cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A248134 Consider a number x as a concatenation of two integers, a and b: x = concat(a,b). 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

14, 19, 21, 28, 42, 47, 63, 84, 105, 126, 147, 149, 168, 189, 199, 298, 323, 497, 646, 795, 911, 969, 1292, 1499, 1822, 1999, 2087, 2733, 2998, 3089, 3248, 3379, 3644, 4555, 4997, 5411, 5466, 6178, 6377, 6496, 7288, 7995, 8199, 9161, 9267, 9744, 10822, 12356
Offset: 1

Views

Author

Paolo P. Lava, Oct 02 2014

Keywords

Comments

If the number x is rewritten as concat(a,b), the problem is to find a value of y such that x = a*F(y) + b*F(y+1), if a < b, or x = b*F(y) + a*F(y+1), if a > b, where F(y) is a Fibonacci number (see values of x, a, b, y, for 1
Similar to A130792 but here the minimum number is deleted since the beginning.
All the listed numbers admit only one concatenation, concat(a,b), that, through the addition process, leads to themselves. Is there any number that admit more than one single concatenation?
Sequence is infinite. Let us consider the numbers 19, 199, 1999, 19...9 and let us divide them as concat(1,9), concat(1,99), concat(1,999), concat(1,9...9). In two steps we have the initial numbers back: 1 + 9 = 10 and 9 + 10 = 19; 1 + 99 = 100 and 99 + 100 = 199, etc.

Examples

			Let us rewrite 5411 as 54 U 11. Then:
11 + 54 = 65;
54 + 65 =  119;
65 + 119 = 184;
119 + 184 = 303;
184 + 303 = 487;
303 + 487 = 790;
487 + 790 = 1277;
790 + 1277 = 2067;
1277 + 2067 = 3344;
2067 + 3344 = 5411, that is 11*F(10) + 54*F(11) = 11*55 + 54*89 = 605 + 4806 = 5411.

Links

Crossrefs

Cf. A000045, A007629, A130792, A246544, A247012, A247013.

Programs

  • Maple
    P:=proc(q,h) local a,b,k,n,t,v; v:=array(1..h);
for n from 1 to q do for k from 1 to ilog10(n) do
a:=n mod 10^k; b:=trunc(n/10^k); if a

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

Views

Author

Paolo P. Lava, May 22 2015

Keywords

Comments

A002827 is a subset of this sequence.
No more terms below 10^8. - Amiram Eldar, Jan 12 2019

Examples

			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.

Links

Crossrefs

Cf. A002827, A034444, A246544, A247012, A247013, A248134.

Programs

  • Maple
    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]
  • Mathematica
    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 *)

Extensions

a(11)-a(12) from Amiram Eldar, Jan 12 2019

A258270 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 the reverse of themselves.

Original entry on oeis.org

6, 75, 133, 1005, 1603, 4258, 5299, 84292, 89944, 170568, 192901, 303003, 695364, 1633303
Offset: 1

Views

Author

Paolo P. Lava, May 25 2015

Keywords

Examples

			Unitary aliquot parts of 6 are 1, 2, 3. We have: 1 + 2 + 3 = 6 that is equal to its reverse.
Unitary aliquot parts of 75 are 1, 3, 25. We have: 1 + 3 + 25 = 29; 3 + 25 + 29 = 57 that is the reverse of 75.
Unitary aliquot parts of 84292 are 1, 4, 13, 52, 1621, 6484, 21073. We have: 1 + 4 + 13 + 52 + 1621 + 6484 + 21073 = 29248 that is the reverse of 84292.

Links

Crossrefs

Cf. A246544, A247012, A258142.

Programs

  • Maple
    with(numtheory): R:=proc(w) local x, y; x:=w; y:=0;while x>0 do
y:=10*y+(x mod 10); x:=trunc(x/10); od: y; end:
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 a:=sort([op(divisors(n))]);
b:=[]; c:=ilog10(n)+1; for k from 1 to nops(a)-1 do if gcd(a[k],n/a[k])=1
then b:=[op(b),a[k]]; fi; od; if nops(b)>1 then
for k from 1 to nops(b) do v[k]:=b[k]; od; t:=nops(b)+1; v[t]:=add(v[k],k=1..nops(b)); if R(v[t])=n then print(n); else
while ilog10(v[t])+1<=c do t:=t+1; v[t]:=add(v[k], k=t-nops(b)..t-1);
if R(v[t])=n then print(n); break; fi; od; fi; fi; fi; od;
end: P(10^9,1000);

A263344 Consider the abundant 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 number of iterations reach a sum equal to themselves.

Original entry on oeis.org

1700, 5950, 155574, 274550, 300894, 715275, 758625, 1365234, 1404172, 1542500, 1661750, 2095250, 2239750, 2673250, 2962250, 3106750, 3395750, 3829250, 4226625, 4262750, 4407250, 4700619, 5398750, 6371092, 8167635, 8560024, 12305620, 13725855, 15497625, 15586263
Offset: 1

Views

Author

Paolo P. Lava, Oct 15 2015

Keywords

Examples

			Aliquot parts of 1700 are 1, 2, 4, 5, 10, 17, 20, 25, 34, 50, 68, 85, 100, 170, 340, 425, 850. The abundant numbers are 20, 100, 340. Therefore:
20 + 100 + 340 = 460;
100 + 340 + 460 = 900;
340 + 460 + 900 = 1700.

Crossrefs

Cf. A005101 (abundant numbers), A027751 (aliquot parts), A246544, A247012, A258142, A258270.

Programs

Extensions

More terms from Amiram Eldar, Mar 20 2019

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

Views

Author

Paolo P. Lava, May 02 2019

Keywords

Examples

			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.

Links

Crossrefs

Cf. A002827, A034444, A246544, A247012, A247013, A248134, A258142.

Programs

  • Maple
    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]
  • Mathematica
    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 *)

Extensions

a(20)-a(31) from Amiram Eldar, May 07 2019
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