|
|
A239686
|
|
Numbers n such that if n = a U b (where U denotes concatenation) then sigma*(a) + sigma*(b) = abs(sigma*(n) - n), where sigma*(n) is the sum of the anti-divisors of n.
|
|
1
|
|
|
47, 118, 205, 846, 898, 1219, 4181, 4236, 4701, 4929, 6014, 6516, 13276, 30445, 59956, 61916, 63216, 67314, 72066, 79554, 90674, 106316, 128998, 129179, 136816, 142486, 143396, 180448, 229914, 284894, 357841, 421318, 483286, 486721, 487618, 500218, 642445
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Neither a or b minor than 2 are considered because numbers 1 and 2 have no anti-divisors.
Similar to A239562 but using anti-divisors instead of divisors.
|
|
LINKS
|
|
|
EXAMPLE
|
Anti-divisors of 4701 are 2, 6, 7, 17, 79, 119, 1343, 553, 3134 and their sum is 5260. Consider 4701 as 4 U 701. Anti-divisors of 4 is 3 and of 701 are 2, 3, 23, 61, 467 whose sum is 556. At the end we have that 5260 - 4701 = 559 = 3 + 556.
|
|
MAPLE
|
with(numtheory);
T:=proc(t) local w, x, y; x:=t; y:=0; while x>0 do x:=trunc(x/10); y:=y+1; od; end:
P:=proc(q) local a, b, c, d, f, g, i, j, k, n;
for n from 1 to q do b:=T(n); k:=0; j:=n; while j mod 2<>1 do k:=k+1; j:=j/2; od;
a:=sigma(2*n+1)+sigma(2*n-1)+sigma(n/2^k)*2^(k+1)-6*n-2;
for i from 1 to b-1 do c:=trunc(n/10^i); d:=n-c*10^i; if c>2 and d>2 then
k:=0; j:=c; while j mod 2<>1 do k:=k+1; j:=j/2; od;
f:=sigma(2*c+1)+sigma(2*c-1)+sigma(c/2^k)*2^(k+1)-6*c-2;
k:=0; j:=d; while j mod 2<>1 do k:=k+1; j:=j/2; od;
g:=sigma(2*d+1)+sigma(2*d-1)+sigma(d/2^k)*2^(k+1)-6*d-2;
if f+g=a-n then print(n); break; fi; fi; od; od; end: P(10^9);
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base,hard
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|