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a(n) = prime(1^4) + prime(2^4) + ... + prime(n^4).
1

%I #22 Mar 05 2023 12:39:21

%S 2,55,474,2093,6730,17357,38748,77621,143308,248037,407558,641437,

%T 973380,1432721,2052922,2874563,3944166,5314265,7045924,9206477,

%U 11874460,15134597,19083406,23826383,29480190,36172177,44039724

%N a(n) = prime(1^4) + prime(2^4) + ... + prime(n^4).

%C Analog of prime(1^2) + prime(2^2) + ... + prime(n^2) (A109724). For a(n) to be prime for n > 1 it is necessary but not sufficient that n == 0 (mod 4).

%H Harvey P. Dale, <a href="/A109796/b109796.txt">Table of n, a(n) for n = 1..100</a>

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

%F a(n) = Sum_{i=1..n} A000040(A000583(i)).

%e a(1) = 2 because prime(1^4) = prime(1) = 2.

%e a(2) = 55 because prime(1^4) + prime(2^4) = prime(1) + prime(16) = 2 + 53.

%e a(3) = 474 because prime(1^4) + prime(2^4) + prime(3^4) = prime(1) + prime(16) + prime(81) = 2 + 53 + 419.

%e a(4) = 2093 because prime(1^4) + prime(2^4) + prime(3^4) + prime(4^4) = 2 + 53 + 419 + prime(256) = 2 + 53 + 419 + 1619.

%e a(8) = 2 + 53 + 419 + 1619 + 4637 + 10627 + 21391 + 38873 = 77621 (which is prime).

%e a(12) = 2 + 53 + 419 + 1619 + 4637 + 10627 + 21391 + 38873 + 65687 + 104729 + 159521 + 233879 = 641437 (which is prime).

%e a(28) = 2 + 53 + 419 + 1619 + 4637 + 10627 + 21391 + 38873 + 65687 + 104729 + 159521 + 233879 + 331943 + 459341 + 620201 + 821641 + 1069603 + 1370099 + 1731659 + 2160553 + 2667983 + 3260137 + 3948809 + 4742977 + 5653807 + 6691987 + 7867547 + 9195889 = 53235613 (which is prime).

%e It is a coincidence that a(1), a(2) and a(3) are all palindromes.

%t Accumulate[Table[Prime[n^4],{n,30}]] (* _Harvey P. Dale_, Feb 02 2019 *)

%o (PARI) A109796(n)={

%o sum(i=1,n,prime(i^4))

%o } /* _R. J. Mathar_, Mar 09 2012 */

%Y First differences are A109791.

%Y Cf. A000040, A000290, A000583, A011757, A109724, A109770.

%K nonn

%O 1,1

%A _Jonathan Vos Post_, Aug 15 2005