A177148 -id:A177148 - cp's OEIS frontend

A118881 Square of sum of decimal digits of n.

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 36, 49, 64, 81, 100, 121, 144

Offset: 0

Giovanni Teofilatto, May 25 2006

a(k) = k iff k = 0, 1, 81; also, the only solution to the double equation a(k) = m and a(m) = k with k < m is (169, 256) (proof in Diophante link, 2ème jonglerie). - Bernard Schott, Mar 08 2021

If the map i->a(i) is iterated starting at i = n, the trajectory will eventually reach one ot the three cycles (1) (if n == +-1 mod 9), (169,256) (if n == 2,4,5,7 mod 9), or (81) (if n == 0 mod 3). - N. J. A. Sloane, Mar 17 2025

			From _R. J. Mathar_, Jul 08 2012: (Start)
Trajectories of the map x->a(x), A177148:
1 ->1 ->1 ->1 ->1 ->1 ->1 ->1 ->1 ->...
2 ->4 ->16 ->49 ->169 ->256 ->169 ->256 ->169 ->...
3 ->9 ->81 ->81 ->81 ->81 ->81 ->81 ->81 ->...
4 ->16 ->49 ->169 ->256 ->169 ->256 ->169 ->256 ->...
5 ->25 ->49 ->169 ->256 ->169 ->256 ->169 ->256 ->...
6 ->36 ->81 ->81 ->81 ->81 ->81 ->81 ->81 ->...
7 ->49 ->169 ->256 ->169 ->256 ->169 ->256 ->169 ->...
8 ->64 ->100 ->1 ->1 ->1 ->1 ->1 ->1 ->... (End)

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Diophante, A1915, Jongleries n°2 avec les chiffres (in French).
Michael Penn, Squaring the sum of digits, YouTube video, 2021.

Cf. A007953.

read("transforms") :
A118881 := proc(n)
        digsum(n)^2 ;
end proc: # R. J. Mathar, Jul 08 2012

Table[Total[IntegerDigits[n]]^2,{n,0,70}] (* Harvey P. Dale, Jul 31 2012 *)

a(n) = sumdigits(n)^2; \\ Michel Marcus, Mar 08 2021

def a(n): return sum(map(int, str(n)))**2
print([a(n) for n in range(67)]) # Michael S. Branicky, Nov 19 2021

a(n) = A007953(n)^2. [R. J. Mathar, Apr 22 2010]

A182128 Number of iterations of the map n -> (sum of the decimal digits of n)^3 before reaching the last number of the cycle.

Original entry on oeis.org

0, 0, 2, 3, 3, 2, 3, 3, 1, 2, 1, 2, 3, 3, 2, 3, 3, 1, 2, 2, 2, 3, 3, 2, 3, 3, 1, 2, 2, 2, 3, 3, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 2, 3, 1, 2, 2, 2, 2, 3, 2, 2, 3, 1, 2, 2, 2, 2, 3, 2

Offset: 0

Michel Lagneau, Apr 13 2012

a(n) is the number of times that the cube of the sum of the digits must be calculated before reaching the last number of the cycle.

			0 is in the sequence twice because 0 -> 0 and 1 -> 1;
a(3) = 3:
    3 ->       3^3 = 27;
   27 ->   (2+7)^3 = 729;
  729 -> (7+2+9)^3 = 18^3 = 5832 is the end of the map because 5832 -> (5+8+3+2)^3 = 18^3 is already in the trajectory. Hence we obtain the map: 3 -> 27 -> 729 -> 5832 with 3 iterations.

Cf. A177148, A178481, A118880.

A182128 := proc(n)
        local traj ,c;
        traj := n ;
        c := [n] ;
        while true do
                traj := A118880(traj) ;
                if member(traj,c) then
                        return nops(c)-1 ;
                end if;
                c := [op(c),traj] ;
        end do:
end proc:
seq(A182128(n),n=0..80) ; # R. J. Mathar, Jul 08 2012

A178481 Number of steps of the map x -> A055566(x), starting at n, before reaching the end of the cycle.

Original entry on oeis.org

0, 0, 5, 3, 4, 4, 2, 3, 2, 2, 1, 3, 2, 4, 2, 1, 2, 1, 2, 2, 5, 3, 3, 1, 1, 3, 1, 1, 0, 1, 3, 1, 2, 1, 1, 0, 0, 1, 3, 1, 3, 2, 2, 2, 1, 1, 0, 3, 2, 3, 4, 2, 4, 2, 1, 2, 3, 1, 5, 4, 2, 4, 1, 2, 2, 3, 1, 4, 4, 1, 4, 1, 2, 2, 3, 2, 3, 4, 2, 4, 2

