|
%I #13 Oct 19 2014 16:03:34
%S 47,118,205,846,898,1219,4181,4236,4701,4929,6014,6516,13276,30445,
%T 59956,61916,63216,67314,72066,79554,90674,106316,128998,129179,
%U 136816,142486,143396,180448,229914,284894,357841,421318,483286,486721,487618,500218,642445
%N 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.
%C Neither a or b minor than 2 are considered because numbers 1 and 2 have no anti-divisors.
%C Similar to A239562 but using anti-divisors instead of divisors.
%e 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.
%p with(numtheory);
%p 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 P:=proc(q) local a, b, c, d, f, g, i, j, k,n;
%p 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;
%p a:=sigma(2*n+1)+sigma(2*n-1)+sigma(n/2^k)*2^(k+1)-6*n-2;
%p 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
%p k:=0; j:=c; while j mod 2<>1 do k:=k+1; j:=j/2; od;
%p f:=sigma(2*c+1)+sigma(2*c-1)+sigma(c/2^k)*2^(k+1)-6*c-2;
%p k:=0; j:=d; while j mod 2<>1 do k:=k+1; j:=j/2; od;
%p g:=sigma(2*d+1)+sigma(2*d-1)+sigma(d/2^k)*2^(k+1)-6*d-2;
%p if f+g=a-n then print(n); break; fi; fi; od; od; end: P(10^9);
%Y Cf. A066272, A066417, A239687.
%K nonn,base,hard
%O 1,1
%A _Paolo P. Lava_, Mar 24 2014
|
