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 1-2 of 2 results.

A362953 Numbers N such that N + the sum of the cubes of its digits is again a third power.

Original entry on oeis.org

0, 34, 352, 540, 1167, 1942, 2176, 3312, 4093, 5454, 8019, 9380, 12025, 12130, 13068, 13158, 15344, 15991, 16279, 16675, 21149, 22699, 22789, 30988, 32257, 32365, 35238, 37883, 37955, 41866, 45549, 54523, 57906, 58530, 62579, 72588, 83692, 83782, 89604, 102952
Offset: 0

Views

Author

Will Gosnell and M. F. Hasler, May 09 2023

Keywords

Examples

			The sum of the cubes of the digits of N = 34 is 3^3 + 4^3 = 27 + 64 = 91; added to the number N itself yields  125 which is again a cube, 5^3. Therefore 34 is in this sequence.
		

Crossrefs

Cf. A000578 (the cubes), A055012 (sum of cubes of decimal digits of n).
Cf. A362954 (the same for 4th powers).

Programs

  • PARI
    select( {is(n,p=3)=ispower(vecsum([d^p|d<-digits(n)])+n,p)}, [0..10^5])
    
  • Python
    aupto = 103000
    A362953 = []
    A000578 = set(cu**3 for cu in range(0, int(aupto**(1/3)+3)))
    for n in range(0,aupto+1):
        if n + sum(int(digit)**3 for digit in str(n)) in A000578: A362953.append(n)
    print(A362953) # Karl-Heinz Hofmann, May 24 2023

A362945 Numbers k such that k + the sum of the fifth powers of its digits is again a fifth power.

Original entry on oeis.org

0, 210, 66374, 76933, 294137, 722571, 741300, 1023716, 2412593, 3959235, 4022940, 5090402, 6351886, 7821198, 9653864, 11808061, 11814811, 20427955, 24190287, 24239133, 24245187, 33518020, 33532714, 33536551, 39060306, 52476271, 52480534, 69255266, 69265142, 79091461, 89980491, 90147222
Offset: 0

Views

Author

M. F. Hasler, Jun 15 2023

Keywords

Crossrefs

Cf. A000584 (4th powers), A055014 (sum of 5th powers of decimal digits).
Cf. A362953 (same for 3rd powers), A362954 (same for 4th powers).

Programs

  • PARI
    is_A362945(n,p=5)=ispower(vecsum([d^p|d<-digits(n)])+n,p)
    for(n=0,1e8, is_A362945(n) && print1(n", "))
    
  • Python
    ispower5 = lambda n: n==round(n**.2)**5
    is_A362945 = lambda n: ispower5(sum(int(d)**5 for d in str(n))+n)
    def A362945(n, A=[0]):
        while len(A) < n:
            A.append(A[-1]+1)
            while not is_A362945(A[-1]): A[-1]+=1
        return A[n]
Showing 1-2 of 2 results.