A147541 -id:A147541 - cp's OEIS frontend

A220418 Express 1 - x - x^2 - x^3 - x^4 - ... as product (1 + g(1)x) (1 + g(2)x^2) (1 + g(3)x^3) ... and use a(n) = - g(n).

1, 1, 2, 3, 6, 8, 18, 27, 54, 84, 186, 296, 630, 1008, 2106, 3711, 7710, 12924, 27594, 48528, 97902, 173352, 364722, 647504, 1340622, 2382660, 4918482, 9052392, 18512790, 33361776, 69273666, 127198287, 258155910, 475568220, 981288906, 1814542704, 3714566310

Offset: 1

Michel Marcus, Dec 14 2012

nonn

This is the PPE (power product expansion) of A153881 (with offset 0).
When p is prime, a(p) = (2^p-2)/p (A064535).
From Petros Hadjicostas, Oct 04 2019: (Start)
This sequence appears as an example in Gingold and Knopfmacher (1995) starting at p. 1223.
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/(1 - x - x^2 - x^3 - x^4 - ...) is given in A290261, which is also studied in Gingold and Knopfmacher (1995, p. 1234).
(End)
The number of terms in the Zassenhaus formula exponent of order n, as computed by the algorithm by Casas, Murua & Nadinic, is equal to a(n) at least for n = 2..24. - Andrey Zabolotskiy, Apr 09 2023

Alois P. Heinz, Table of n, a(n) for n = 1..2000
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.
Fernando Casas, Ander Murua and Mladen Nadinic, Efficient computation of the Zassenhaus formula, Computer Physics Communications, 183 (2012), 2386-2391; arXiv:1204.0389 [math-ph], 2012.
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. A064535, A147541, A153881, A157162, A170908, A170909, A170910, A170911, A170912, A170913, A170914, A170915, A170916, A170917, A220420, A273866, A290261.

b:= proc(n, i) option remember; `if`(n=0 or i<1, 1,
      b(n, i-1)+a(i)*b(n-i, min(n-i, i)))
    end:
a:= proc(n) option remember; 2^n-b(n, n-1) end:
seq(a(n), n=1..40);  # Alois P. Heinz, Jun 22 2018

b[n_, i_] := b[n, i] = If[n == 0 || i < 1, 1, b[n, i - 1] + a[i]*b[n - i, Min[n - i, i]]];
a[n_] := a[n] = 2^n - b[n, n - 1] ;
Array[a, 40] (* Jean-François Alcover, Jul 09 2018, after Alois P. Heinz *)

a(m) = {default(seriesprecision, m+1); gk = vector(m); pol = 1 + sum(n=1, m, -x^n); gk[1] = polcoeff( pol, 1); for (k=2, m, pol = taylor(pol/(1+gk[k-1]*x^(k-1)), x); gk[k] = polcoeff(pol, k, x);); for (k=1, m, print1(-gk[k], ", "););}

g(1) = -1 and for k > 1, g(k) satisfies Sum_{d|k} (1/d)*(-g(k/d))^d = (2^k - 1)/k, where a(k) = -g(k). - Gevorg Hmayakyan, Jun 05 2016 [Corrected by Petros Hadjicostas, Oct 04 2019. See p. 1224 in Gingold and Knopfmacher (1995).]
From Petros Hadjicostas, Oct 04 2019: (Start)
a(2*n - 1) = A290261(2*n - 1) for n >= 1 because A290261 gives the PPE of 1/(1 - x - x^2 - x^3 - ...) = (1 - x)/(1 - 2*x).
Define (A(m,n): n,m >= 1) by A(m=1,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)

Name edited by Petros Hadjicostas, Oct 04 2019

A147655 a(n) is the coefficient of x^n in the polynomial given by Product_{k>=1} (1 + prime(k)*x^k).

Original entry on oeis.org

1, 2, 3, 11, 17, 40, 86, 153, 283, 547, 1069, 1737, 3238, 5340, 9574, 17251, 27897, 45845, 78601, 126725, 207153, 353435, 550422, 881454, 1393870, 2239938, 3473133, 5546789, 8762663, 13341967, 20676253, 31774563, 48248485, 74174759, 111904363, 170184798

Offset: 0

Neil Fernandez, Nov 09 2008

nonn

Sum of all squarefree numbers whose prime indices sum to n. A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798. - Gus Wiseman, May 09 2019

