A293518 Number of surviving even nodes at generation n in the binary tree of persistently squarefree numbers (see A293230).

0, 1, 1, 2, 2, 2, 3, 6, 6, 8, 12, 16, 20, 31, 34, 56, 63, 88, 112, 150, 208, 287, 379, 511, 690, 908, 1239, 1637, 2252, 2945, 4052, 5348, 7203, 9681, 12974, 17432, 23470, 31419, 42254, 57026, 76182, 102845, 137764, 185271, 249065, 334864, 449586, 604164, 811709, 1089661, 1465433, 1968592

Offset: 0

Author

Antti Karttunen, Oct 16 2017

Keywords

• nonn

Examples

			a(3) = 2 because in the binary tree illustrated in A293230, there are two even nodes at the level 3 (namely, the nodes 10 and 14) that spawn just one offspring each.
		

Crossrefs

Cf. A008966, A293230, A293233, A293517, A293519, A293521.

Programs

PARI

\\ Compute the sequences A293441, A293518 and A293519 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); }
writeA293441etc_counts(n,tops) = { my(os=0, es=0, k=0); for(i=1,#tops, if((tops[i]%2), k++; if(!issquarefree(1+(2*tops[i])), os++), if(issquarefree(1+(2*tops[i])), es++));); write("b293441.txt", n, " ", k); write("b293518.txt", n, " ", es); write("b293519.txt", n, " ", os); print1(k, ", ");}
tops_of_tree = [1];
write("b293441.txt", 0, " ", 1);
write("b293518.txt", 0, " ", 0);
write("b293519.txt", 0, " ", 0);
print1(1, ", ");
for(n=1,51,tops_of_tree = nextA293230generation(tops_of_tree); writeA293441etc_counts(n,tops_of_tree););

Scheme

(define (A293518 n) (add (lambda (k) (* (if (and (= 0 (A008966 (+ k k))) (= 1 (A008966 (+ 1 k k)))) 1 0) (abs (A293233 k)))) (A000079 n) (+ -1 (A000079 (+ 1 n)))))
;; Implements sum_{i=lowlim..uplim} intfun(i)
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))

Formula

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

a(n) + A293519(n) = A293521(n).

a(n) - A293519(n) = A293517(n).