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.

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A379397 Numbers that can be written in exactly two different ways as a sum of at most nine positive third powers.

Original entry on oeis.org

8, 9, 16, 27, 28, 29, 30, 32, 33, 34, 36, 37, 40, 41, 42, 43, 44, 48, 49, 51, 54, 55, 57, 58, 59, 60, 61, 62, 63, 66, 69, 71, 73, 74, 76, 77, 78, 79, 80, 85, 87, 88, 90, 95, 101, 102, 103, 104, 106, 109, 111, 114, 115, 116, 117, 122, 123, 239
Offset: 1

Views

Author

Patrick De Geest, Dec 22 2024

Keywords

Comments

The 'nine' is not arbitrary. Waring stated that every natural number can be expressed as a sum of at most nine cubes (cf. A002804).

Examples

			29 is in the sequence since 1^3 + 1^3 + 3^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3.
123 is in the sequence since 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3.

Links

Crossrefs

Cf. A002804, A025453, A379396, A379398, A379399, A379400.

Programs

  • PARI
    upto(n) = my(v=vector(n), maxb=sqrtnint(n,3)); forvec(x=vector(9,i,[0,maxb]), s=sum(i=1,9,x[i]^3); if(0x==2,v,1) \\ David A. Corneth, Dec 23 2024

A379398 Numbers that can be written in exactly three different ways as a sum of at most nine positive third powers.

Original entry on oeis.org

35, 56, 64, 65, 67, 68, 70, 75, 81, 82, 83, 84, 86, 89, 92, 93, 94, 96, 97, 98, 99, 100, 105, 107, 108, 110, 112, 113, 118, 119, 120, 121, 124, 125, 127, 130, 141, 142, 143, 148, 149, 150, 151, 167, 169, 174, 175, 176, 177, 178, 183, 186, 188, 202, 204, 212, 213, 214, 240, 247, 303
Offset: 1

Views

Author

Patrick De Geest, Dec 22 2024

Keywords

Comments

The 'nine' is not arbitrary. Waring stated that every natural number can be expressed as a sum of at most nine cubes. (Cf. A002804)

Examples

			67 is in the sequence since 1^3+1^3+1^3+4^3 = 2^3+2^3+2^3+2^3+2^3+3^3 = 1^3+1^3+1^3+1^3+1^3+2^3+3^3+3^3.

Links

Crossrefs

Cf. A002804, A025453, A379396, A379397, A379399, A379400

Programs

  • PARI
    upto(n) = my(v=vector(n), maxb=sqrtnint(n,3)); forvec(x=vector(9,i,[0,maxb]), s=sum(i=1,9,x[i]^3); if(0x==3,v,1)) \\ David A. Corneth, Dec 23 2024

A379399 Numbers that can be written in exactly four different ways as a sum of at most nine positive third powers.

Original entry on oeis.org

72, 91, 126, 128, 129, 131, 132, 134, 135, 136, 137, 138, 139, 140, 144, 146, 147, 154, 155, 156, 157, 158, 162, 164, 165, 166, 168, 170, 171, 172, 173, 179, 180, 181, 184, 185, 187, 191, 193, 194, 195, 199, 203, 205, 206, 207, 210, 211, 215, 221, 228, 229, 230, 231, 232, 241, 242, 266, 267, 293, 295, 319, 330, 338, 366, 455
Offset: 1

Views

Author

Patrick De Geest, Dec 23 2024

Keywords

Comments

The 'nine' is not arbitrary. Waring stated that every natural number can be expressed as a sum of at most nine cubes (cf. A002804).
Conjecture: this sequence is finite and a(66) = 455 is the last term. Verified up to 10^8. - Charles R Greathouse IV, Dec 28 2024

Examples

			215 is in the sequence since 1^3+2^3+3^3+3^3+3^3+5^3 = 1^3+1^3+2^3+2^3+2^3+4^3+5^3 = 2^3+2^3+3^3+3^3+3^3+3^3+3^3+4^3 = 1^3+2^3+2^3+2^3+2^3+3^3+3^3+4^3+4^3.

Links

Crossrefs

Cf. A002804, A025453, A379396, A379397, A379398, A379400.

Programs

  • PARI
    upto(n) = my(v=vector(n), maxb=sqrtnint(n, 3)); forvec(x=vector(9, i, [0, maxb]), s=sum(i=1, 9, x[i]^3); if(0x==4, v, 1) \\ David A. Corneth, Dec 23 2024

A379400 Numbers that can be written in exactly five different ways as a sum of at most nine positive third powers.

Original entry on oeis.org

133, 145, 152, 153, 159, 160, 161, 163, 182, 190, 192, 196, 197, 198, 200, 201, 209, 218, 220, 223, 225, 226, 233, 234, 235, 236, 237, 238, 246, 248, 249, 255, 258, 265, 268, 269, 274, 277, 279, 292, 300, 302, 304, 305, 310, 311, 312, 329, 331, 340, 356, 364, 365, 429
Offset: 1

Views

Author

Patrick De Geest, Dec 23 2024

Keywords

Comments

The 'nine' is not arbitrary. Waring stated that every natural number can be expressed as a sum of at most nine cubes (A002804).