Offset: 0

Michel Lagneau, May 28 2010

a(n) is the number of times taking the 5th powers of the sums of digits before reaching a sum seen before (reaching the last number of the cycle).
Example:
6 -> 6^5 = 7776 -> (7+7+7+6)^5 = 27^5.
27^5 = 14348907 -> (1+4+3+4+8+9+0+7)^5 = 36^5.
36^5 = 60466176, last number of the cycle because (6+0+4+6+6+1+7+6)^5 = 36^5 = 60466176 belongs to the list.
Generalization for the k-th powers and conjecture: For each k >= 1, iteration of taking the k-th powers of digit sums reaches a cycle.
Example with k = 17; start with 3.
3^17 = 129140163, sum = 27,
27^17 = 2153693963075557766310747, sum = 117,
117^17 = 144264558065210807467328187211661877, sum = 153,
153^17 = 13796036156758195415808856807283698713, sum = 189,
189^17 = 501014933601411817143935347829544613629, sum = 153 is already in the set.
[It remains unclear whether the author wanted to define iterations of (sumofdigits of n)^5, compatible with A177148 and A182128, or sumofdigits(n^5) here. I've taken the latter to be more compliant with the first terms of the original submission. - R. J. Mathar, Jul 08 2012]

			a(0) = 0 and a(1) = 0 because 0 -> 0 and 1 -> 1.
a(15) = 1 because 15^5 = 759375 -> (7+5+9+3+7+5) = 36,
36 ^5 = 60466176 -> (6+0+4+6+6+1+7+6) = 36.

Cf. A177148, A182128.

A178481 := proc(n)
        local traj ,c;
        traj := n ;
        c := [n] ;
        while true do
                traj := A055566(traj) ;
                if member(traj,c) then
                        return nops(c)-1 ;
                end if;
                c := [op(c),traj] ;
        end do:
end proc:
seq(A178481(n),n=0..80) ; # R. J. Mathar, Jul 08 2012

A182129 Number of iterations of the orbit n -> (sum of the decimal digits of n)^n starting with n, needed to stabilize.

Original entry on oeis.org

0, 5, 3, 3, 5, 3, 3, 5, 5, 1, 6, 6, 2, 6, 9, 5, 2, 6, 2, 7, 5, 5, 6, 6, 6, 3, 5, 2, 9, 7, 6, 13, 12, 9, 5, 9, 2, 10, 9, 7, 15, 9, 7, 4, 7, 2, 6, 3, 7, 12, 6, 9, 9, 5, 2, 10, 12, 10, 14, 7, 8, 8, 11, 2, 13, 10, 5, 9, 8, 15, 9, 6, 2, 17, 13, 8, 9, 5, 15, 12

Offset: 1

Michel Lagneau, Apr 13 2012

a(n) is the number of times you form the n-power of the sum of the digits before reaching the last number of the cycle.
Generalization and conjecture: Let k be a positive integer. The number of iterations of the orbit k -> (sum of the decimal digits of k)^n is finite for any exponent n and any starting value k.
Example with n = 17; start with k = 3.
3^17 = 129140163, sum of the decimal digits = 27,
27^17 = 2153693963075557766310747, sum of the decimal digits = 117,
117^17 = 144264558065210807467328187211661877, sum of the decimal digits = 153,
153^17 = 13796036156758195415808856807283698713, sum of the decimal digits = 189,
189^17 = 501014933601411817143935347829544613629, sum of the decimal digits = 153 is already in the trajectory.

			0 is in the sequence 1^1 -> 1;
For the power 2, a(2) = 5:
    2 ->       2^2 =   4;
    4 ->       4^2 =  16;
   16 ->   (1+6)^2 =  49;
   49 ->   (4+9)^2 = 169;
  169 -> (1+6+9)^2 = 256 is the end of the cycle because 256 -> (2+5+6)^2 = 169 is already in the trajectory. Hence we obtain the map: 2 -> 4 -> 16 -> 49 -> 169 -> 256 with 5 iterations.

Cf. A177148, A182128.

with(numtheory) : T :=array(1..500) :W:=array(1..500):for n from 1 to 80 do : k:=0:nn:=n:for it from 1 to 50 do:T :=convert(nn,base,10) :l:=nops(T):s1:=sum(T[i],i=1..l):s:=s1^n:k:=k+1:W[k]:=s:nn:=s:od: z:= [seq(W[i],i=1..k)]:V:=convert(z,set):n1:=nops(V): printf(`%d, `,n1):od:

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A118881 Square of sum of decimal digits of n.

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A182128 Number of iterations of the map n -> (sum of the decimal digits of n)^3 before reaching the last number of the cycle.

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A178481 Number of steps of the map x -> A055566(x), starting at n, before reaching the end of the cycle.

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A182129 Number of iterations of the orbit n -> (sum of the decimal digits of n)^n starting with n, needed to stabilize.

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Maple