A046197 -id:A046197 - cp's OEIS frontend

A226064 Largest fixed point in base n for the sum of the fourth power of its digits.

1, 1, 243, 419, 1, 1, 273, 1824, 9474, 10657, 1, 8194, 1, 53314, 47314, 36354, 1, 246049, 53808, 378690, 170768, 185027, 1, 247507, 1, 1002324, 722739, 278179, 301299, 334194, 1004643, 959859, 1, 1538803, 1798450, 1, 4168450, 2841074, 1, 1877793, 5556355

Offset: 2

Kevin L. Schwartz and Christian N. K. Anderson, May 24 2013

All fixed points in base n have at most 5 digits. Proof: In order to be a fixed point, a number with d digits in base n must meet the condition n^d <= d*(n-1)^4, which is only possible for d < 5.

For 5-digit numbers vwxyz in base n, only numbers where v*n^4 + n^3 - 1 <= v^4 + 3*(n-1)^4 or v*n^4 + n^4 - 1 <= v^4 + 4*(n-1)^4 are possible fixed points. v <= 2 for n <= 250.

			The fixed points in base 8 are {1,16,17,256,257,272,273}, because in base 8, these are written as {1,20,21,400,401,420,421} and 1^4 = 1, 2^4 + 0^4 = 16, 2^4 + 1^4 = 17, 4^4 + 0^4 + 0^4 = 256, etc. The largest of these is 273 = a(8).

Christian N. K. Anderson, Table of n, a(n) for n = 2..250
Christian N. K. Anderson, Table of base, largest fixed point, number of fixed points, and a list of all fixed points in base 10 and base n for n = 1..250

Cf. A226063 (number of fixed points).
Cf. A052455 (fixed points in base 10).
Cf. A023052, A046074, A046197.

for(b in 2:50) {
    fp=c()
    for(w in 1:b-1) for(x in 1:b-1) if((v1=w^4+x^4)<=(v2=w*b^3+x*b^2))
        for(y in 1:b-1) if((u1=v1+y^4)<=(u2=v2+y*b) & u1+b^4>u2+b-1) {
            z=which(u1+(1:b-1)^4==u2+(1:b-1))-1
            if(length(z)) fp=c(fp,u2+z)
        }
    cat("Base",b,":",fp[-1],"\n")
}

A346630 Smallest number which reaches the narcissistic number 153 after n steps when repeatedly summing the cubes of its digits.

Original entry on oeis.org

153, 135, 18, 3, 9, 12, 33, 114, 78, 126, 6, 117, 669, 177, 12558, 44499999999999999999

Offset: 0

Jörg Zurkirchen, Jul 25 2021

All the terms a(n) as well as the intermediate results will be multiples of 3:
x^3 mod 3 = x mod 3 [0^3 = 0; 1^3 = 1; (-1)^3 = -1].
Therefore (sum of cubes of digits) mod 3 = (sum of digits) mod 3.
Because the only multiple of 3 in A046197 is 153, every number which is a multiple of 3 will end up at 153.
Some other terms (not dealt with here) may reach a cycle of length > 1:
Elizabeth Todd has shown that only numbers (1 mod 3) and (2 mod 3) may reach a cycle, and the only possible cycles are {55, 230, 130}, {136, 244}, {160, 217, 352}, {919, 1459}. That means that numbers (0 mod 3) never reach a cycle but just a single number, namely 153.
Shyam Sunder Gupta tested all the multiples of 3 less than 10^5. He found that they all reach 153, in accordance with the above statements.
The values a(n) for n>15 are really too big to be fully written out (and so are missing in the list), as Jon E. Schoenfield calculated for n=16 and n=17:
a(16) = 3.777999...999*10^61042524005486970; it has one 3, three 7's, and 61042524005486967 9's, so the sum of the cubes of its digits is 1*3^3 + 3*7^3 + 61042524005486967*9^3 = 44499999999999999999 = a(15).
a(17) consists of the digit string 45888 followed by a very, very long string of 9's. The number of 9's in that string is (a(16) - 1725)/729, which is a 61042524005486968-digit number consisting of the digit 5 followed by 753611407475147 copies of the 81-digit string 182441700960219478737997256515775034293552812071330589849108367626886145404663923 followed by a single instance of the 60-digit string 182441700960219478737997256515775034293552812071330589849106.

			a(3) = 3, for 3^3 = 27, 2^3 + 7^3 = 351, 3^3 + 5^3 + 1^3 = 153.
a(13) = 177, for 177 -> 687 -> 1071 -> 345 -> 216 -> 225 -> 141 -> 66 -> 432 -> 99 -> 1458 -> 702 -> 351 -> 153 (13 = longest chain for numbers up to 10^4).
The process ends because 153 = 1^3 + 5^3 + 3^3.