			Form a product from the primes: (1 + 2*x) * (1 + 3*x^2) * (1 + 5*x^3) * ...* (1 + prime(n)*x^n) * ... Multiplying out gives 1 + 2*x + 3*x^2 + 11*x^3 + ..., so the sequence begins 1, 2, 3, 11, ....
From _Petros Hadjicostas_, Apr 10 2020: (Start)
Let f(m) = prime(m). Using the strict partitions of n (see A000009), we get:
a(1) = f(1) = 2,
a(2) = f(2) = 3,
a(3) = f(3) + f(1)*f(2) = 5 + 2*3 = 11,
a(4) = f(4) + f(1)*f(3) = 7 + 2*5 = 17,
a(5) = f(5) + f(1)*f(4) + f(2)*f(3) = 11 + 2*7 + 3*5 = 40,
a(6) = f(6) + f(1)*f(5) + f(2)*f(4) + f(1)*f(2)*f(3) = 13 + 2*11 + 3*7 + 2*3*5 = 86,
a(7) = f(7) + f(1)*f(6) + f(2)*f(5) + f(3)*f(4) + f(1)*f(2)*f(4) = 17 + 2*13 + 3*11 + 5*7 + 2*3*7 = 153. (End)

Alois P. Heinz, Table of n, a(n) for n = 0..9000 (first 1001 terms from Harvey P. Dale)

Row sums of A246867 and A258323.

Cf. A000009,A005117, A015723, A022629, A056239, A066189, A112798, A145519, A147541, A325504, A325506, A325537.

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
      b(n, i-1) +`if`(i>n, 0, b(n-i, i-1)*ithprime(i))))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..50);  # Alois P. Heinz, Sep 05 2014

nn=40;Take[Rest[CoefficientList[Expand[Times@@Table[1+Prime[n]x^n,{n,nn}]],x]],nn] (* Harvey P. Dale, Jul 01 2012 *)

a(n) = [x^n] Product_{k>=1} 1+prime(k)*x^k. - Alois P. Heinz, Sep 05 2014

a(n) = Sum_{(b_1,...,b_n)} f(1)^b_1 * f(2)^b_2 * ... * f(n)^b_n, where f(m) = prime(m), and the sum is taken over all lists (b_1,...,b_n) with b_j in {0,1} and Sum_{j=1..n} j*b_j = n. - Petros Hadjicostas, Apr 10 2020

More terms from Harvey P. Dale, Jul 01 2012
a(0)=1 inserted by Alois P. Heinz, Sep 05 2014
Name edited by Petros Hadjicostas, Apr 10 2020

A147654 Result of using the positive integers 1,2,3,... as coefficients in an infinite polynomial series in x and then expressing this series as Product_{k>=1} (1+a(k)x^k).

Original entry on oeis.org

1, 2, 1, 3, 0, -2, 0, 9, 0, -6, 0, 4, 0, -18, 0, 93, 0, -54, 0, 72, 0, -186, 0, 232, 0, -630, 0, 1020, 0, -2106, 0, 10881, 0, -7710, 0, 13824, 0, -27594, 0, 49440, 0, -97902, 0, 191844, 0, -364722, 0, 590800, 0, -1340622, 0, 2656920, 0, -4918482, 0, 9791784, 0, -18512790

Offset: 1

Neil Fernandez, Nov 09 2008

sign

			From the positive integers 1,2,3,..., construct the series 1+x+2x^2+3x^3+4x^4+... a(1) is always the coefficient of x, here 1. Divide by (1+a(1)x), i.e. here (1+x), to get the quotient (1+a(2)x^2+...), which here gives a(2)=2. Then divide this quotient by (1+a(2)x^2), i.e. here (1+2x^2), to get (1+a(3)x^3+...), giving a(3)=1.

Seiichi Manyama, Table of n, a(n) for n = 1..500

Cf. A028310, A147541, A147559.

Product_{k>=1} (1+a(k)*x^k) = 1 + Sum_{k>=1} k*x^k. - Seiichi Manyama, Jun 24 2018

More terms from Seiichi Manyama, Jun 23 2018

A147542 Product(1 + a(n)x^n, n=1..infinity) = sum(F(k+1)x^k, k=1..infinity) = 1/(1-x-x^2), where F(n) = A000045(n) (Fibonacci numbers).

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 4, 18, 8, 8, 18, 17, 40, 50, 88, 396, 210, 296, 492, 690, 1144, 1776, 2786, 3545, 6704, 10610, 16096, 25524, 39650, 63544, 97108, 269154, 236880, 389400, 589298, 956000, 1459960, 2393538, 3604880, 5739132, 9030450, 14777200

Offset: 1

Neil Fernandez, Nov 06 2008

nonn

A formal infinite product representation for the Fibonacci numbers (A000045(n+1)).

