A247012 -id:A247012 - cp's OEIS frontend

A247013 Consider the prime factors, with multiplicity, in ascending order, of a composite number not ending in 0. 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.

4, 9, 14, 94, 194, 371, 1887, 1994, 11282, 25656, 61081, 66691, 112082, 394407, 582225, 4284191, 5681778, 9317913, 9361072, 9493615, 19120874, 75519134, 92688481

Offset: 1

Paolo P. Lava, Sep 09 2014

Similar to A212875 but reading the sum backwards.

If numbers ending in 0 are allowed, 1200 has factors 2,2,2,2,3,5,5 which add up to 21. - Chai Wah Wu, Sep 12 2014

			Prime factors of 1994 are 2 and 997;
2 + 997 = 999;
997 + 999 = 1996;
999 + 1996 = 2995;
1996 + 2995 = 4991 that is the reverse of 1994.
Prime factors of 25656 are 2^3, 3 and 1069;
2 + 2 + 2 + 3 + 1069 = 1078;
2 + 2 + 3 + 1069 + 1078 = 2154;
2 + 3 + 1069 + 1078 + 2154 = 4306;
3 + 1069 + 1078 + 2154 + 4306 = 8610;
1069 + 1078 + 2154 + 4306 + 8610 = 17217;
1078 + 2154 + 4306 + 8610 + 17217 = 33365;
2154 + 4306 + 8610 + 17217 + 33365 = 65652 that is the reverse of 25656.

Cf A097090, A212875, A247012.

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,d,j,k,n,t,v; v:=array(1..h);
for n from 2 to q do if not isprime(n) then a:=ifactors(n)[2];
b:=nops(a); c:=ilog10(n)+1; t:=0; d:=[];
for k from 1 to b do for j from 1 to a[k,2] do d:=[op(d),a[k,1]]; od;
od; d:=sort(d); for k from 1 to nops(d) do v[k]:=d[k]; od; a:=nops(d);
t:=a; t:=t+1; v[t]:=add(v[k],k=1..t-1); 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-a..t-1);
if R(v[t])=n then print(n); break; fi; od; fi; fi; od; end:
P(10^6, 1000);

A247013 = {};
For[n = 4, n <= 1000000, n++,
 If[Mod[n, 10] == 0 || PrimeQ[n], Continue[]];
 r = IntegerReverse[n];
 a = Flatten[Map[Table[#[[1]], {#[[2]]}] &, FactorInteger[n]]];
 sum = Total[a];
 While[sum < r, sum = Total[a = Join[Rest[a], {sum}]]];
 If[sum == r, AppendTo[A247013, n]];
]; A247013 (* Robert Price, Sep 08 2019 *)

from itertools import chain
from sympy import isprime, factorint
A247013_list = []
for n in range(2,10**8):
    m = int(str(n)[::-1])
    if n % 10 and not isprime(n):
        x = sorted(chain.from_iterable([p]*e for p,e in factorint(n).items()))
        y = sum(x)
        while y < m:
            x, y = x[1:]+[y], 2*y-x[0]
        if y == m:
            A247013_list.append(n) # Chai Wah Wu, Sep 12 2014

More terms and definition edited by Chai Wah Wu, Sep 12 2014

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

Paolo P. Lava, Oct 02 2014

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.

			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.

Paolo P. Lava, Table of n, a(n) for n = 1..410
Paolo P. Lava, List of [x,a,b,y] in the equation x = a*F(y) + b*F(y+1), for 1< x <10^6

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

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

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Author

Paolo P. Lava, May 22 2015

Keywords

nonn
more

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

Eric Weisstein's World of Mathematics, Unitary Divisor

Eric Weisstein's World of Mathematics, Unitary Divisor Function

Eric Weisstein's World of Mathematics, Unitary Perfect Number

Wikipedia, Unitary divisor

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

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Author

Paolo P. Lava, May 25 2015

Keywords

nonn
more
base

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

Eric Weisstein's World of Mathematics, Unitary Divisor

Eric Weisstein's World of Mathematics, Unitary Divisor Function

Eric Weisstein's World of Mathematics, Unitary Perfect Number

Wikipedia, Unitary divisor

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

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Author

Paolo P. Lava, Oct 15 2015

Keywords

nonn

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

Maple

with(numtheory):P:=proc(q,h) local a,b,k,t,v; global n; 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 sigma(b[k])>2*b[k] 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

seqQ[n_] := Module[{d = Select[Most[Divisors[n]], DivisorSigma[1, #] > 2 # &]}, Switch[Length[d], ?(# < 1 &), False, ?(# == 1 &), d[[1]] == n, , k = 0; While[k < n, k = Total[d]; d = Rest[AppendTo[d, k]]]; k == n]]; seq = {}; Do[ If[seqQ[n], AppendTo[seq, n]], {n, 2, 10^6}]; seq (* _Amiram Eldar, Mar 20 2019 *)

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

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Author

Paolo P. Lava, May 02 2019

Keywords

nonn

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

Eric Weisstein's World of Mathematics, Unitary Divisor

Wikipedia, Unitary divisor

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

Showing 1-6 of 6 results.

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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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

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.

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Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

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.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

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.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

Extensions

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.

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Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Extensions