A002804 -id:A002804 - cp's OEIS frontend

A065622 Numerator of 1 - (3/4)^n - frac((3/2)^n), where frac(x) = x - floor(x).

0, -1, 3, 13, 159, 173, 1767, 12789, 17759, 126237, 292183, 1930245, 3724303, 23940141, 14206087, 99585429, 640559295, 12562430525, 7042526903, 43417422885, 813747135599, 494896655693, 3000760993767, 18098709141429, 249612172740383

Offset: 0

Henry Bottomley, Dec 03 2001

The presumption that the fraction is positive for n > 1 underlies the presumed solution to Waring's problem.

			a(3) = 13 since 1 - (3/4)^3 - frac((3/2)^3) = 1 - 27/64 - frac(27/8) = 1 - 27/64 - 3/8 = (64 - 27 - 24)/64 = 13/64.

Harry J. Smith, Table of n, a(n) for n = 0..200
Eric Weisstein's World of Mathematics, Waring's Problem

Denominator is A000302. Cf. A002804.

Table[1 - (3/4)^n - FractionalPart[(3/2)^n], {n, 0, 24}] // Numerator (* Jean-François Alcover, Apr 26 2016 *)

{ for (n=0, 200, a=numerator(1 - (3/4)^n - frac((3/2)^n)); write("b065622.txt", n, " ", a) ) } \\ Harry J. Smith, Oct 24 2009

a(n) = 4^n*(1 + floor((3/2)^n)) - 3^n - 6^n = A005061(n) - A002380(n)*A000079(n) = A000302(n)*(1 + A002379(n)) - A000244(n) - A000400(n).

A252487 Smallest k such that n^7 = a_1^7 + ... + a_k^7 and all a_i are positive integers less than n.

Original entry on oeis.org

128, 28, 66, 39, 28, 26, 21, 20, 18, 22, 22, 22, 20, 21, 14, 17, 14, 14, 17, 16, 17, 14, 16, 13, 15, 13, 12, 15, 13, 15, 13, 14, 13, 14, 13, 13, 14, 12, 12, 12, 13, 12, 12, 12, 11, 13, 13, 12, 12, 13, 12, 12, 11, 12, 11, 11, 12, 12, 11, 12, 9, 12, 11, 11, 11

Offset: 2

M. F. Hasler, Dec 17 2014

nonn

Inspired by Fermat's Last Theorem: 2 never occurs in this sequence.
No n is known for which a(n)<7, according to the MathWorld page. The values 7, 8, 9, ... occur first at indices 568, 102, 62, ...
I conjecture that the sequence is bounded by the initial term a(2)=128. Probably even a(4)=66, a(5)=39, a(6)=28 and some more are followed only by smaller terms.
I've uploaded two scripts; one to compute the b-file and one to generate an IP file. For the first script, a parameter kmax can be set to gain a speedup but more memory is used. The other one (which also works with large integers now) should be used in case someone has a good IP-solver. Higher terms might be computable faster with a good IP solver. - Manfred Scheucher, Aug 14 2015
From results on Waring's problem, it is known that all a(n) <= A002804(7) = 143, and a(n) <= 33 for all sufficiently large n. - Robert Israel, Aug 16 2015

Giovanni Resta, Table of n, a(n) for n = 2..200
Jean-Charles Meyrignac, Computing Minimal Equal Sums Of Like Powers
Manfred Scheucher, Sage Script for IP-generation
Manfred Scheucher, Sage Script for b-file generation
Eric Weisstein's World of Mathematics, Diophantine Equation--7th Powers
Eric Weisstein's World of Mathematics, Waring's Problem

Cf. A002804, A161882, A161883, A161884, A161885, A252486.