For references see A147541. [R. J. Mathar, Mar 12 2009]

Jean-François Alcover, Table of n, a(n) for n = 1..200
Wolfdieter Lang Two recurrences for the general problem.
R. J. Mathar, Re: polynomial-to-product transform, Maple code (2008). [From _R. J. Mathar_, Mar 12 2009]

Cf. A000045, A137852, A006973, A157159.

m = 200;
sol = Thread[CoefficientList[Sum[Log[1 + a[n] x^n], {n, 1, m}] - Log[1/(1 - x - x^2)] + O[x]^(m + 1), x] == 0] // Solve // First;
Array[a, m] /. sol (* Jean-François Alcover, Oct 22 2019 *)

From Wolfdieter Lang, Mar 06 2009: (Start)
Recurrence I: With FP(n,m) the set of partitions of n with m distinct parts (which could be called fermionic partitions (fp)):
a(n)= F(n+1) - sum(sum(product(a(k[j]),j=1..m),fp from FP(n,m)),m=2..maxm(n)), with maxm(n):=A003056(n) and the distinct parts k[j], j=1,...,m, of the partition fp of n, n>=3. Inputs a(1)=F(2)=1, a(2)=F(3)=2. See the array A008289(n,m) for the cardinality of the set FP(n,m).
Recurrence II: With the definition of FP(n,m) from the above recurrence I, P(n,m) the general set of partitions of n with m parts, and the multinomial numbers M_0 (given for every partition under A048996):
a(n) = sum((d/n)*(-a(d)^(n/d)),d|n with 1=2; a(1)=F(2)=1. The exponents e(j)>=0 satisfy sum(j*e(j),j=1..n)=n and sum(e(j),j=1..m). The M_0 numbers are m!/product(e(j)!,j=1..n).
Example of recurrence I: a(4) = F(5) - a(1)*a(3) = 5 - 1*1 = 4.
Example of recurrence II: a(4)= 2*(-1)^2 + (1*F(5)-(1/2)*(2*F(2)*F(4) + 1*F(3)^2) + (1/3)*3*F(2)^2*F(3)) = 4. (End)

More terms and revised description from Wolfdieter Lang Mar 06 2009
Edited by N. J. A. Sloane, Mar 11 2009 at the suggestion of Vladeta Jovovic
More terms from R. J. Mathar, Mar 12 2009

A147557 Result of using the primes as coefficients in an infinite polynomial series in x and then expressing this series as (1+a(1)x)(1+a(2)x^2)(1+a(3)x^3)...

Original entry on oeis.org

2, 3, -1, 9, -4, 0, -16, 89, -52, 60, -182, 214, -620, 966, -2142, 10497, -7676, 13684, -27530, 48288, -98372, 190928, -364464, 619496, -1341508, 2649990, -4923220, 9726940, -18510902, 37055004, -69269976, 213062855, -258284232, 527143794

Offset: 1

Neil Fernandez, Nov 07 2008

sign

			From the primes, construct the series 1+2x+3x^2+5x^3+7x^4+... a(1) is always the coefficient of x, here 2. Divide by (1+2x) to get the quotient (1+a(2)x^2+...), which here gives a(2)=3. Then divide this quotient by (1+a(2)x^2), i.e. here (1+3x^2), to get (1+a(3)x^3+...), giving a(3)=-1.

Cf. A147541

ser=1+Sum[Prime[i]x^i,{i,110}];ss=1+2x;Do[ser=Normal[Series[ser/(Take[ser,2]),{x,0,105}]];ss+=ser[[2]],{100}];A147557=CoefficientList[ss,x] (* Zak Seidov, Nov 10 2008 *)

Corrected and extended by Zak Seidov, Nov 10 2008

A220420 Express the Sum_{n>=0} p(n)x^n, where p(n) is the partition function, as a product Product_{k>=1} (1 + a(k)x^k).

Original entry on oeis.org

1, 2, 1, 4, 1, 0, 1, 14, 1, -4, 1, -8, 1, -16, 1, 196, 1, -54, 1, -92, 1, -184, 1, 144, 1, -628, 1, -1040, 1, -2160, 1, 41102, 1, -7708, 1, -12932, 1, -27592, 1, 54020, 1, -98496, 1, -173720, 1, -364720, 1, 853624, 1, -1341970, 1, -2383916, 1, -4918536, 1

Offset: 1

Michel Marcus, Dec 14 2012

sign

This is the PPE (power product expansion) of A000041.
When n is odd, a(n) = 1.
When n is even, a(n) = 2, 4, 0, 14, -4, -8, -16, 196, -54, -92, -184, 144, -628, -1040, -2160, 41102, ...
Alkauskas (2016, Problem 3, p. 3) conjectured that a(8*k+2), a(8*k+4), and a(8*k+6) are all negative, and a(8*k) is positive for k >= 1. [This statement is not wholly true for k = 0.] - Petros Hadjicostas, Oct 07 2019

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.
Giedrius Alkauskas, Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems, arXiv:1609.09842 [math.NT], 2016.
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, Canadian Journal of Mathematics 47(6) (1995), 1219-1239.
W. Lang, Recurrences for the general problem.