Table of n, a(n) for n = 0..15.
Shyam Sunder Gupta, Curious Properties of 153.
Shyam Sunder Gupta, Beauty of Number 153, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 15, 399-410.
Elizabeth Todd, Happy numbers, p. 11.

Cf. A055012 (sum of cubes of digits), A182111 (number of steps to a cycle), A165330 (cycle end), A046156.
Cf. A046197 (proving that {0, 1, 153, 370, 371, 407} are the only possible fixed points for all numbers of any size when repeatedly summing the cubes of its digits).
Cf. A346789 (concluding the number from the sum of the cubes of its digits).

Table[k=0;While[Last[s=NestList[Total[IntegerDigits@#^3]&,k,n]]!=153||Count[s,153]!=1,k=k+3];k,{n,0,14}] (* Giorgos Kalogeropoulos, Jul 30 2021 *)

A065138 Numbers that are equal to the sum of their digits plus the sum of cubes of their digits.

Original entry on oeis.org

0, 12, 30, 666, 870, 960, 1998

Offset: 1

Henry Bottomley, Oct 15 2001

			12 is included because 12 = (1 + 2) + (1^3 + 2^3).

Underwood Dudley, Numerology or, what Pythagoras wrought, Spectrum Series, Chapter 7 - Beastly Curiosities, page 71, Mathematical Association of America, Washington, D.C., 1997. [From Robert G. Wilson v, Apr 27 2009]

Terry Trotter, Trotter Math. News, 1998[broken link]

Cf. A046197, A065137.

fQ[n_] := Block[{id = IntegerDigits@ n}, Plus @@ id + Plus @@ (id^3) == n]; Select[ Range[0, 2000], fQ@# &] (* Robert G. Wilson v, Apr 27 2009 *)

A165332 Duplicate of A031179.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 17, 18, 19, 20, 21, 23, 24, 26, 27, 29, 30, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 58, 59, 60, 62, 63, 65, 66, 67, 68, 69, 70

Offset: 1

Reinhard Zumkeller, Sep 17 2009

dead

Original title: Numbers that eventually reach a fixed point under "x -> sum of cubes of digits of x" (see A055012, A046197).

Table of n, a(n) for n = 1..57

A307417 Numbers that can be expressed in a base in such a way that the sum of cubes of their digits in this base equals the original number.

Original entry on oeis.org

8, 9, 16, 17, 27, 28, 29, 35, 43, 54, 55, 62, 64, 65, 72, 91, 92, 99, 118, 125, 126, 127, 128, 133, 134, 152, 153, 189, 190, 216, 217, 224, 243, 244, 250, 251, 280, 307, 341, 342, 343, 344, 351, 370, 371, 407, 432, 433, 468, 469, 512, 513, 514, 520, 539, 559

Offset: 1

César Eliud Lozada, Apr 07 2019

There are infinitely many such numbers (proof in the second Johnson link).

			a(1) = 8 = [2, 0] (base 4) =  2^3 + 0^3
a(2) = 9 = [2, 1] (base 4) =  2^3 + 1^3
a(3) = 16 = [2, 2] (base 7) =  2^3 + 2^3
a(4) = 17 = [1, 2, 2] (base 3) =  1^3 + 2^3 + 2^3

César Eliud Lozada, Table of n, a(n) for n = 1..114
Allan Wm. Johnson Jr., Crux Mathematicorum, Vol. 5, No. 1, January 1979, problem 407, 16.
Allan Wm. Johnson Jr., Crux Mathematicorum, Vol. 5, No. 9, November 1979, solution to problem 407, 273-277.