M:= 10^8:
R:= Vector(M,144, datatype=integer[4]):
for p from 1 to floor(M^(1/7)) do
  p7:= p^7;
  if p > 1 then A[p]:= R[p7] fi;
  R[p7]:= 1;
  for j from p7+1 to M do
    R[j]:= min(R[j],1+R[j - p7]);
  od
od:
F:= proc(n,k,ub)
   local lb, m, bestyet, res;
   if ub <= 0 then return -1 fi;
   if n <= M then
     if n = 0 then return 0
     elif R[n] > ub then return -1
     else return R[n]
     fi
   fi;
   lb:= floor(n/k^7);
   if lb > ub then return -1 fi;
   bestyet:= ub;
   for m from lb to 0 by -1 do
     res:= procname(n-m*k^7, k-1, bestyet-m);
     if res >= 0 then
       bestyet:= res+m;
     fi
   od:
   return bestyet
end proc:
for n from floor(M^(1/7))+1 to 50 do
   A[n]:= F(n^7,n-1,144)
od:
seq(A[n],n=2..50); # Robert Israel, Aug 17 2015

a(n,verbose=0,m=7)={N=n^m;for(k=3,999,forvec(v=vector(k-1,i,[1,n\sqrtn(k+1-i,m)]),ispower(N-sum(i=1,k-1,v[i]^m),m,&K)&&K>0&&!(verbose&&print1("/*"n" "v"*/"))&&return(k),1))}

More terms from Manfred Scheucher, Aug 15 2015

a(50)-a(66) from Giovanni Resta, Aug 17 2015

A259942 Numbers that are larger than or equal to the sum of the cubes of their prime factors (with multiplicity).

Original entry on oeis.org

1, 64, 96, 108, 128, 144, 162, 192, 216, 240, 243, 256, 270, 288, 300, 320, 324, 360, 384, 400, 405, 432, 448, 450, 480, 486, 500, 504, 512, 540, 560, 567, 576, 600, 625, 630, 640, 648, 672, 675, 700, 720, 729, 750, 756, 768, 784, 800, 810, 840, 864, 875, 882, 896, 900, 945, 960, 972, 980, 1000

Offset: 1

Francesco Di Matteo, Nov 08 2015

nonn

This sequence is analogous to A166319 but with cubes instead of squares.

			64 = 2*2*2*2*2*2 >= 6 * 2^3, so 64 is in the sequence.
96 = 3*2*2*2*2*2 >= 3^3 + 5 * 2^3, so 96 is in the sequence.
256 = 4*4*4*4 >= 4*4^3, so 256 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Waring's_problem
Wikipedia, Smooth number

Cf. A002804, A166319.

isA259942 := proc(n)
    local ifa;
    ifa := ifactors(n)[2] ;
    return (n >= add( op(2,p)*op(1,p)^3,p=ifa)) ;
end proc:
for n from 0 to 1000 do
    if isA259942(n) then
        printf("%d,",n);
    end if;
end do: # R. J. Mathar, Nov 27 2015

scpfQ[n_]:=n>=Total[Flatten[Table[#[[1]],{#[[2]]}]&/@ FactorInteger[ n]]^3]; Select[Range[1000],scpfQ] (* Harvey P. Dale, Dec 25 2015 *)

isok(n) = {my(f = factor(n)); n >= sum(k=1, #f~, f[k,2]*f[k,1]^3);} \\ Michel Marcus, Nov 28 2015

from sympy import factorint
for j in range(1, 1001):
    k =  factorint(j)
    it = list(k.keys())
    va = list(k.values())
    alfa = 0
    for l in range(0,len(k)):
        alfa = alfa + va[l]*(it[l]**3)
    if alfa <=j:
        print(j)

A284641 Expansion of (Sum_{k>=0} x^(k^2*(k+1)^2/4))^12.

