A147541 Result of using the primes as coefficients in an infinite polynomial series in x and then expressing this series as (1+x)(1+a(1)*x)(1+a(2)*x^2) ... .

1, 2, 1, 3, 2, -4, 2, 5, 4, -6, 4, 4, 10, -36, 18, 45, 34, -72, 64, -24, 124, -358, 258, 170, 458, -1260, 916, 148, 1888, -4296, 3690, 887, 7272, -17616, 14718, -5096, 29610, -67164, 58722, -26036, 119602, -244496, 242256, -104754, 487352, -1029384

Offset: 1

Author

Neil Fernandez, Nov 06 2008

Keywords

• sign

Comments

This is the PPE (power product expansion) of A036467. - R. J. Mathar, Feb 01 2010

Examples

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

Links

• H. Gingold, A note on reduction of operations via power product approximations, Utilitas Math. 37 (1990), 79-89. [From _R. J. Mathar_, Nov 10 2008]
• H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 1219-1239. [From _R. J. Mathar_, Nov 10 2008]
• H. Gingold, A. Knopfmacher and D. Lubinsky, The zero distribution of the partial products of power product expansions, Analysis 13 (1993), 133-157. [From _R. J. Mathar_, Nov 10 2008]

Crossrefs

Cf. A000040, A147542.

Programs

Maple

From R. J. Mathar, Feb 01 2010: (Start)
# Partition n into a set of distinct positive integers, the maximum one
# being m.
# Example: partitionsQ(7,5) returns [[2,5],[3,4],[1,2,4]] ;
# Richard J. Mathar, 2008-11-10
partitionsQ := proc(n,m)
local p,t,rec,q;
p := [] ;
# take 't' of the n and recursively determine the partitions of
# what has been left over.
for t from min(m,n) to 1 by -1 do
# Since we are only considering partitions into distinct parts,
# the triangular numbers set a lower bound on the t.
if t*(t+1)/2 >= n then
rec := partitionsQ(n-t,t-1) ;
if nops(rec) = 0 then
p := [op(p),[t]] ;
else
for q in rec do
p := [op(p),[op(q),t]] ;
end do:
end if;
end if;
end do:
RETURN(p) ;
end proc:
# Power product expansion of L.
# L is a list starting with 1, which is considered L[0].
# Returns the list [a(1),a(2),..] such that
# product_(i=1,2,..) (1+a(i)x^i) = sum_(j=0,1,2,...) L[j]x^j.
# Richard J. Mathar, 2008-11-10
ppe := proc(L)
local pro,i,par,swithi,snoti,m,p,k ;
pro := [] ;
for i from 1 to nops(L)-1 do
par := partitionsQ(i,i) ;
swithi := 0 ;
snoti := 0 ;
for p in par do
if i in p then
m := 1 ;
for k from 1 to nops(p)-1 do
m := m*op(op(k,p),pro) ;
end do;
swithi := swithi+m ;
else
snoti := snoti+mul( op(k,pro),k=p) ;
end if;
end do:
pro := [op(pro), (op(i+1,L)-snoti)/swithi] ;
end do:
RETURN(pro) ;
end proc:
read("transforms") ;
A147541 := proc(nmax)
local L,L1,L2 ;
L := [1,seq(ithprime(n),n=1..nmax)] ;
L1 := [seq((-1)^n,n=0..nmax+10)] ;
A036467 := CONV(L,L1) ;
ppe(A036467) ;
end:
A147541(47) ;
(End)

Extensions

Extended by R. J. Mathar, Feb 01 2010