%I A053699
%S A053699 1,5,31,121,341,781,1555,2801,4681,7381,11111,16105,22621,30941,41371,
%T A053699 54241,69905,88741,111151,137561,168421,204205,245411,292561,346201,
%U A053699 406901,475255,551881,637421,732541,837931,954305,1082401,1222981
%N A053699 a(n) = n^4+n^3+n^2+n+1.
%C A053699 a(n) = 11111 in base n.
%C A053699 Let Phi_k(x) be the k-th cyclotomic polynomial and form the sequence 
               Phi_k(0), Phi_k(1), Phi_k(2), ... This gives A000027 (k=2), A002061 
               (k=3), A002522 (k=4), A053699 (k=5), A002061 (k=6), A053716 (k=7), 
               A002523 (k=8), A060883 (k=9), A060884 (k=10), A060885 (k=11), A060886 
               (k=12), A060887 (k=13), A060888 (k=14), A060889 (k=15), A060890 (k=16), 
               A060891 (k=18), A060892 (k=20), A060893 (k=24), A060894 (k=30), A060895 
               (k=32), A060896 (k=36).
%F A053699 a(n) =n^4+n^3+n^2+n+1 =(n^5-1)/(n-1)
%e A053699 a(3)=121 because 11111 base 3 =81+27+9+3+1=121
%Y A053699 Sequence in context: A024399 A099083 A096944 this_sequence A152122 A041303 
               A077719
%Y A053699 Adjacent sequences: A053696 A053697 A053698 this_sequence A053700 A053701 
               A053702
%K A053699 nonn
%O A053699 0,2
%A A053699 Henry Bottomley (se16(AT)btinternet.com), Mar 23 2000