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Number of partitions of n into parts with digital root = 6.
11

%I #16 Sep 11 2024 11:37:44

%S 0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,1,0,0,2,0,0,1,0,0,3,0,0,2,0,

%T 0,3,0,0,3,0,0,4,0,0,4,0,0,6,0,0,5,0,0,7,0,0,7,0,0,9,0,0,9,0,0,12,0,0,

%U 11,0,0,15,0,0,15,0,0,18,0,0,19,0,0,23,0,0,23,0,0,29,0,0,29,0,0,35,0,0,37

%N Number of partitions of n into parts with digital root = 6.

%C a(n) = A114102(n) - A116371(n) - A116372(n) - A116373(n) - A116374(n) - A116375(n) - A116377(n) - A116378(n) - A114099(n).

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DigitalRoot.html">Digital Root</a>

%F a(n) = A035386(floor(n/3))*0^(n mod 3).

%F G.f.: Product_{j>=0} 1/(1 - x^(6+9*j)). - _Robert Israel_, Apr 13 2015

%e a(30) = #{24+6, 15+15, 6+6+6+6+6} = 3.

%p N:= 1000: # to get a(1) to a(N)

%p g:= mul(1/(1-x^(6+9*j)), j=0..floor((N-6)/9)):

%p S:= series(g, x, N+1):

%p seq(coeff(S,x,j),j=1..N); # _Robert Israel_, Apr 13 2015

%Y Cf. A010888.

%Y Cf. A147706.

%K nonn,base

%O 1,24

%A _Reinhard Zumkeller_, Feb 12 2006