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.

Showing 1-4 of 4 results.

A247012 Consider the 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, 133, 172, 841, 1005, 1603, 4258, 5299, 192901, 498906, 1633303, 5307589, 16333303, 20671542, 41673714, 42999958, 73687923
Offset: 1

Views

Author

Paolo P. Lava, Sep 09 2014

Keywords

Comments

A072234 is a subset of this sequence.
a(18) > 2*10^8. - Tyler Busby, Mar 19 2023

Examples

			Aliquot parts of 1005 are 1, 3, 5, 15, 67, 201 and 335:
  1 + 3 + 5 + 15 + 67 + 201 + 335 = 627;
  3 + 5 + 15 + 67 + 201 + 335 + 627 = 1253;
  5 + 15 + 67 + 201 + 335 + 627 + 1253 = 2503;
  15 + 67 + 201 + 335 + 627 + 1253 + 2503 = 5001 that is the reverse of 1005.
Aliquot parts of 1603 are 1, 7 and 229:
  1 + 7 + 229 = 237;
  7 + 229 + 237 = 473;
  229 + 237 + 473 = 939;
  237 + 473 + 939 = 1649;
  473 + 939 + 1649 = 3061 that is the reverse of 1603;

Crossrefs

Cf. A072234, A097090, A246544, A247013.

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 2 to q do if not isprime(n) then
a:=sort([op(divisors(n))]); b:=nops(a)-1; c:=ilog10(n)+1;
for k from 1 to b do v[k]:=a[k]; od;
t:=b+1; v[t]:=add(v[k], k=1..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-b..t-1);
if R(v[t])=n then print(n); break; fi; od; fi; fi; od;
end: P(10^9, 1000);
  • Mathematica
    A247012 = {};
For[n = 4, n <= 1000000, n++,
 If[PrimeQ[n], Continue[]];
 r = IntegerReverse[n];
 a = Most[Divisors[n]];
 sum = Total[a];
 While[sum < r, sum = Total[a = Join[Rest[a], {sum}]]];
 If[sum == r, AppendTo[A247012, n]];
]; A247012 (* Robert Price, Sep 08 2019 *)
  • Python
    from sympy import isprime, divisors
A247012_list = []
for n in range(2,10**9):
    m = int(str(n)[::-1])
    if not isprime(n):
        x = divisors(n)
        x.pop()
        y = sum(x)
        while y < m:
            x, y = x[1:]+[y], 2*y-x[0]
        if y == m:
            A247012_list.append(n) # Chai Wah Wu, Sep 12 2014

Extensions

a(9), a(11)-a(17) from Chai Wah Wu, Sep 13 2014

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.

Original entry on oeis.org

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

Views

Author

Paolo P. Lava, Sep 09 2014

Keywords

Comments

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

Examples

			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.

Crossrefs

Cf A097090, A212875, A247012.

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,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);
  • Mathematica
    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 *)
  • Python
    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

Extensions

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

A107099 G.f. satisfies A(A(x)) = x + 4*x^3, where A(x) = Sum_{n>=0} a(n)*x^(2*n+1).

Original entry on oeis.org

1, 2, -6, 36, -266, 2028, -13596, 50088, 566694, -16598580, 232284876, -1912070088, 631155132, 239439857272, -2781218767224, -17362458802992, 795693633448710, -458070639409908, -335724554310292548, 4520379769156382616, 109439050270732883028, -3828757746830590219608
Offset: 0

Views

Author

Paul D. Hanna, May 13 2005

Keywords

Comments

Coefficients [x^n] A(x) = 0 (mod 3) except at n = 3^k (conjecture).

Examples

			A(x) = 1*x + 2*x^3 - 6*x^5 + 36*x^7 - 266*x^9 + 2028*x^11 - 13596*x^13 +-...

Links

Crossrefs

Cf. A027436, A097090.

Programs

  • PARI
    b(n) = local(A,B,F);F=x+4*x^3+x*O(x^n);A=F;if(n==0,0, for(i=0,n,B=serreverse(A);A=(A+subst(B,x,F))/2);polcoeff(A,n,x));
a(n) = b(2*n+1);

A372580 Expansion of g.f. A(x) satisfying A( A(x) - 4*A(x)^2 + 4*A(x)^3 ) = x.

Original entry on oeis.org

1, 2, 10, 66, 498, 4056, 34644, 305310, 2749110, 25142172, 232728588, 2176116348, 20532197196, 195344525540, 1872680305544, 18073069864926, 175419949070118, 1710976713480396, 16761489153049788, 164888041322062428, 1628416166697339324, 16136415431311552992, 160333972547949898584
Offset: 1

Views

Author

Paul D. Hanna, Jun 27 2024

Keywords

Comments

a(38) = -19599187894639322176080463718044944 is the first negative term.

Examples

			G.f.: A(x) = x + 2*x^2 + 10*x^3 + 66*x^4 + 498*x^5 + 4056*x^6 + 34644*x^7 + 305310*x^8 + 2749110*x^9 + 25142172*x^10 + 232728588*x^11 + 2176116348*x^12 + ...
where A( A(x)*(1 - 2*A(x))^2 ) = x.
RELATED SERIES.
Let B(x) = A(A(x)), then B(x)/x is the g.f. of A369510:
B(x) = x + 4*x^2 + 28*x^3 + 240*x^4 + 2288*x^5 + 23296*x^6 + 248064*x^7 + 2728704*x^8 + 30764800*x^9 + ... + 2^(n-1)*binomial(3*n-2,n-1)/n * x^n + ...
where B(x)*(1 - 2*B(x))^2 = x.
Let R(x) be the series reversion, R(A(x)) = x, then
R(x) = A(x)*(1 - 2*A(x))^2 = x - 2*x^2 - 2*x^3 - 6*x^4 - 22*x^5 - 80*x^6 - 228*x^7 - 18*x^8 + 6694*x^9 + ... + (-1)^(n-1)*A097090(n)*x^n + ...
where R(R(x)) = x*(1 - 2*x)^2.

Links

Crossrefs

Cf. A097090, A369510, A006013.

Programs

  • PARI
    {a(n) = my(A=x); for(k=2,n+1, A=truncate(A); A = (A + serreverse( A*(1 - 2*A)^2 +x*O(x^k)))/2 ); polcoeff(A,n)}
for(n=1,30, print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.
(1) x = A( A(x)*(1 - 2*A(x))^2 ).
(2) x = A(A( x*(1 - 2*x)^2 )).
(3) x = B(x)*(1 - 2*B(x))^2 where B(x) = A(A(x)) (A369510).
(4) A(x) = Series_Reversion( A(x)*(1 - 2*A(x))^2 ).
Showing 1-4 of 4 results.

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