| Displaying 1-5 of 5 results found.
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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
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
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.
PROG
(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)
Number of ways to write n as sum of cubes>1.
+10
13
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
COMMENTS
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.
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 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)
EXAMPLE
a(160)=4: 160 = 20*2^3 = 4^3+12*2^3 = 2*4^3+4*2^3 = 5^3+3^3+2^3.
Numbers which can be written as sum of squares>1.
+10
12
4, 8, 9, 12, 13, 16, 17, 18, 20, 21, 22, 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, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82
COMMENTS
Numbers which can be written as a sum of squares of primes. - Hieronymus Fischer, Nov 11 2007 Equivalently, numbers which can be written as a sum of squares of 2 and 3. Proof for numbers m>=24: if m=4*(k+6), k>=0, then m=(k+6)*2^2; if m=4*(k+6)+1 than m=(k+4)*2^2+3^2; if m=4*(k+6)+2 then m=(k+2)*2^2+2*3^2; if m=4*(k+6)+3 then m=k*2^2+3*3^2. Clearly, the numbers a(n)<24 can also be written as sums of squares of 2 and 3. Explicit representation as a sum of squares of 2 and 3 for numbers m>23: m=c*2^2+d*3^2, where c:=((floor(m/4) - 2*(m mod 4))>=0 and d:=m mod 4. - Hieronymus Fischer, Nov 11 2007
MATHEMATICA
Join[{4, 8, 9, 12, 13, 16, 17, 18, 20, 21, 22}, Range[24, 82]] (* Jean-François Alcover, Aug 01 2018 *)
PROG
(PARI) a(n)=if(n>11, n+12, [4, 8, 9, 12, 13, 16, 17, 18, 20, 21, 22][n]) \\ Charles R Greathouse IV, Aug 21 2011
CROSSREFS
Cf. A000290, A078136, A078131, A001597, A025475, A078134, A078135, A078139, A090677, A134600, A134605, A134608, A134612, A134616, A134618, A134620.
Numbers which cannot be written as sum of cubes > 1.
+10
6
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
COMMENTS
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
EXAMPLE
181 is not in the list since 181 = 7*2^3 + 5^3.
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]];
Numbers having exactly one representation as sum of cubes>1.
+10
5
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
COMMENTS
Conjecture: the sequence is finite; is a(63)=218 the last entry?
EXAMPLE
72 is not a term, as 72 = 8+8+8+8+8+8+8+8+8 = 8+64.
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