Original entry on oeis.org

1, 12, 66, 220, 495, 792, 924, 792, 495, 232, 198, 672, 1981, 3960, 5544, 5544, 3960, 1980, 726, 792, 2982, 7920, 13860, 16632, 13860, 7920, 2970, 880, 2046, 7920, 18480, 27720, 27720, 18480, 7920, 1980, 727, 4092, 14520, 29700, 38610, 33264, 19404, 7920, 2475, 1584, 6996, 22584, 43560, 55440, 49896

Offset: 0

Ilya Gutkovskiy, May 06 2017

nonn

Number of ways to write n as an ordered sum of 12 squares of triangular numbers (A000537).
Every number is the sum of three triangular numbers (Fermat's polygonal number theorem).
Conjecture: a(n) > 0 for all n.
Extended conjecture: every number is the sum of at most 12 squares of triangular numbers (or partial sums of cubes).
Is there a solution, in analogy with Waring's problem (see A002804), for the partial sums of k-th powers?

Ilya Gutkovskiy, Extended graphical example
Index to sequences related to polygonal numbers

Cf. A000217, A000537, A014787, A282173, A282288.

nmax = 50; CoefficientList[Series[Sum[x^(k^2 (k + 1)^2/4), {k, 0, nmax}]^12, {x, 0, nmax}], x]

G.f.: (Sum_{k>=0} x^(k^2*(k+1)^2/4))^12.

A098821 a(n) = (n-2) * 2^(n-1) + 5.

Original entry on oeis.org

4, 4, 5, 9, 21, 53, 133, 325, 773, 1797, 4101, 9221, 20485, 45061, 98309, 212997, 458757, 983045, 2097157, 4456453, 9437189, 19922949, 41943045, 88080389, 184549381, 385875973, 805306373, 1677721605, 3489660933, 7247757317

Offset: 0

Parthasarathy Nambi, Oct 08 2004

			a(5) = 3*2^4 + 5 = 53.

G. H. Hardy and J. E. Littlewood, "Some problems of partitio numerorum (VI): Further researches in Waring's Problem", Math. Z. vol. 23, 1-37, (1925)
T. D. Wooley, "Large improvements in Waring's Problem", Ann. Math. vol. 135, 131-164 (1992)

Eric Weisstein, Waring's problem
Index entries for linear recurrences with constant coefficients, signature (5, -8, 4).

Cf. A018889, A002804.

Table[(n - 2)*2^(n - 1) + 5, {n, 0, 30}] (* Stefan Steinerberger, Mar 06 2006 *)
LinearRecurrence[{5,-8,4},{4,4,5},40] (* Harvey P. Dale, Feb 22 2013 *)

a(n)=(n-2)<<(n-1)+5 \\ Charles R Greathouse IV, Jul 23 2015

From Colin Barker, Jan 28 2012: (Start)
G.f.: (4-16*x+17*x^2)/(1-5*x+8*x^2-4*x^3).
a(n)=5*a(n-1)-8*a(n-2)+4*a(n-3). (End)

More terms from Stefan Steinerberger, Mar 06 2006

A098823 a(n) = 16(8prime(n) + 7).

Original entry on oeis.org

368, 496, 752, 1008, 1520, 1776, 2288, 2544, 3056, 3824, 4080, 4848, 5360, 5616, 6128, 6896, 7664, 7920, 8688, 9200, 9456, 10224, 10736, 11504, 12528, 13040, 13296, 13808, 14064, 14576, 16368, 16880, 17648, 17904, 19184, 19440, 20208, 20976

Offset: 1

Parthasarathy Nambi, Oct 08 2004

nonn

			4^2 * (8*2 + 7) = 368 when p=2.

Eric Weisstein's World of Mathematics, Waring's Problem

Cf. A018889, A002804.

Table[16*(8*Prime[n] + 7), {n, 1, 40}] (* Stefan Steinerberger, Mar 06 2006 *)

main(m)=forprime(p=2,m,print1(16 * (8 * p + 7),", ")) \\ Anders Hellström, Aug 26 2015

More terms from Stefan Steinerberger, Mar 06 2006

A206375 Vinogradov's constants arising in enumeration of solutions to Waring's problem in the evil numbers (A001969).

