A372556 a(n) = largest number k <= A130249(n) for which A372555(n-A001045(k)) = A372555(n)-1, where A372555(n) is the least number of Jacobsthal numbers that add up to n.

0, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 0

Author

Antti Karttunen, May 07 2024

Keywords

• nonn

Comments

An auxiliary sequence for computing A372555 with a mutually recursive algorithm.

Differs from A130249 for the first time at n=63, 84, 191, 212, 255, etc. See A372558.

Conjecture: For all n, either a(n) = A130249(n) or a(n) = A130249(n)-1. In other words, there is always a minimal solution (in number of summands) for representing n as a sum of Jacobsthal numbers that its largest summand is either A001045(A130249(n)) [same as obtained with a greedy algorithm A265745], or the next smaller Jacobsthal number. - Antti Karttunen, May 10 2024

Links

• Antti Karttunen, Table of n, a(n) for n = 0..87381

Crossrefs

Cf. A001045, A130249, A147612, A372556, A372558.

Programs

PARI

up_to = 87381; \\ = A001045(18).
A001045(n) = (2^n - (-1)^n) / 3;
A130249(n) = (#binary(3*n+1)-1);
A372555_or_556list(up_to_n,return_556_instead) = { my(v372555 = vector(up_to_n), v372556 = vector(up_to_n)); v372555[1] = 1; v372556[1] = 2; for(n=2,#v372556, my(m=-1,mk=-1,s=A130249(n)); if(A001045(s)==n, v372555[n] = 1; v372556[n] = s, forstep(k=s, 1, -1, my(c=v372555[n-A001045(k)]); if(m<0 || cA001045(mk)])); if(return_556_instead,v372556,v372555); };
v372556 = A372555_or_556list(up_to,1);
A372556(n) = if(!n,n,v372556[n]);

Scheme

;; Use the program given in A372555.