A293520 Number of dying nodes (withering branches) at generation n in the binary tree of persistently squarefree numbers (A293230).

0, 0, 0, 0, 1, 2, 2, 1, 2, 3, 3, 4, 9, 6, 13, 11, 21, 32, 40, 52, 60, 64, 90, 129, 169, 242, 321, 434, 549, 808, 1026, 1395, 1929, 2551, 3405, 4578, 6131, 8275, 11196, 14814, 20198, 26823, 36295, 48840, 65337, 87634, 118138, 158324, 212870, 287014

Offset: 0

Author

Antti Karttunen, Oct 12 2017

Keywords

• nonn

Comments

Provided that A293441 is strictly growing, then certainly a(n) < A293441(n), because only even nodes may die and A293441(n-1) gives the number of even nodes at level n.

Examples

			a(4) = 1 because in the binary tree illustrated in A293230, it is the only node 22 at the level 4 that does not generate any new buds as both 2*22 = 44 and 1+(2*22) = 45 are nonsquarefree numbers.
		

Crossrefs

Cf. A293230, A293441, A293521, A293522.

Programs

PARI

\\ Compute sequences A293230, A293520, A293521, A293522 at the same time:
allocatemem(2^30);
next_living_bud_or_zero(n) = if(issquarefree(n),n,0);
nextA293230generation(tops) = { my(new_tops = vecsort(vector(2*#tops,i,next_living_bud_or_zero((2*tops[(i+1)\2])+((i+1)%2))),,8)); if(0==new_tops[1], vector(#new_tops-1,i,new_tops[1+i]), new_tops); }
write_counts(n,tops) = { my(w=0, s=0, b=0, k); for(i=1,#tops,if((tops[i]%2), if(issquarefree(1+(2*tops[i])), b++, s++), if(issquarefree(1+(2*tops[i])), s++, w++)); ); write("b293520.txt", n, " ", w); write("b293521.txt", n, " ", s); write("b293522.txt", n, " ", b); write("b293230.txt", n, " ", k=length(tops)); print1(k, ", ");}
tops_of_tree = [1];
write("b293230.txt", 0, " ", 1);
write("b293520.txt", 0, " ", 0);
write("b293521.txt", 0, " ", 0);
write("b293522.txt", 0, " ", 1);
print1(1, ", ");
for(n=1,52,tops_of_tree = nextA293230generation(tops_of_tree); write_counts(n,tops_of_tree););

Formula

a(n) = Sum_{k=(2^n)..(2^(1+n)-1)} abs(A293233(k))*[0 == A008966(2*k)+A008966(1+2*k)].