, 30, 42, 42, 30, 15,  5,  1,  0;
  1,  6, 21, 48, 75, 81, 60, 30, 10,  2,  0,  0;

Cf. A220418, A273866, A289501, A290261, A290262.

nn=12;Solve[Table[Expand[SeriesCoefficient[Product[1/(1+c[k]x^k),{k,n}],{x,0,n}]]==-t,{n,nn}],Table[c[n],{n,nn}]][[1,All,2]]

cp's OEIS Frontend

A290261 Write 1 - x/(1-x) as an inverse power product 1/(1 + a(1)*x) * 1/(1 + a(2)*x^2) * 1/(1 + a(3)*x^3) * 1/(1 + a(4)*x^4) * ...

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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).

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A295632 Write 1/Product_{n > 1}(1 - 1/n^s) in the form Product_{n > 1}(1 + a(n)/n^s).

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A295635 Write 2 - Zeta(s) in the form 1/Product_{n > 1}(1 + a(n)/n^s).

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A295636 Write 2 - Zeta(s) in the form Product_{n > 1}(1 - a(n)/n^s).

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A290320 Write 1 - t * x/(1-x) as an inverse power product 1/(1+c(1)x) * 1/(1+c(2)x^2) * 1/(1+c(3)x^3) * ... The sequence is a regular triangle where T(n,k) is the coefficient of t^k in c(n), 1 <= k <= n.

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Mathematica

A290261 Write 1 - x/(1-x) as an inverse power product 1/(1 + a(1)x) 1/(1 + a(2)x^2) 1/(1 + a(3)x^3) 1/(1 + a(4)x^4) ...

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