Cf. A005188, A023052, A046197.

sqn:= []; lis:=[];
for n to 1000 do
  b := 2;
  while b < n do #needs to be adjusted
    q := convert(n, base, b);
    s := convert(map(proc (X) options operator, arrow; X^3 end proc, q), `+`);
    if evalb(s = n) then
      sqn := [op(sqn), n];
      lis := [op(lis), [n, b, ListTools[Reverse](q)]];
      break
    end if;
    b := b+1
  end do
end do;
lis := lis; #list of decompositions [number, base, conversion]
sqn := sqn; #sequence

A383347 Numbers that have the same set of digits as the sum of the cubes of their digits.

Original entry on oeis.org

0, 1, 135, 137, 153, 173, 307, 315, 317, 351, 370, 371, 407, 470, 513, 531, 703, 704, 713, 730, 731, 740, 3007, 3070, 3700, 4007, 4070, 4700, 7003, 7004, 7030, 7040, 7300, 7400, 11112, 11113, 11121, 11131, 11211, 11311, 12111, 12599, 12959, 12995, 13111, 15299

Offset: 1

Jean-Marc Rebert, Apr 24 2025

Contains 3*10^j + 7*10^k and 4*10^j + 7*10^k if j <> k and max(j,k) >= 2. - Robert Israel, Apr 25 2025

			135 and 1^3 + 3^3 + 5^3 = 153 have the same set of digits {1,3,5}, so 135 is a term.

Cf. A055012, A249515.

Cf. A046197 (a subsequence).

filter:= proc(n) local L,t;
  L:= convert(n,base,10);
  convert(L,set) = convert(convert(add(t^3,t=L),base,10),set)
end proc:
select(filter, [$0 .. 10^5]); # Robert Israel, Apr 25 2025

q[k_] := Module[{d = IntegerDigits[k]}, Union[d] == Union[IntegerDigits[Total[d^3]]]]; Select[Range[0, 16000], q] (* Amiram Eldar, Apr 24 2025 *)

isok(k) = my(d=digits(k)); Set(d) == Set(digits(sum(i=1, #d, d[i]^3))); \\ Michel Marcus, Apr 24 2025

def ok(n): return set(s:=str(n)) == set(str(sum(int(d)**3 for d in s)))
print([k for k in range(2*10**4) if ok(k)]) # Michael S. Branicky, Apr 24 2025

A334601 Positive integers m such that sum of cubes of the digits of m, t=A055012(m), is a multiple of m (m/A055012(m) is an integer >= 1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 24, 27, 37, 48, 153, 370, 371, 407, 459

Offset: 1

Zak Seidov, May 07 2020 and May 12 2020

Corresponding values of t: 1, 8, 27, 64, 125, 216, 343, 512, 729, 72, 351, 370, 576, 153, 370, 371, 407, 918 (first 9 terms are all cubes).
Corresponding values of t/m: 1, 4, 9, 16, 25, 36, 49, 64, 81, 3, 13, 10, 12, 1, 1, 1, 1, 2 (first 9 terms are all squares).
The subsequence of numbers m such that sum of cubes of its digits is equal to m is A046197 \ {0}. - Bernard Schott, May 11 2020

			m = 459, t = 4^3 + 5^3 + 9^3 = 918, t/m = 2.

Cf. A005188, A046197, A046459, A055012, A055642, A101337, A306354, A306360, A306361, A329291.

Select[Range[500], Divisible[Plus @@ (IntegerDigits[#]^3), #] &] (* Amiram Eldar, May 11 2020 *)

isok(m) = my(d=digits(m)); sum(k=1, #d, d[k]^3) % m == 0; \\ Michel Marcus, May 14 2020

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A226064 Largest fixed point in base n for the sum of the fourth power of its digits.

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A346630 Smallest number which reaches the narcissistic number 153 after n steps when repeatedly summing the cubes of its digits.

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A065138 Numbers that are equal to the sum of their digits plus the sum of cubes of their digits.

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A165332 Duplicate of A031179.

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A307417 Numbers that can be expressed in a base in such a way that the sum of cubes of their digits in this base equals the original number.

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A383347 Numbers that have the same set of digits as the sum of the cubes of their digits.

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A334601 Positive integers m such that sum of cubes of the digits of m, t=A055012(m), is a multiple of m (m/A055012(m) is an integer >= 1).

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