Examples

			133 is in the sequence since 2^3+5^3 = 1^3+1^3+1^3+1^3+1^3+4^3+4^3 = 1^3+2^3+2^3+2^3+3^3+3^3+3^3+3^3 = 1^3+1^3+1^3+1^3+1^3+1^3+1^3+1^3+5^3 = 1^3+1^3+2^3+2^3+2^3+2^3+2^3+3^3+4^3.

Links

Crossrefs

Cf. A002804, A025453, A379396, A379397, A379398, A379399.

Programs

  • PARI
    upto(n) = my(v=vector(n), maxb=sqrtnint(n, 3)); forvec(x=vector(9, i, [0, maxb]), s=sum(i=1, 9, x[i]^3); if(0x==5, v, 1) \\ David A. Corneth, Dec 23 2024

A267826 Numbers not of the form w^3 + 2*x^3 + 3*y^3 + 4*z^3, where w, x, y and z are nonnegative integers.

Original entry on oeis.org

18, 22, 39, 60, 63, 74, 76, 77, 100, 103, 106, 107, 117, 126, 178, 180, 201, 215, 228, 230, 245, 271, 289, 291, 295, 315, 341, 356, 357, 393, 413, 419, 420, 480, 481, 523, 559, 606, 616, 671, 673, 705, 854, 855, 963, 980, 981, 998, 1103, 1121, 1130, 1298, 1484, 1510, 1643, 1729, 1849, 1916, 1934, 1946
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 07 2016

Keywords

Comments

Conjecture: The sequence has exactly 122 terms the last of which is a(122) = 41405.
We have verified that there are no terms between 41406 and 2*10^5.
The conjecture implies that {P(v)+w^3+2*x^3+3*y^3+4*z^3: w,x,y,z = 0,1,2,...} = {0,1,2,...} whenever P(v) is among the polynomials a*v^3 (a = 1,5,6,7,9,10,12,15,18), b*v^4 (b = 1,2,3,5,6,12,18), c*v^5 (c = 1,2,5,12) and d*v^k (d = 5,12; k = 6,7). Moreover, it also implies that {8*t+w^3+2*x^3+3*y^3+4*z^3: t = 0,1; w,x,y,z = 0,1,2,...} = {0,1,2,...}. If a,b,c,d and m are positive integers with {m*t+a*w^3+b*x^3+c*y^3+d*z^3: t = 0,1; w,x,y,z = 0,1,2,...} = {0,1,2,...}, then we must have m = 8 and {a,b,c,d} = {1,2,3,4}.

Examples

			a(1) = 18 since it is the first nonnegative integer not in the set {w^3 + 2*x^3 + 3*y^3 + 4*z^3: w,x,y,z = 0,1,2,...}.

Links

Crossrefs

Cf. A000578, A002804, A022566, A267861, A271099, A271169, A271237.

Programs

  • Mathematica
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
n=0;Do[Do[If[CQ[m-4*z^3-3y^3-2x^3],Goto[aa]],{z,0,(m/4)^(1/3)},{y,0,((m-4z^3)/3)^(1/3)},{x,0,((m-4z^3-3y^3)/2)^(1/3)}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,0,1946}]

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

A282288 Expansion of (Sum_{k>=0} x^(k^4))^19.

Original entry on oeis.org

1, 19, 171, 969, 3876, 11628, 27132, 50388, 75582, 92378, 92378, 75582, 50388, 27132, 11628, 3876, 988, 513, 2926, 15505, 58140, 162792, 352716, 604656, 831402, 923780, 831402, 604656, 352716, 162792, 58140, 15504, 3078, 3249, 23275, 116280, 406980, 1058148, 2116296, 3325608, 4157010, 4157010, 3325608
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 12 2017

Keywords

Comments

Number of ways to write n as an ordered sum of 19 fourth powers (A000583).
a(n) > 0 for all n >= 0.
Every natural number is the sum of at most 19 fourth powers (Balasubramanian, 1986).

Links

Crossrefs

Cf. A000583, A002377, A002804, A014110, A173682.

Programs

  • Mathematica
    nmax = 42; CoefficientList[Series[Sum[x^k^4, {k, 0, nmax}]^19, {x, 0, nmax}], x]

Formula

G.f.: (Sum_{k>=0} x^(k^4))^19.

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.

A174420 Let A(n) = floor((3/2)^n), B(n)=3^n-2^n*A(n); then a(n)=2^n-A(n)-B(n)-2.

