A003341 - OEIS

%I #30 Jul 31 2021 17:57:41

%S 7,22,37,52,67,82,87,97,102,112,117,132,147,162,167,177,182,197,212,

%T 227,242,247,262,277,292,307,322,327,337,342,352,357,372,387,402,407,

%U 417,422,437,452,467,482,487,502,517,532,547,562,567,577,582,592,597,612,627

%N Numbers that are the sum of 7 positive 4th powers.

%H Robert Israel, <a href="/A003341/b003341.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BiquadraticNumber.html">Biquadratic Number</a>.

%e From _David A. Corneth_, Aug 04 2020: (Start)

%e 5971 is in the sequence as 5971 = 3^4 + 3^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4.

%e 12022 is in the sequence as 12022 = 1^4 + 2^4 + 7^4 + 7^4 + 7^4 + 7^4 + 7^4.

%e 16902 is in the sequence as 16902 = 1^4 + 1^4 + 3^4 + 6^4 + 7^4 + 9^4 + 9^4. (End)

%p N:= 1000:

%p S1:= {seq(i^4,i=1..floor(N^(1/4)))}:

%p S2:= select(`<=`,{seq(seq(i+j,i=S1),j=S1)},N):

%p S4:= select(`<=`,{seq(seq(i+j,i=S2),j=S2)},N):

%p S6:= select(`<=`,{seq(seq(i+j,i=S2),j=S4)},N):

%p sort(convert(select(`<=`,{seq(seq(i+j,i=S1),j=S6)},N),list)); # _Robert Israel_, Jul 21 2019

%o (Python)

%o from itertools import combinations_with_replacement as mc

%o def aupto(limit):

%o     qd = [k**4 for k in range(1, int(limit**.25)+2) if k**4 + 6 <= limit]

%o     ss = set(sum(c) for c in mc(qd, 7))

%o     return sorted(s for s in ss if s <= limit)

%o print(aupto(630)) # _Michael S. Branicky_, Jul 22 2021

%Y Cf. A000583, A003330, A003340, A003342, A003352, A047718, A345568, A345823.

%K nonn,easy

%O 1,1

%A _N. J. A. Sloane_