A258142
Consider the unitary aliquot parts, in ascending order, of a composite number. Take their sum and repeat the process deleting the minimum number and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.
Original entry on oeis.org
6, 21, 60, 85, 90, 261, 976, 2009, 87360, 97273, 4948133, 68353213
Offset: 1
Divisors of 85 are 1, 5, 17, 85. Unitary aliquot parts are 1, 5, 17.
We have:
1 + 5 + 17 = 23;
5 + 17 + 23 = 45;
17 + 23 + 45 = 85.
Divisors of 2009 are 1, 7, 41, 49, 287, 2009.
Unitary aliquot parts are 1, 41, 49. We have:
1 + 41 + 49 = 91;
41 + 49 + 91 = 181;
49 + 91 + 181 = 321;
91 + 181 + 321 = 593;
181 + 321 + 593 = 1095;
321 + 593 + 1095 = 2009.
-
with(numtheory):P:=proc(q,h) local a,b,k,n,t,v; v:=array(1..h);
for n from 1 to q do if not isprime(n) then b:=sort([op(divisors(n))]); a:=[];
for k from 1 to nops(b)-1 do if gcd(b[k],n/b[k])=1 then a:=[op(a),b[k]]; fi; od;
a:=sort(a); b:=nops(a); if b>1 then for k from 1 to b do v[k]:=a[k]; od;
t:=b+1; v[t]:=add(v[k], k=1..b); while v[t]
-
aQ[n_] := Module[{s = Most[Select[Divisors[n], GCD[#, n/#] == 1 &]]}, If[Length[s] == 1, False, While[Total[s] < n, AppendTo[s, Total[s]]; s = Rest[s]]; Total[s] == n]]; Select[Range[2, 10^8], aQ] (* Amiram Eldar, Jan 12 2019 *)
A289868
Consider the digit reverse of a number x. Take the sum of these digits and repeat the process deleting the first addend and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to x.
Original entry on oeis.org
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 17, 21, 25, 42, 63, 84, 143, 286, 2355, 5821, 6618, 11709, 12482, 33747, 39571, 129109, 466957, 1162248, 1565166, 1968084, 3636638, 3853951, 4898376, 13443745, 13933175, 17118698, 22421197, 24153462, 147440984, 209989875, 245742153
Offset: 0
Digit reverse of 17 is 71 and 7 + 1 = 8, 1 + 8 = 9, 8 + 9 = 17;
Digit reverse of 286 is 682 and 6 + 8 + 2 = 16, 8 + 2 + 16 = 26, 2 + 16 + 26 = 44, 16 + 26 + 44 = 86, 26 + 44 + 86 = 156, 44 + 86 + 156 = 286.
-
P:=proc(q) local a,b,k,n; for n from 0 to q do a:=convert(n,base,10); b:=convert(a,`+`); while b
-
Select[Range[10^6], Function[n, Total@ NestWhile[Append[Drop[#, 1], Total@ #] &, Reverse@ IntegerDigits@ n, Total@ # < n &] == n]] (* Michael De Vlieger, Jul 20 2017 *)
-
is(n) = {my(d = Vecrev(digits(n))); while(vecsum(d)David A. Corneth, Jul 20 2017
A307859
Consider the non-unitary aliquot parts, in ascending order, of a composite number. Take their sum and repeat the process deleting the minimum number and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.
Original entry on oeis.org
24, 112, 189, 578, 1984, 2125, 3993, 5043, 9583, 19197, 32512, 126445, 149565, 175689, 225578, 236883, 1589949, 1862935, 1928125, 3171174, 5860526, 6149405, 11442047, 16731741, 60634549, 75062535, 134201344, 177816209, 1162143369, 4474779517, 10369035821
Offset: 1
Divisors of 578 are 1, 2, 17, 34, 289, 578. Non-unitary aliquot parts are 17 and 34.
We have:
17 + 34 = 51;
34 + 51 = 85;
51 + 85 = 136;
85 + 136 = 221;
136 + 221 = 357;
221 + 357 = 578.
-
with(numtheory):P:=proc(q,h) local a,b,c,k,n,t,v; v:=array(1..h);
for n from 1 to q do if not isprime(n) then b:=sort([op(divisors(n))]);
a:=[]; for k from 2 to nops(b)-1 do if gcd(b[k],n/b[k])>1 then
a:=[op(a),b[k]]; fi; od; b:=nops(a); if b>1 then c:=0;
for k from 1 to b do v[k]:=a[k]; c:=c+a[k]: od;
t:=b+1; v[t]:=c; while v[t]
-
aQ[n_] := CompositeQ[n] && Module[{s = Select[Divisors[n], GCD[#, n/#] != 1 &]}, If[Length[s] < 2, False, While[Total[s] < n, AppendTo[s, Total[s]]; s = Rest[s]]; Total[s] == n]]; Select[Range[10^4], aQ] (* Amiram Eldar, May 07 2019 *)
Showing 1-3 of 3 results.
Comments