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Numbers that can be represented in more than one way as the sum of cubes of three distinct positive numbers in arithmetic progression.
1

%I #17 Dec 16 2022 09:37:15

%S 5643,12384,31977,45144,99072,123849,152361,153792,255816,259776,

%T 269739,274968,334368,361152,477576,500445,705375,792576,863379,

%U 912339,928017,950931,990792,1090584,1218888,1230336,1548000,1629144,1700424,1737252,1799523,1813512,1935549,1941192,2046528,2078208

%N Numbers that can be represented in more than one way as the sum of cubes of three distinct positive numbers in arithmetic progression.

%C Numbers k such that there are at least two pairs of positive numbers (a,d) such that k = a^3 + (a+d)^3 + (a+2d)^3.

%C The first term that has three such representations is 255816 = 8^3 + 34^3 + 60^3 = 18^3 + 38^3 + 58^3 = 43^3 + 44^3 + 45^3.

%C 346380489216 has four such representations: 1188^3 + 3888^3 + 6588^3, 1728^3 + 4104^3 + 6480^3, 4248^3 + 4824^3 + 5400^3 and 4665^3 + 4864^3 + 5063^3. It might not be the first.

%H Robert Israel, <a href="/A359055/b359055.txt">Table of n, a(n) for n = 1..2500</a>

%H R. Israel et al, <a href="https://math.stackexchange.com/questions/4599457/sum-of-cubes-of-three-positive-integers-in-arithmetic-progression-in-four-ways">Sum of cubes of three positive integers in arithmetic progression in four ways?</a>, Mathematics StackExchange, Dec. 2022.

%e a(1) = 5643 is a term because 5643 = 1^3 + (1+8)^3 + (1+2*8)^3 = 6^3 + (6+5)^3 + (6+2*5)^3.

%p N:= 10^7: # to get terms <= N

%p S:= {}: S2:= {}:

%p for a from 1 while a^3 + (a+1)^3 + (a+2)^3 <= N do

%p for d from 1 do

%p x:= a^3 + (a+d)^3 + (a+2*d)^3;

%p if x > N then break fi;

%p if member(x,S) then S2:= S2 union {x} fi;

%p S:= S union {x}

%p od od:

%p sort(convert(S,list));

%Y Cf. A306213.

%K nonn

%O 1,1

%A _Robert Israel_, Dec 14 2022