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Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=(k-r)*(k-r+1)/2, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=((k-r)*(k-r+1)/2)*a(j).
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%I #18 Jul 27 2014 12:22:23

%S 1,1,3,1,1,3,6,1,1,1,3,3,3,6,10,1,1,1,3,1,1,3,6,3,3,3,9,6,6,10,15,1,1,

%T 1,3,1,1,3,6,1,1,1,3,3,3,6,10,3,3,3,9,3,3,9,18,6,6,6,18,10,10,15,21,1,

%U 1,1,3,1,1,3,6,1,1,1,3,3,3,6,10,1,1,1,3,1,1,3,6,3,3,3,9,6,6,10,15

%N Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=(k-r)*(k-r+1)/2, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=((k-r)*(k-r+1)/2)*a(j).

%C See A245196 for a list of other sequences produced by this type of recurrence.

%C It follows from the definition that the final entries in the blocks are triangular numbers.

%e Arranged into blocks:

%e 1,

%e 1, 3,

%e 1, 1, 3, 6,

%e 1, 1, 1, 3, 3, 3, 6, 10,

%e 1, 1, 1, 3, 1, 1, 3, 6, 3, 3, 3, 9, 6, 6, 10, 15,

%e 1, 1, 1, 3, 1, 1, 3, 6, 1, 1, 1, 3, 3, 3, 6, 10, 3, 3, 3, 9, 3, 3, 9, 18, 6, 6, 6, 18, 10, 10, 15, 21,

%e ...

%p G:=[seq(n,n=0..30)];

%p m:=1;

%p f:=proc(n) option remember; global m,G; local k,r,j,np;

%p k:=1+floor(log[2](n)); np:=2^k-n;

%p if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;

%p if j=0 then G[k-r]; else m*G[k-r]*f(j); fi;

%p end;

%p [seq(f(n),n=1..120)];

%Y Cf. A245196, A245547.

%K nonn,tabf

%O 1,3

%A _N. J. A. Sloane_, Jul 26 2014