A192279 Anti-hypersigma(n): sum of the anti-divisors of n plus the recursive sum of the anti-divisors of the anti-divisors until 2 is reached.

2, 5, 7, 9, 19, 17, 17, 40, 33, 37, 45, 40, 67, 49, 89, 96, 65, 88, 71, 134, 127, 91, 189, 120, 187, 170, 91, 166, 151, 219, 243, 164, 261, 140, 315, 392, 233, 310, 247, 374, 245, 150, 461, 280, 285, 347, 407, 468, 215, 538, 515, 234, 565, 422, 609, 532, 495

Offset: 3

Author

Paolo P. Lava, Jul 13 2011

Keywords

• nonn
• easy

Comments

Similar to A191150 but using anti-divisors. The recursion is stopped when 2 is reached because 2 has no anti-divisors.

Examples

			n=14 -> anti-divisors are 3,4,9. We start with 3+4+9=16.
Now for 3, 4 and 9 we repeat the procedure:
3-> 2 -> no anti-divisors. To add: 2.
4-> 3 -> 2 -> no anti-divisors. To add: 3+2=5.
9-> 2,6. To add: 2+6=8.
--- 2 -> no anti-divisors.
--- 6 -> 4 -> 3 -> 2 -> no anti-divisors. To add: 4+3+2=9.
Total is 16+2+5+8+9=40.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 3..5000

Crossrefs

Cf. A066272, A191150.

Programs

Maple

with(numtheory);
P:=proc(n)
local a,b,c,k,s;
a:={};
for k from 2 to n-1 do if abs((n mod k)- k/2) < 1 then a:=a union {k}; fi;
od;
b:=nops(a); c:=op(a); s:=0;
if b>1 then
  for k from 1 to b do s:=s+c[k]; od;
else s:=c;
fi;
b:=nops(a); c:=(sort([op(a)]));
for k from 1 to b do if c[k]>2 then s:=s+P(c[k]); fi; od;
s;
end:
Antihps:=proc(i)
local n;
for n from 1 to i do print(P(n)); od;
end: