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A110567 a(n) = n^(n+1) + 1.

%I A110567 #21 Sep 08 2022 08:45:20
%S A110567 1,2,9,82,1025,15626,279937,5764802,134217729,3486784402,100000000001,
%T A110567 3138428376722,106993205379073,3937376385699290,155568095557812225,
%U A110567 6568408355712890626,295147905179352825857,14063084452067724991010
%N A110567 a(n) = n^(n+1) + 1.
%C A110567 For n >= 2, a(n) = the n-th positive integer such that a(n) (base n) has a block of exactly n consecutive zeros.
%C A110567 Comments from _Alexander Adamchuk_, Nov 12 2006 (Start)
%C A110567 (2n+1)^2 divides a(2n). a(2n)/(2n+1)^2 = {1,1,41,5713,1657009,826446281,633095889817,691413758034721,...} = A081215(2n).
%C A110567 p divides a(p-1) for prime p. a(p-1)/p = {1,3,205,39991,9090909091,8230246567621,...} = A081209(p-1) = A076951(p-1).
%C A110567 p^2 divides a(p-1) for an odd prime p. a(p-1)/p^2 = {1,41,5713,826446281,633095889817,1021273028302258913,1961870762757168078553, 14199269001914612973017444081,...} = A081215(p-1).
%C A110567 Prime p divides a((p-3)/2) for p = {13,17,19,23,37,41,43,47,61,67,71,89, 109,113,137,139,157,163,167,181,191,...}.
%C A110567 Prime p divides a((p-5)/4) for p = {29,41,61,89,229,241,281,349,421,509,601,641,661,701,709,769,809,821,881,...} = A107218(n) Primes of the form 4x^2+25y^2.
%C A110567 Prime p divides a((p-7)/6) for p = {79,109,127,151,313,421,541,601,613,751,757,787,...}.
%C A110567 Prime p divides a((p-9)/8) for p = {41,337,401,521,569,577,601,857,929,937,953,977,...} A subset of A007519(n) Primes of form 8n+1.
%C A110567 Prime p divides a((p-11)/10) for p = {41,181,331,601,761,1021,1151,1231,1801,...}.
%C A110567 Prime p divides a((p-13)/12) for p = {313,337,433,1621,1873,1993,2161,2677,2833,...}. (End)
%H A110567 G. C. Greubel, <a href="/A110567/b110567.txt">Table of n, a(n) for n = 0..385</a>
%F A110567 a(n) = A007778(n) + 1.
%F A110567 a(n) = A110567(n) for n > 1. - _Georg Fischer_, Oct 20 2018
%e A110567 Examples illustrating the Comment:
%e A110567 a(2) = 9 because the first positive integer (base 2) with a block of 2 consecutive zeros is 100 (base 2) = 4, and the 2nd is 1001 (base 2) = 9 = 1 + 2^3.
%e A110567 a(3) = 82 because the first positive integer (base 3) with a block of 3 consecutive zeros is 1000 (base 3) = 81, the 2nd is 2000 (base 3) = 54 and the 3rd is 10001 (base 3) = 82 = 1 + 3^4.
%e A110567 a(4) = 1025 because the first positive integer (base 4) with a block of 4 consecutive zeros is 10000 (base 4) = 256, the 2nd is 20000 (base 4) = 512, the 3rd is 30000 (base 4) = 768 and the 4th 100001 (base 4) = 1025 = 1 + 4^5. and the 2nd is 1001 (base 2) = 9 = 1 + 2^3.
%t A110567 Table[n^(n+1)+1,{n,0,30}] (* _Harvey P. Dale_, Oct 30 2015 *)
%o A110567 (PARI) for(n=0,25, print1(1 + n^(n+1), ", ")) \\ _G. C. Greubel_, Aug 31 2017
%o A110567 (Magma) [n^(n+1) + 1: n in [0..25]]; // _G. C. Greubel_, Oct 16 2017
%Y A110567 Cf. A007778: n^(n+1); A000312: n^n; A014566: Sierpinski numbers of the first kind: n^n + 1.
%Y A110567 Cf also A081209, A076951, A081215. A110567.
%K A110567 easy,nonn
%O A110567 0,2
%A A110567 _Jonathan Vos Post_, Sep 12 2005
%E A110567 Entry revised by _N. J. A. Sloane_, Oct 20 2018 at the suggestion of _Georg Fischer_.