A239562
Numbers n such that n = concatenate(a, b) and sigma(a) + sigma(b) = sigma(n) - n.
Original entry on oeis.org
39, 119, 253, 581, 1875, 2077, 14477, 15879, 17823, 100637, 160529, 232477, 251189, 286437, 506587, 552739, 605729, 806179, 1170695, 3272257, 3295289, 4085129, 4201441, 4657133, 4844701, 5625173, 8106509, 12430289, 23943721, 33857009, 41782973, 64012513
Offset: 1
For n = 232477 we can consider 232477 = 2 U 32477 and sigma(232477) = 265696, sigma(2) = 3, sigma(32477) = 33216 and 265696 - 232477 = 33219 = 3 + 33216.
For n = 251189 we can consider 251189 = 25 U 1189 and sigma(251189) = 252480, sigma(25) = 31, sigma(1189) = 1260 and 252480 - 251189 = 1291 = 31 + 1260.
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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,i,n; for n from 1 to q do a:=sigma(n); b:=T(n);
for i from 1 to b-1 do c:=trunc(n/10^i); d:=n-c*10^i;
if sigma(c)+sigma(d)=a-n then print(n); break; fi; od; od; end: P(10^9);
A239687
Numbers n such that if n = a U b (where U denotes concatenation) then abs(sigma*(a) - sigma*(b)) = abs(sigma*(n) - n), where sigma*(n) is the sum of the anti-divisors of n.
Original entry on oeis.org
54, 436, 2014, 2466, 3365, 4143, 4965, 7922, 9332, 15426, 17554, 24006, 32874, 33574, 39476, 44296, 49976, 54118, 83726, 116174, 137635, 163964, 164824, 177546, 203514, 220789, 235434, 379096, 420716, 476475, 597741, 600354, 604986, 680266, 736306, 748966
Offset: 1
Anti-divisors of 4143 are 2, 5, 6, 1657, 2762 and their sum is 4432. Consider 4143 as 4 U 143. Anti-divisors of 4 is 3 and of 143 are 2, 3, 5, 7, 15, 19, 22, 26, 41, 57, 95 whose sum is 292. At the end we have that 4432 - 4143 = 289 = 292 - 3.
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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 abs(f-g)=abs(a-n) then print(n); break; fi; fi; od; od; end: P(10^9);
Showing 1-2 of 2 results.
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