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

1, 2, 9, 82, 1025, 15626, 279937, 5764802, 134217729, 3486784402, 100000000001, 3138428376722, 106993205379073, 3937376385699290, 155568095557812225, 6568408355712890626, 295147905179352825857, 14063084452067724991010

Offset

0,2

Comments

For n >= 2, a(n) = the n-th positive integer such that a(n) (base n) has a block of exactly n consecutive zeros.

Comments from Alexander Adamchuk, Nov 12 2006 (Start)

(2n+1)^2 divides a(2n). a(2n)/(2n+1)^2 = {1,1,41,5713,1657009,826446281,633095889817,691413758034721,...} = A081215(2n).

p divides a(p-1) for prime p. a(p-1)/p = {1,3,205,39991,9090909091,8230246567621,...} = A081209(p-1) = A076951(p-1).

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).

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,...}.

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.

Prime p divides a((p-7)/6) for p = {79,109,127,151,313,421,541,601,613,751,757,787,...}.

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.

Prime p divides a((p-11)/10) for p = {41,181,331,601,761,1021,1151,1231,1801,...}.

Prime p divides a((p-13)/12) for p = {313,337,433,1621,1873,1993,2161,2677,2833,...}. (End)

Links

G. C. Greubel, Table of n, a(n) for n = 0..385

Formula

a(n) = A007778(n) + 1.

a(n) = A110567(n) for n > 1. - Georg Fischer, Oct 20 2018

Example

Examples illustrating the Comment:
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.
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.
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.

Mathematica

Table[n^(n+1)+1, {n, 0, 30}] (* Harvey P. Dale, Oct 30 2015 *)

Prog

(PARI) for(n=0, 25, print1(1 + n^(n+1), ", ")) \\ G. C. Greubel, Aug 31 2017
(Magma) [n^(n+1) + 1: n in [0..25]]; // G. C. Greubel, Oct 16 2017

Crossrefs

Cf. A007778: n^(n+1); A000312: n^n; A014566: Sierpinski numbers of the first kind: n^n + 1.

Cf also A081209, A076951, A081215. A110567.

Sequence in context: A301861 A112670 A117581 * A123570 A006040 A067309

Adjacent sequences: A110564 A110565 A110566 * A110568 A110569 A110570

Keyword

easy,nonn

Author

Jonathan Vos Post, Sep 12 2005

Extensions

Entry revised by N. J. A. Sloane, Oct 20 2018 at the suggestion of Georg Fischer.

Status

approved