Cf. A000041, A147541, A170908, A170909, A170910, A170911, A170912, A170913, A170914, A170915, A170916, A170917, A220418, A290261.

terms = 55; sol[0] = {};
sol[m_] := sol[m] = Join[sol[m - 1], If[OddQ[m], {a[m] -> 1}, First @ Solve[Thread[Table[PartitionsP[n], {n, 0, m}] == CoefficientList[ (Product[1 + a[n]*x^n, {n, 1, m}] /. sol[m - 1]) + O[x]^(m + 1), x]]]]];
Array[a, terms] /. sol[terms] (* Jean-François Alcover, Dec 06 2018, corrected Oct 03 2019 *)
(* Second program: *)
A[m_, n_] := A[m, n] = Which[m == 1, PartitionsP[n], 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] (* Jean-François Alcover, Oct 03 2019, using the formula given by Petros Hadjicostas *)

a(m) = {default(seriesprecision, m+1); ak = vector(m); pol = 1 / eta(x + x * O(x^m)); ak[1] = polcoeff(pol, 1); for (k=2, m, pol = taylor(pol / (1+ak[k-1]*x^(k-1)), x); ak[k] = polcoeff(pol, k, x);); for (k=1, m, print1(ak[k], ", "););}

From Petros Hadjicostas, Oct 04 2019: (Start)
Define (A(m,n): n,m >= 1) by A(m=1,n) = p(n) = A000041(n) 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).]
a(n) = Sum_{s|n} s/n + Sum_{s|n, s > 1} (-a(n/s))^s/s. [Eq. (1) in Alkauskas (2008, 2009).]
(End)

A305882 -1 + Product_{n>=1} 1/(1 + a(n)*x^n) = g.f. of A000040 (prime numbers).

Original entry on oeis.org

-2, 1, 1, 4, 4, 13, 16, 44, 52, 112, 182, 411, 620, 1318, 2142, 5148, 7676, 15228, 27530, 58660, 98372, 207392, 364464, 763263, 1341508, 2773990, 4923220, 10470948, 18510902, 37546152, 69269976, 148419094, 258284232, 534761242, 981480012, 2004302204

Offset: 1

Ilya Gutkovskiy, Jun 13 2018

sign

			1/((1 - 2*x) * (1 + x^2) * (1 + x^3) * (1 + 4*x^4) * (1 + 4*x^5) * ... * (1 + a(n)*x^n) * ...) =  1 + 2*x + 3*x^2 + 5*x^3 + 7*x^4 + 11*x^5 + ... + A000040(k)*x^k + ...

Cf. A000040, A030010, A030018, A145519, A147541, A147557, A305871, A305881.

Product_{n>=1} 1/(1 + a(n)*x^n) = 1 + Sum_{k>=1} prime(k)*x^k.

Product_{n>=1} (1 + a(n)*x^n) = Sum_{k>=0} A030018(k)*x^k.

A147544 Coefficient of x^n in product (1+x)Product_{j>=1} (1 + prime(j)x^j).

Original entry on oeis.org

3, 5, 14, 28, 57, 126, 239, 436, 830, 1616, 2806, 4975, 8578, 14914, 26825, 45148, 73742, 124446, 205326, 333878, 560588, 903857, 1431876, 2275324, 3633808, 5713071, 9019922, 14309452, 22104630, 34018220, 52450816, 80023048, 122423244, 186079122, 282089161, 431595854, 647808336, 966099832, 1442708500

Offset: 1

Neil Fernandez, Nov 06 2008

nonn

			(1+x)*(1+2*x)*(1+3*x^2)*(1+5*x^3)*(1+7*x^4)*(1+11*x^5) ... = 1 + 3*x + 5*x^2 + 14*x^3 + 28*x^4 + 57*x^5 + 126*x^6 + 239*x^7 + 436*x^8 + 830*x^9 + 1616*x^10 + ..., so the sequence begins 3, 5, 14, ...

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A000040, A147541.

