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 1-6 of 6 results.

A078135 Numbers which cannot be written as a sum of squares > 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 19, 23
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 19 2002

Keywords

Comments

Numbers such that A078134(n)=0.
"Numbers which cannot be written as sum of squares > 1" is equivalent to "Numbers which cannot be written as sum of squares of primes." Equivalently, numbers which can be written as the sum of nonzero squares can also be written as sum of the squares of primes." cf. A090677 = number of ways to partition n into sums of squares of primes. - Jonathan Vos Post, Sep 20 2006
The sequence is finite with a(12)=23 as last member. Proof: When k=a^2+b^2+..., k+4 = 2^2+a^2+b^2+... If k can be written as sum of the squares of primes, k+4 also has this property. As 24,25,26,27 have the property, by induction, all numbers > 23 can be written as sum of squares>1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Apr 07 2007
Also, numbers which cannot be written as sum of squares of 2 and 3 (see A078137 for the proof). Explicit representation as sum of squares of primes, or rather of squares of 2 and 3, for numbers m>23: we have m=c*2^2+d*3^2, where c:=(floor(m/4) - 2*(m mod 4))>=0, d:=m mod 4. For that, the finiteness of the sequence is proved constructively. - Hieronymus Fischer, Nov 11 2007
Also numbers n such that every integer partition of n contains a squarefree number. For example, 21 does not belong to the sequence because there are integer partitions of 21 containing no squarefree numbers, namely: (12,9), (9,8,4), (9,4,4,4). - Gus Wiseman, Dec 14 2018

Crossrefs

Programs

  • Mathematica
    nn=100;
    ser=Product[If[SquareFreeQ[n],1,1/(1-x^n)],{n,nn}];
    Join@@Position[CoefficientList[Series[ser,{x,0,nn}],x],0]-1 (* Gus Wiseman, Dec 14 2018 *)

Formula

A090677(a(n)) = 0. - Jonathan Vos Post, Sep 20 2006 [corrected by Joerg Arndt, Dec 16 2018]
A033183(a(n)) = 0. - Reinhard Zumkeller, Nov 07 2009

A078128 Number of ways to write n as sum of cubes > 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 1, 0, 2, 1, 0, 1, 0, 0, 1, 0, 2, 1, 0, 2, 0, 0, 1, 0, 2, 1, 0, 2, 0, 0, 1, 0, 2, 1
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 19 2002

Keywords

Comments

a(A078129(n))=0; a(A078130(n))=1; a(A078131(n))>0;
Conjecture (lower bound): for all k exists b(k) such that a(n)>k for n>b(k); see b(0)=A078129(83)=154 and b(1)=A078130(63)=218.

Examples

			a(160)=4: 160 = 20*2^3 = 4^3+12*2^3 = 2*4^3+4*2^3 = 5^3+3^3+2^3.
		

Crossrefs

Programs

Formula

a(n) = 1/n*Sum_{k=1..n} (b(k)-1)*a(n-k), a(0) = 1, where b(k) is sum of cube divisors of k. - Vladeta Jovovic, Nov 20 2002
From Vaclav Kotesovec, Jan 05 2017: (Start)
a(n) = A003108(n) - A003108(n-1).
a(n) ~ exp(4*(Gamma(1/3) * Zeta(4/3))^(3/4) * n^(1/4) / 3^(3/2)) * (Gamma(1/3) * Zeta(4/3))^(3/2) / (8 * 3^(5/2) * Pi^2 * n^2).
(End)

A078130 Numbers having exactly one representation as sum of cubes > 1.

