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 1-5 of 5 results.

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)

A078136 Numbers having exactly one representation as sum of squares>1.

Original entry on oeis.org

4, 8, 9, 12, 13, 17, 18, 21, 22, 26, 27, 30, 31, 35, 39
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 19 2002

Keywords

Comments

A078134(a(n))=1.
The sequence is finite with a(15)=39 as last term, since numbers m>39 can be represented as sums of squares>1 (even of squares of primes and even of squares of 2, 3 and 4 and even of squares of 2, 3 and 5) in at least two ways. Proof: if m=40+4k, k>=0, then m=(k+10)*2^2=(k+1)*2^2+4*3^2; if m=41+4k, then m=(k+8)*2^2+3^2=(k+4)*2^2+5^2; if m=42+4k, then m=(k+6)*2^2+2*3^2=(k+2)*2^2+3^2+5^2; if m=43+4k, then m=(k+4)*2^2+3*3^2=k*2^2+2*3^2+5^2. - Hieronymus Fischer, Nov 11 2007

Crossrefs

A078129 Numbers which cannot be written as sum of cubes > 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 25, 26, 28, 29, 30, 31, 33, 34, 36, 37, 38, 39, 41, 42, 44, 45, 46, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 65, 66, 68, 69, 71, 73, 74, 76, 77, 79, 82, 84, 85, 87, 90, 92, 93, 95, 98, 100, 101, 103, 106, 109, 111, 114, 117, 119, 122, 127, 130, 138, 146, 154
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 19 2002

Keywords

Comments

A078128(a(n))=0.
The sequence is finite because every number greater than 181 can be represented using just 8 and 27. - Franklin T. Adams-Watters, Apr 21 2006
More generally, the numbers which are not the sum of k-th powers larger than 1 are exactly those in [1, 6^k - 3^k - 2^k] but not of the form 2^k*a + 3^k*b + 5^k*c with a,b,c nonnegative. This relies on the following fact applied to m=2^k and n=3^k: if m and n are relatively prime, then the largest number which is not a linear combination of m and n with positive integer coefficients is mn - m - n. - Benoit Jubin, Jun 29 2010

Examples

			181 is not in the list since 181 = 7*2^3 + 5^3.
		

Crossrefs

Programs

  • Mathematica
    terms = 83; A078131 = (Exponent[#, x]& /@ List @@ Normal[1/Product[1-x^j^3, {j, 2, Ceiling[(3 terms)^(1/3)]}] + O[x]^(3 terms)])[[2 ;; terms+1]];
    Complement[Range[Max[A078131]], A078131] (* Jean-François Alcover, Aug 04 2018 *)

Extensions

Sequence completed by Franklin T. Adams-Watters, Apr 21 2006
Edited by R. J. Mathar and N. J. A. Sloane, Jul 06 2010

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

A122612 Duplicate of A078131.

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, 141
Offset: 1

Views

Author

Jonathan Vos Post, Sep 20 2006

Keywords

Comments

Previous name was: Sums of cubes of primes.
Starts out identical to A078130 (numbers having exactly one representation as sum of cubes>1), until 72. It seems that 154 is the largest integer which cannot be represented as the sum of cubes of primes.
154 is the largest integer that cannot be represented as the sum of cubes of primes. Indeed, every number greater than 154 can be represented as a sum of multiples of 2^3, 3^3, and 5^3. - Giovanni Resta, Jun 16 2016

Crossrefs

Cf. A000040 (primes), A030078 (cubes of primes), A078130.

Programs

  • Python
    from sympy import primerange, integer_nthroot as iroot
    def ok(n):
        cands = [p**3 for p in primerange(2, iroot(n, 3)[0]+1) if p**3 <= n]
        return n in cands or any(ok(n-c) for c in cands)
    print(list(filter(ok, range(142)))) # Michael S. Branicky, Aug 16 2021

Formula

{A030078} UNION {A030078 + A030078} UNION {A030078 + A030078 + A030078}... = a*8 + b*27 + c*125 + d*343 + e*1331 + f*2197 = a*(p(1))^3 + b*(p(2))^3 + c*(p(3))^3 + d*(p(4))^3 + e*(p(5))^3 + ... where p(i) = A000040(i) and a, b, c, d, e, f, ... are nonnegative integers.
Showing 1-5 of 5 results.