R:=PowerSeriesRing(Integers(), 51); Coefficients(R!( (1+x)*(&*[1+NthPrime(j)*x^j: j in [1..52]])-1 )); // G. C. Greubel, Oct 28 2022

With[{m=51}, Rest@CoefficientList[Series[(1+x)*Product[(1+Prime[j]*x^j), {j,m+2}], {x,0,m}], x]] (* G. C. Greubel, Oct 28 2022 *)

def A147544_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1+x)*product(1+nth_prime(j)*x^j for j in range(1,53)) ).list()
a=A147544_list(51); a[1:]  # G. C. Greubel, Oct 28 2022

G.f.: -1 + (1+x)*Product_{j>=1} (1 + prime(j)*x^j).

Terms a(11) onward corrected by G. C. Greubel, Oct 28 2022

A342223 Product_{n>=1} 1 + a(n)*x^n = Sum_{n=-oo..oo} x^(n^2) = theta_3(x).

Original entry on oeis.org

2, 0, 0, 2, -4, 8, -16, 32, -54, 108, -184, 368, -628, 1296, -2160, 4610, -7708, 15848, -27592, 58316, -98496, 207576, -364720, 756872, -1341970, 2778300, -4918536, 10443152, -18512788, 37698416, -69273664, 145105952, -258224544, 534996900, -981494752, 2020011290, -3714566308, 7614288360

Offset: 1

Neil Bickford, Mar 06 2021

sign

Coefficients in the power product expansion for theta_3(x), the third Jacobi theta function, described in A000122, also denoted theta_3(0, x) or theta_00(1, x).
See A147541 for additional references.
a(9) = -54 is the first term whose absolute value is not a power of 2.

			This gives 1 + 2x + 2x^4 + 2x^9 + ... = (1+2x)*(1+2x^4)*(1-4x^5)*(1+8x^6)*...
To compute this sequence's terms, start with the series expansion 1 + 2x + 2x^4 + 2x^9 + ...; this gives a(1) = 2, then divide by 1 + a(1)*x to get 1 + 2x^4 - 4x^5 + 8x^6 - 16x^7 ...; this gives a(2) = a(3) = 0 and a(4) = 2, then divide by 1 + a(4)*x to get 1 - 4x^5 + 8x^6 - 16x^7 ...; this gives a(5) = -4, then divide by 1 + a(5)*x to get 1 + 8x^6 - 16x^7 ...

Bill Gosper and Joerg Arndt, Discussions in Math-Fun Mailing List, circa Feb 25 2021 - Mar 2 2021.

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

Cf. A000122, A147541, A147542.

FoldPairList[{Coefficient[#1, q^#2], #1/(1 + q^#2 Coefficient[#1, q^#2])} &, #, Range[#[[5]] - 1]] &[Series[EllipticTheta[3, 0, q], {q, 0, 100}]] (* based on code from Bill Gosper, Feb 25 2021 *)

cp's OEIS Frontend

A220418 Express 1 - x - x^2 - x^3 - x^4 - ... as product (1 + g(1)*x) * (1 + g(2)*x^2) *(1 + g(3)*x^3) * ... and use a(n) = - g(n).

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A147655 a(n) is the coefficient of x^n in the polynomial given by Product_{k>=1} (1 + prime(k)*x^k).

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A147654 Result of using the positive integers 1,2,3,... as coefficients in an infinite polynomial series in x and then expressing this series as Product_{k>=1} (1+a(k)x^k).

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A147542 Product(1 + a(n)*x^n, n=1..infinity) = sum(F(k+1)*x^k, k=1..infinity) = 1/(1-x-x^2), where F(n) = A000045(n) (Fibonacci numbers).

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A147557 Result of using the primes as coefficients in an infinite polynomial series in x and then expressing this series as (1+a(1)x)(1+a(2)x^2)(1+a(3)x^3)...

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A220420 Express the Sum_{n>=0} p(n)*x^n, where p(n) is the partition function, as a product Product_{k>=1} (1 + a(k)*x^k).

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A305882 -1 + Product_{n>=1} 1/(1 + a(n)*x^n) = g.f. of A000040 (prime numbers).

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A147544 Coefficient of x^n in product (1+x)*Product_{j>=1} (1 + prime(j)*x^j).

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A220418 Express 1 - x - x^2 - x^3 - x^4 - ... as product (1 + g(1)x) (1 + g(2)x^2) (1 + g(3)x^3) ... and use a(n) = - g(n).

A147542 Product(1 + a(n)x^n, n=1..infinity) = sum(F(k+1)x^k, k=1..infinity) = 1/(1-x-x^2), where F(n) = A000045(n) (Fibonacci numbers).

A220420 Express the Sum_{n>=0} p(n)x^n, where p(n) is the partition function, as a product Product_{k>=1} (1 + a(k)x^k).

A147544 Coefficient of x^n in product (1+x)Product_{j>=1} (1 + prime(j)x^j).