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Revision History for A307649

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G.f. A(x) satisfies: (1 + x) = A(x)*A(x^2)^2*A(x^3)^3*A(x^4)^4* ... *A(x^k)^k* ...
(history; published version)
#5 by Susanna Cuyler at Sat Apr 20 08:11:56 EDT 2019
STATUS

proposed

approved

#4 by Ilya Gutkovskiy at Fri Apr 19 22:35:06 EDT 2019
STATUS

editing

proposed

#3 by Ilya Gutkovskiy at Fri Apr 19 22:12:44 EDT 2019
NAME

allocated for Ilya Gutkovskiy

G.f. A(x) satisfies: (1 + x) = A(x)*A(x^2)^2*A(x^3)^3*A(x^4)^4* ... *A(x^k)^k* ...

DATA

1, 1, -2, -5, 0, 4, 9, 2, -10, -21, 29, 15, -18, -80, 50, 59, 207, -228, -244, -315, 868, 103, 360, -1907, 752, -151, 3802, -5032, 965, -5279, 13742, -6049, 9107, -33835, 25398, -15098, 63365, -79614, 51752, -117194, 196980, -156321, 209085, -435223, 463497, -441950, 871202, -1146187, 1023944, -1704179

OFFSET

0,3

COMMENTS

Weigh transform of A055615.

FORMULA

G.f.: Product_{k>=1} (1 + x^k)^(mu(k)*k).

EXAMPLE

G.f.: A(x) = 1 + x - 2*x^2 - 5*x^3 + 4*x^5 + 9*x^6 + 2*x^7 - 10*x^8 - 21*x^9 + 29*x^10 + 15*x^11 - 18*x^12 - 80*x^13 + ...

MATHEMATICA

terms = 49; CoefficientList[Series[Product[(1 + x^k)^(MoebiusMu[k] k), {k, 1, terms}], {x, 0, terms}], x]

terms = 49; A[_] = 1; Do[A[x_] = (1 + x)/Product[A[x^k]^k, {k, 2, terms}] + O[x]^(terms + 1) // Normal, terms + 1]; CoefficientList[A[x], x]

CROSSREFS
KEYWORD

allocated

sign

AUTHOR

Ilya Gutkovskiy, Apr 19 2019

STATUS

approved

editing

#2 by Ilya Gutkovskiy at Fri Apr 19 22:12:44 EDT 2019
NAME

allocated for Ilya Gutkovskiy

KEYWORD

recycled

allocated

#1 by Russ Cox at Sun Jan 27 08:30:53 EST 2019
KEYWORD

recycled

STATUS

approved