A243970 Smallest positive integer m such that n can be expressed as a partial sum of the divisors of m taken in decreasing order.

1, 1, 2, 2, 3, 5, 4, 4, 7, 6, 10, 6, 6, 9, 8, 8, 16, 10, 10, 19, 15, 14, 12, 14, 14, 12, 26, 12, 12, 29, 16, 16, 21, 18, 34, 20, 18, 37, 18, 18, 27, 20, 20, 43, 24, 30, 46, 33, 32, 28, 24, 34, 39, 28, 24, 28, 28, 24, 58, 24, 24, 30, 32, 32, 64, 65, 30, 67, 51

Offset: 0

Author

Michel Marcus, Jun 16 2014

Keywords

• nonn

Comments

Sequence is similar to A167485, but here, the partial sums are evaluated in decreasing order starting from the highest divisor of n, n, down to the smallest one, 1. Thus for any n>0, a(n) exists and is at most equal to n: the highest divisor of n.

Examples

			From n=1 to 2, these partial sums are: 1; 2, 3. So 3 has appeared in the partial divisors sums of 2. Hence a(3)=2.
a(11) = 6 as 11 can be written as 6 + 3 + 2 where 6, 3 and 2 are some divisors of 6 in decreasing order starting at 6. No smaller number than 6 lets us write 11 like so, proving a(11) = 6. - _David A. Corneth_, Nov 29 2024
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..20000

Crossrefs

Cf. A167485.

Programs

PARI

ps(n) = {vps = []; d = divisors(n); ips = 0; forstep (i=#d, 1, -1, ips += d[i]; vps = concat(vps, ips);); vps;}
a(n) = {if (n==0, return (1)); i=1; found=0; while (! found, v = ps(i); if (vecsearch(v, n), found=1, i++);); i;}

PARI

first(n) = {
	n--;
	my(res = [1..n]);
	for(i = 1, n,
		d = divisors(i);
		s = 0;
		forstep(j = #d, 1, -1,
			s+=d[j];
			if(s <= n,
				res[s] = min(res[s], i);
			,
				next(2);
			);
		);
	);
	concat(1, res);
} \\ David A. Corneth, Nov 29 2024