A002377 Least number of 4th powers needed to represent n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 5, 1, 2, 3

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• nice

Comments

No terms are greater than 19, see A002804. - Charles R Greathouse IV, Aug 01 2013

Seven values of n need the maximum of 19 fourth powers. These form the arithmetic progression {79, 159, 239, 319, 399, 479, 559} each term being congruent to 79 mod 80. For n < 625 the available fourth powers are congruent to 1 or 16 mod 80, requiring 4*16 + 15*1 to sum to 79. However, 625 = 5^4 is congruent to 65 and 1*65 + 14*1 = 79. So for n > 625 and congruent to 79, only 15 fourth powers are needed to satisfy the mod 80 arithmetic. - Peter Munn, Apr 12 2017

References

• D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 82.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• T. D. Noe, Table of n, a(n) for n = 1..14000
• C. A. Bretschneider, Zerlegung der Zahlen bis 4100 in Biquadrate, J. Reine Angew. Math., 46 (1853), 1-28.
• Eric Weisstein's World of Mathematics, Biquadratic Number

Crossrefs

Cf. A002828, A002376, A188462, A374012.

Cf. A046049, A046050, A099591.

Programs

Mathematica

Cnt4[n_] := Module[{k = 1}, While[Length[PowersRepresentations[n, k, 4]] == 0, k++]; k]; Array[Cnt4, 100] (* T. D. Noe, Apr 01 2011 *)
seq[n_] := Module[{v = Table[0, {n}], s, p}, s = Sum[x^(k^4), {k, 1, n^(1/4)}] + O[x]^(n+1); p=1; For[k=1, k <= 19, k++, p *= s; For[i=1, i <= n, i++, If[v[[i]]==0 && Coefficient[p, x, i] != 0, v[[i]] = k]]]; v];
seq[100] (* Jean-François Alcover, Sep 28 2019, after Andrew Howroyd *)

PARI

seq(n)={my(v=vector(n), s=sum(k=1, sqrtint(sqrtint(n)), x^(k^4)) + O(x*x^n), p=1); for(k=1, 19, p*=s; for(i=1, n, if(!v[i] && polcoeff(p,i), v[i]=k))); v} \\ Andrew Howroyd, Jul 06 2018

Python

from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A002377(n):
    if n == 1: return 1
    for k in count(1):
        try:
            next(power_representation(n,4,k))
        except:
            continue
        return k # Chai Wah Wu, Jun 25 2024

Extensions

More terms from Arlin Anderson (starship1(AT)gmail.com)