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 A110567 a(n) = n^(n+1) + 1. 4
 1, 2, 9, 82, 1025, 15626, 279937, 5764802, 134217729, 3486784402, 100000000001, 3138428376722, 106993205379073, 3937376385699290, 155568095557812225, 6568408355712890626, 295147905179352825857, 14063084452067724991010 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified January 26 20:11 EST 2020. Contains 331288 sequences. (Running on oeis4.)