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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.
3
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
0,2
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
PROG
(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 || c<m, m=c; mk=k)); v372556[n] = mk; v372555[n] = 1+v372555[n-A001045(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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 07 2024
STATUS
approved