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%I #14 Nov 28 2015 19:34:22
%S 0,0,0,0,1,1,0,0,0,0,0,1,0,0,1,0,1,1,0,0,0,0,2,0,0,0,0,1,0,1,0,0,1,0,
%T 0,1,0,0,0,0,0,1,0,0,1,0,2,0,1,0,0,1,0,0,0,0,0,2,1,0,0,0,1,0,0,1,0,0,
%U 0,1,1,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,1,0,0,0,0,1,1,0,0,0,0,1,0,0,1,2,0,0,1,0,0,0,1,1,1,0
%N a(n) = number of leaf-children n has in the tree generated by edge-relation A049820(child) = parent.
%C a(n) = number of such terms k in A045765 for which k - d(k) = n [where d(k) is the number of divisors of k, A000005(k)].
%H Antti Karttunen, <a href="/A262900/b262900.txt">Table of n, a(n) for n = 0..65538</a>
%F a(n) = Sum_{k = A082284(n) .. A262686(n)} [A049820(k) = n] * [A060990(k) = 0].
%F In the above formula [ ] stands for Iverson bracket, giving in the first instance as its result 1 only when A049820(k) = n (that is, when k is really a child of n), and 0 otherwise, and in the second instance 1 only when A060990(k) = 0 (that is, when k itself has no children), and 0 otherwise. - Comment corrected by _Antti Karttunen_, Nov 27 2015
%e a(4) = 1, as there is only one such term k in A045765 which satisfies the condition A049820(k) = 4, namely 8 (8 - d(8) = 4).
%e a(5) = 1, as the only term in A045765 satisfying the condition is 7, as 7 - d(7) = 5.
%e a(22) = 2, as there are exactly two terms in A045765 satisfying the condition, namely 25 and 28, as 25 - d(25) = 28 - d(28) = 22.
%o (Scheme)
%o (define (A262900 n) (let loop ((s 0) (k (A262686 n))) (cond ((<= k n) s) ((= n (A049820 k)) (loop (+ s (if (zero? (A060990 k)) 1 0)) (- k 1))) (else (loop s (- k 1))))))
%Y Cf. A000005, A045765, A049820, A060990, A082284, A262686.
%Y Cf. A262901 (indices of nonzero terms), A262902.
%K nonn
%O 0,23
%A _Antti Karttunen_, Oct 06 2015
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