cp's OEIS Frontend

A064630 Number of parts if 4^n is partitioned into parts of size 3^n as far as possible into parts of size 2^n as far as possible and into parts of size 1^n.

%I A064630 #12 Jun 21 2018 02:00:25
%S A064630 2,5,5,16,25,15,66,121,146,771,1220,3641,8093,15843,28359,50236,33366,
%T A064630 36709,145250,137776,548024,2186496,1066102,4251976,16984368,28678103,
%U A064630 13620614,205950171,100716646,381399635,1397934923,3826001641
%N A064630 Number of parts if 4^n is partitioned into parts of size 3^n as far as possible into parts of size 2^n as far as possible and into parts of size 1^n.
%C A064630 Corresponds to the only solution of the Diophantine equation 4^n = x*3^n + y*2^n + z*1^n with constraints 0 <= y < 3^n/2^n, 0 <= z < 2^n.
%C A064630 Binary order (cf. A029837) of a(n) is close to n.
%H A064630 Harry J. Smith, <a href="/A064630/b064630.txt">Table of n, a(n) for n = 1..250</a>
%F A064630 a(n) = A064628(n) + floor(A064629(n)/2^n) + (A064629(n) mod 2^n) = floor(4^n/3^n) + floor((4^n mod 3^n)/2^n) + ((4^n mod 3^n) mod 2^n)
%e A064630 4^6 = 4096 = 729 + 729 + 729 + 729 + 729 + 64 + 64 + 64 + 64 + 64 + 64 + 64 + 1 + 1 + 1 = 5*3^6 + 7*2^6 + 3*1^6, so a(6) = 5 + 7 + 3 = 15.
%o A064630 (PARI) {for(n=1,32,a=divrem(4^n,3^n); b=divrem(a[2],2^n); print1(a[1]+b[1]+b[2],","))}
%o A064630 (PARI) { f=t=w=1; for (n=1, 250, f*=4; t*=3; w*=2; a=divrem(f, t); b=divrem(a[2], w); write("b064630.txt", n, " ", a[1]+b[1]+b[2]) ) } \\ _Harry J. Smith_, Sep 20 2009
%Y A064630 Cf. A064628, A064629, A060692, A029837.
%K A064630 nonn
%O A064630 1,1
%A A064630 _Labos Elemer_, Oct 01 2001
%E A064630 Edited by _Klaus Brockhaus_, May 24 2003