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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.
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0
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
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.
EXAMPLE
60 is in the sequence because 60 = 5^3 - 4^3 - 1^3, with 5 = 4 + 1.
PROG
(PARI) ok(n) = {my(i=1, a=0, m=0, j); if(n%6==0, while(a<=n&&m==0, j=1; while(j<i&&m==0, a=3*i*j*(i-j); if(a==n, m=1); j+=1); i+=1)); m}
Numbers which can be written as sum of cubes > 1.
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5
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
EXAMPLE
35 is in the sequence because 35 = 27 + 8.
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 *)
Number of ways to write n as sum of cubes>1.
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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 cannot be written as a sum of squares > 1.
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20
1, 2, 3, 5, 6, 7, 10, 11, 14, 15, 19, 23
COMMENTS
"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
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 *)
Primes which cannot be written as sum of cubes>1.
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4
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 61, 71, 73, 79, 101, 103, 109, 127
Numbers having exactly one representation as sum of cubes>1.
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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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