Original entry on oeis.org

-2, -2, -1, 0, 8, 4, 26, 98, 68, 245, 284, 941, 908, 2921, 866, 3038, 9773, 95842, 26864, 82811, 776048, 235984, 715436, 2157533, 14878043, 27882168, 16575521, 116892244, 82326503, 515542801, 1009949246, 882651721, 500902958, 1503356036, 4511038850, 13534572662, 40605902153, 53101505973, 21870478820, 65618808017
Offset: 0

Views

Author

N. J. A. Sloane, Nov 27 2010

Keywords

Comments

Related to Waring's problem.
a(n) is conjectured to be >= 0 for all n >= 3. If this were proved it would settle Waring's problem (see A002804). It is known that a(n) >= 0 for 3 <= n <= 471600000.
If we rewrite the formula as (2^n-1)*ceiling((3/2)^n) - 3^n - 1, we see more clearly a comparison between 3^n and the product of an undervaluation of 2^n and an overvaluation of (3/2)^n. If the undervaluation is proportionately smaller than the ceiling overvaluation, a(n) is nonnegative. 2^n grows faster than (3/2)^n, so for a negative value to occur the target difference between (3/2)^n and ceiling((3/2)^n) gets smaller as n gets larger, and the sum of these target differences (for n > 0) is finite. - Peter Munn, Dec 08 2022

References

  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 337.
  • R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, pp. 285-324 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.

Links

Crossrefs

Cf. A002804.

Programs

  • Mathematica
    a[n_] := Floor[(3/2)^n]; b[n_] := 3^n - 2^n*a[n]; Table[2^n - a[n] - b[n] - 2, {n, 0, 39}] (* Jean-François Alcover, Apr 05 2011 *)
  • Python
    def A174420(n): return ((m:=3**n)&-(k:=1<>n)-2 # Chai Wah Wu, Jun 25 2024

A267861 Number of ways to write n as 2*t + u^4 + v^4 + 2*w^4 + 3*x^4 + 4*y^4 + 6*z^4, where t is 0 or 1, and u, v, w, x, y, z are nonnegative integers with u <= v and v > 0.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 5, 5, 6, 5, 5, 5, 4, 3, 4, 3, 3, 4, 3, 4, 5, 5, 5, 6, 5, 5, 5, 4, 3, 3, 3, 2, 4, 2, 4, 4, 5, 5, 6, 5, 5, 6, 4, 4, 3, 3, 2, 4, 2, 4, 4, 4, 5, 6, 5, 6, 6, 4, 4, 4, 3, 2, 4, 2, 4, 5, 6, 5, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 07 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 1, 2, 111, 127, 143, 158, 221, 223, 240, 460, 463, 480, 545, 560, 561, 1455, 1695, 1776, 2175. Moreover, any integer n > 10^4 not among 10543, 17935, 37583, 40383, 78543 can be written as u^4 + v^4 + 2*w^4 + 3*x^4 + 4*y^4 + 6*z^4 with u,v,w,x,y,z nonnegative integers.
If a(1),...,a(7) are positive integers with a(1) <= a(2) <= ... <= a(7) and a(1)+...+a(7) = g(4) = 19 such that {a(1)*x(1)^4+...+a(7)*x(7)^4: x(1),...,x(7) = 0,1,2,...} = {0,1,2,...}, then the tuple (a(1),...,a(7)) must be (1,1,2,2,3,4,6) or (1,1,2,2,3,3,7). Similarly, if a(1),...,a(8) are positive integers with a(1) <= a(2) <= ... <= a(8) and a(1)+...+a(8) = g(5) = 37 such that {a(1)*x(1)^5+...+a(8)*x(8)^5: x(1),...,x(8) = 0,1,2,...} = {0,1,2,...}, then (a(1),...,a(8)) must be (1,1,2,3,4,6,8,12) or (1,1,2,3,4,5,7,14).

Examples

			a(111) = 1 since 111 = 2*1 + 2^4 + 3^4 + 2*1^4 + 3*0^4 + 4*1^4 + 6*1^4.
a(240) = 1 since 240 = 2*0 + 2^4 + 2^4 + 2*0^4 + 3*2^4 + 4*2^4 + 6*2^4.
a(1776) = 1 since 1776 = 2*0 + 4^4 + 5^4 + 2*3^4 + 3*3^4 + 4*1^4 + 6*3^4.
a(2175) = 1 since 2175 = 2*1 + 0^4 + 4^4 + 2*2^4 + 3*5^4 + 4*1^4 + 6*1^4.

Links

Crossrefs

Cf. A000583, A000584, A002804, A267826, A271099, A271169, A271237.

Programs

  • Mathematica
    QQ[n_]:=QQ[n]=n>0&&IntegerQ[n^(1/4)]
Do[r=0;Do[If[QQ[n-2t-6*z^4-4y^4-3x^4-2w^4-u^4],r=r+1],{t,0,Min[1,n/2]},{z,0,((n-2t^8)/6)^(1/4)},{y,0,((n-2t-6z^4)/4)^(1/4)},{x,0,((n-2t-6z^4-4y^4)/3)^(1/4)},
{w,0,((n-2t-6z^4-4y^4-3x^4)/2)^(1/4)},{u,0,((n-2t-6z^4-4y^4-3x^4-2w^4)/2)^(1/4)}];Print[n," ",r];Continue,{n,1,70}]
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