A244314 Nonnegative integers m such that the Semigreedy Catalan representation A244159(m) contains at least one zero.

0, 2, 5, 6, 7, 14, 15, 16, 17, 18, 19, 20, 21, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157

Offset: 0

Author

Antti Karttunen, Jun 25 2014

Keywords

• nonn

Comments

Starting offset is zero because A244159(0) = 0 is a borderline case (either one zero, or no zeros if leading zeros are discarded).

From a(1)=2 onward the positions of zeros in A244233.

After zero consists of successive subsequences containing terms from A000108(k) to (A000108(k)+A014138(k-2)-1) computed from k >= 2 onward, as: [2], [5,6,7], [14 .. 21], [42 .. 63], [132 .. 195], [429 .. 624], [1430 .. 2054], [4862 .. 6916], etc.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..9891

Crossrefs

Subsequence of A244217.

Cf. A000108, A014138, A014143, A244233, A244159, A244217, A244317, A081291, A081293.

Programs

Scheme

(define (A244314 n) (cond ((zero? n) 0) ((= 1 n) 2) (else (+ -1 n (- (A000108 (+ 1 (A244317 n))) (A014143 (- (A244317 n) 2)))))))

Formula

a(0) = 0, a(1) = 2, and for n >= 2, a(n) = n + A000108(1+A244317(n)) - A014143(A244317(n)-2) - 1.