Original entry on oeis.org

8, 16, 32, 64, 128, 256, 512, 1024, 1758, 2128, 2536, 2982, 3466, 3988, 4550, 5152, 5792, 6474, 7194, 7956, 8758, 9600, 10484, 11408, 12376, 13384, 14432, 15524, 16658, 17834, 19052, 20314, 21618, 22964, 24354, 25786, 27262, 28780

Offset: 3

Jonathan Vos Post, Feb 07 2012

nonn

From Lemma 2, p. 2, of Eminyan.

			a(11) = 2*floor( 11^2*(log(11) + log(log(11)) + 4) ) = 2*floor(879.970885...) = 2*879 = 1758.

K. M. Eminyan, Waring's problem in the natural numbers with binary expansions of a special type, arXiv:1202.1211v1 [math.NT], Feb 06 2012.

Cf. A001969, A002804.

For 3 <= n <= 10 then 2^n; else if n > 10 then a(n) = 2*floor( n^2*(log(n) + log(log(n)) + 4) ).

A240720 Successive record-setters in Waring's problem Diophantine inequality.

Original entry on oeis.org

2, 5, 14, 46, 58, 105, 157, 163, 455, 1060, 1256, 2677, 8093, 28277, 33327, 49304

Offset: 1

Jean-François Alcover, Apr 11 2014

Eric Weisstein's MathWorld, Waring's Problem.
Wikipedia, Waring's Problem.

Cf. A002804.

Reap[For[record = 1; n = 2, n < 10^9, n++, r = 1 - (3/4)^n - FractionalPart[(3/2)^n]; If[r < record, record = r; Print[{n, record // N}]; Sow[n]]]][[2, 1]]

A356037 Conjecturally, a(n) is the smallest number m such that every natural number is a sum of at most m n-simplex numbers.

Original entry on oeis.org

1, 3, 5, 8, 10, 13, 15, 15, 19, 24

Offset: 1

Mohammed Yaseen, Jul 24 2022

n-simplex numbers are {binomial(k,n); k>=n}.
This problem is the simplex number analog of Waring's problem.
a(2) = 3 was proposed by Fermat and proved by Gauss, see A061336.
Pollock conjectures that a(3) = 5. Salzer and Levine prove this for numbers up to 452479659. See A104246 and A000797.
Kim gives a(4)=8, a(5)=10, a(6)=13 and a(7)=15 (not proved).

			2-simplex numbers are {binomial(k,2); k>=2} = {1,3,6,10,...}, the triangular numbers. 3 is the smallest number m such that every natural number is a sum of at most m triangular numbers. So a(2)=3.
3-simplex numbers are {binomial(k,3); k>=3} = {1,4,10,20,...}, the tetrahedral numbers. 5 is presumed to be the smallest number m such that every natural number is a sum of at most m tetrahedral numbers. So a(3)=5.

Hyun Kwang Kim, On regular polytope numbers, Proc. Amer. Math. Soc. 131 (2003), p. 65-75.

Cf. A002804, A079611.
Minimal number of x-simplex numbers whose sum equals n: A061336 (x=2), A104246 (x=3), A283365 (x=4), A283370 (x=5).
x-simplex numbers: A000217 (x=2), A000292 (x=3), A000332 (x=4), A000389 (x=5), A000579 (x=6), A000580 (x=7), A000581 (x=8), A000582 (x=9).

cp's OEIS Frontend

A065622 Numerator of 1 - (3/4)^n - frac((3/2)^n), where frac(x) = x - floor(x).

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A252487 Smallest k such that n^7 = a_1^7 + ... + a_k^7 and all a_i are positive integers less than n.

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Maple

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A259942 Numbers that are larger than or equal to the sum of the cubes of their prime factors (with multiplicity).

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A284641 Expansion of (Sum_{k>=0} x^(k^2*(k+1)^2/4))^12.

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A098821 a(n) = (n-2) * 2^(n-1) + 5.

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A098823 a(n) = 16*(8*prime(n) + 7).

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A206375 Vinogradov's constants arising in enumeration of solutions to Waring's problem in the evil numbers (A001969).

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A098823 a(n) = 16(8prime(n) + 7).