Original entry on oeis.org

8, 16, 24, 27, 32, 35, 40, 43, 48, 51, 54, 56, 59, 62, 67, 70, 75, 78, 81, 83, 86, 89, 94, 97, 102, 105, 108, 110, 113, 116, 121, 124, 125, 129, 132, 133, 135, 137, 140, 141, 143, 148, 149, 151, 156, 157, 159, 162, 164, 165, 167, 170, 173, 175, 178, 181, 183, 186, 191, 194, 202, 210, 218
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 19 2002

Keywords

Comments

A078128(a(n))=1.
Conjecture: the sequence is finite; is a(63)=218 the last entry?
Yes. An argument similar to that in A078136 can be made, based on the identity m = 8*k = k*2^3 = 4^3 + (k-8)*2^3 which enables trading 4^3 for 8 repeats of 2^3. Then, the remaining residue classes m = 8*k+r for r=1..7 can be handled by known representations for m = 145, 226, 91, 172, 189, 118, and 199, respectively. - Sean A. Irvine, Jun 17 2025

Examples

			72 is not a term, as 72 = 8+8+8+8+8+8+8+8+8 = 8+64.
		

Crossrefs

Extensions

More terms from Sean A. Irvine, Jun 17 2025

A078131 Numbers which can be written as sum of cubes > 1.

Original entry on oeis.org

8, 16, 24, 27, 32, 35, 40, 43, 48, 51, 54, 56, 59, 62, 64, 67, 70, 72, 75, 78, 80, 81, 83, 86, 88, 89, 91, 94, 96, 97, 99, 102, 104, 105, 107, 108, 110, 112, 113, 115, 116, 118, 120, 121, 123, 124, 125, 126, 128, 129, 131, 132, 133, 134, 135, 136, 137, 139, 140
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 19 2002

Keywords

Comments

A078128(a(n)) > 0.

Examples

			35 is in the sequence because 35 = 27 + 8.
		

Crossrefs

Cf. A000578, A078129 (complement of this sequence), A078130, A078137.

Programs

  • Maple
    a:=proc(n) local h, hser: h:=1/product(1-x^(j^3),j=2..30): hser:=series(h,x=0,150): if coeff(hser,x^n)>0 then n else fi end: seq(a(n),n=1..140); # Emeric Deutsch, Mar 30 2006
  • Mathematica
    terms = 60;
    (Exponent[#, x]& /@ List @@ Normal[1/Product[1-x^j^3, {j, 2, Ceiling[(3 terms)^(1/3)]}] + O[x]^(3 terms)])[[2 ;; terms+1]] (* Jean-François Alcover, Aug 04 2018, after Emeric Deutsch *)

Formula

a(n) = n + 83 for n >= 72. - Robert Israel, Apr 05 2020

A078133 Primes which cannot be written as sum of cubes>1.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 61, 71, 73, 79, 101, 103, 109, 127
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 19 2002

Keywords

Comments

Sequence is finite, see A078129.

Crossrefs

A333821 Numbers k that can be represented in the form k = p^3 - q^3 - r^3, where p, q, r are positive integers satisfying p = q + r.

Original entry on oeis.org

6, 18, 36, 48, 60, 90, 126, 144, 162, 168, 210, 216, 252, 270, 288, 330, 360, 378, 384, 396, 468, 480, 486, 540, 546, 594, 630, 720, 750, 792, 816, 858, 918, 924, 972, 990, 1008, 1026, 1140, 1152, 1170, 1260, 1296, 1344, 1386, 1404, 1518, 1530, 1560, 1620, 1638, 1656, 1680, 1728, 1800
Offset: 1

Views

Author

Antonio Roldán, Apr 06 2020

Keywords

Comments

An alternative representation of k is k = 3*q*r*(q+r), with q, r positive integers, then k is a multiple of 6.

Examples

			60 is in the sequence because 60 = 5^3 - 4^3 - 1^3, with 5 = 4 + 1.
		

Crossrefs

Programs

  • PARI
    ok(n) = {my(i=1, a=0, m=0, j); if(n%6==0, while(a<=n&&m==0, j=1; while(j
    				

Formula

a(n) = 6 * A121741(n).
Showing 1-6 of 6 results.