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%I #10 Jun 13 2016 08:33:54
%S 14,45,76,107,138,169,200,231,256,262,287,293,318,324,349,355,380,386,
%T 411,417,442,448,473,498,504,529,535,560,566,591,597,622,628,653,659,
%U 684,715,740,746,771,777,802,808,833,839,864,870,895,926,957,982,988
%N Sum of 14 positive 5th powers.
%C Up to 417 = 13*(2^5) + 1 this sequence is identical to x+2 for x in A003357 Sum of 12 positive 5th powers. Primes in this sequence (107, 293, 349, 653, ...) are A123300. As proved by J.-R. Chen in 1964, g(5) = 37, so every positive integer can be written as the sum of no more than 37 positive 5th powers. G(5) <= 17, bounding the least integer G(5) such that every positive integer beyond a certain point (i.e., all but a finite number) is the sum of G(5) 5th powers.
%H Giovanni Resta, <a href="/A123295/b123295.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/WaringsProblem.html">Waring's Problem</a>.
%F Sumset {A000584 + A000584 + A000584 + A000584 + A000584 + A000584 + A000584 + A000584 + A000584 + A000584 + A000584 + A000584 + A000584 + A000584}.
%e a(1) = 14 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5.
%e a(2) = 45 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 2^5.
%e a(9) = 256 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 3^5.
%e a(11) = 287 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 2^5 + 3^5
%t up = 1000; q = Range[up^(1/5)]^5; a ={0}; Do[b = Select[ Union@ Flatten@ Table[e + a, {e, q}], # <= up &]; a=b, {k, 14}]; a (* _Giovanni Resta_, Jun 12 2016 *)
%Y Cf. A000584, A003336, A003347, A003349, A003350, A003351, A003352, A003353, A003354, A003355, A003356, A003357, A123294-A123297, A008480, A123299, A123300.
%K easy,nonn
%O 1,1
%A _Jonathan Vos Post_, Sep 24 2006
%E 5 missing terms and more terms from _Giovanni Resta_, Jun 12 2016
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