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 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

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

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)

A274459 Least number of perfect powers that add up to n.

Original entry on oeis.org

1, 2, 3, 1, 2, 3, 4, 1, 1, 2, 3, 2, 2, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 1, 2, 1, 2, 2, 3, 2, 1, 2, 2, 2, 1, 2, 3, 3, 2, 2, 3, 2, 2, 2, 3, 3, 2, 1, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 3, 2, 1, 2, 3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 1, 2, 3, 3, 2
Offset: 1

Views

Author

Sergio Pimentel, Jun 23 2016

Keywords

Comments

Least number of perfect powers (A001597) needed to add up to n.
This sequence is close to but not exactly equal to A063274.
a(n) is at most 4 since any number can be written as a sum of 4 squares (Lagrange's theorem), but it is possible that for a sufficiently large n, a(n) < 4.
a(n) <= a(i) + a(n-i) for 1 <= i <= n-1. (for computational ease, the maximum value for i can be chosen as floor(n/2)). a(1991) = 4. for 1992 <= k <= 20000, there is no k such that a(k) = 4. - David A. Corneth, Jun 24 2016 [Next such k is 25887, see A113505. - Vaclav Kotesovec, Jun 25 2016]

Examples

			a(31) = 2 since 31 can be written as the sum of two (31 = 3^3 + 2^2 = 27 + 4) but no fewer than two perfect powers.

Links

Crossrefs

Cf. A001597, A002376, A002377, A002828, A002804, A079611, A063274, A188462, A225926.
Cf. A063275 (indices for which a(n)=3), A113505 (indices for which a(n)=4).

Programs

  • Mathematica
    nn = 72; t = Select[Range@ nn, # == 1 || GCD @@ FactorInteger[#][[All, 2]] > 1 &]; Table[Min@ Map[Length, Select[IntegerPartitions@ n, AllTrue[#, MemberQ[t, #] &] &]], {n, nn}] (* Michael De Vlieger, Jun 23 2016, after Ant King at A001597 *)
  • PARI
    lista(n) = {my(v = vector(n)); for(i = 2,sqrtint(n), for(j = 2, logint(n, i), v[i^j] = 1)); v[1]=1; v[2]=2; for(i=3, #v, if(v[i]==0, v[i] = vecmin(vector( i\2, k,v[k] + v[i-k]))));v} \\ David A. Corneth, Jun 24 2016; corrected by Peter Schorn, Jun 09 2022

Extensions

More terms from Michael De Vlieger, Jun 23 2016
Terms from a(74) from David A. Corneth, Jun 24 2016

A374012 Least number of 6th powers needed to represent n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16
Offset: 1

Views

Author

Seiichi Manyama, Jun 25 2024

Keywords

Comments

a(703) = 73.

References

  • Pillai, S. S. (1940) On Waring’s problem g(6) = 73. Proc. Indian Acad. Sci. 12A: 30-40

Links

Crossrefs

Cf. A002828, A002376, A002377, A188462.
Cf. A002804, A018886.

Programs

  • PARI
    a_vector(n, k=6) = my(v=vector(n), cnt=0, d=0, p=1, s=sum(j=1, sqrtnint(n, k), x^j^k)+x*O(x^n)); while(cnt
  • Python
    from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A374012(n):
    if n == 1: return 1
    for k in count(1):
        try:
            next(power_representation(n,6,k))
        except:
            continue
        return k # Chai Wah Wu, Jun 25 2024

Formula

a(n) <= 73.
Showing 1-3 of 3 results.

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