

A130713


a(0)=a(2)=1, a(1)=2, a(n)=0 for n > 2.


3



1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
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OFFSET

0,2


COMMENTS

Selfconvolution of A019590. Up to a sign the convolutional inverse of the natural numbers sequence.  Tanya Khovanova, Jul 14 2007
Iterated partial sums give the chain A130713 > A113311 > A008574 > A001844 > A005900 > A006325 > A033455 > A259181, up to index. The kth term of the nth partial sums is (n^27n+14 + 4k(k+n4))(k+n4)!/(k1)!/(n1)!, for k > 3n. Iterating partial sums in reverse (nth differences with n zeros prepended) gives row (n+3) of A182533, modulo signs and trailing zeros.  Travis Scott, Feb 19 2023


LINKS



FORMULA

G.f.: 1 + 2*x + x^2.


MAPLE



MATHEMATICA



CROSSREFS



KEYWORD

easy,nonn


AUTHOR



STATUS

approved



