cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A076952 n-th power associated with A076951, or 0 if no such number exists.

Original entry on oeis.org

0, 1, 8, 0, 1024, 0, 279936, 0, 512, 59049, 100000000000, 0, 106993205379072, 0, 155568095557812224, 0, 295147905179352825856, 0, 708235345355337676357632, 0, 476837158203125, 0, 7511413302012830262726227918848, 0, 1125899906842624
Offset: 1

Views

Author

Amarnath Murthy, Oct 20 2002

Keywords

Crossrefs

Cf. A076951.

Formula

a(n) = n*A076951(n) - 1.

Extensions

More terms from R. J. Mathar, Jul 01 2007

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

Original entry on oeis.org

1, 2, 9, 82, 1025, 15626, 279937, 5764802, 134217729, 3486784402, 100000000001, 3138428376722, 106993205379073, 3937376385699290, 155568095557812225, 6568408355712890626, 295147905179352825857, 14063084452067724991010
Offset: 0

Views

Author

Jonathan Vos Post, Sep 12 2005

Keywords

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)

Examples

			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.

Links

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.

Programs

  • Magma
    [n^(n+1) + 1: n in [0..25]]; // G. C. Greubel, Oct 16 2017
  • Mathematica
    Table[n^(n+1)+1,{n,0,30}] (* Harvey P. Dale, Oct 30 2015 *)
  • PARI
    for(n=0,25, print1(1 + n^(n+1), ", ")) \\ G. C. Greubel, Aug 31 2017

Formula

a(n) = A007778(n) + 1.
a(n) = A110567(n) for n > 1. - Georg Fischer, Oct 20 2018

Extensions

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

A081215 a(n) = (n^(n+1)+(-1)^n)/(n+1)^2.

Original entry on oeis.org

1, 0, 1, 5, 41, 434, 5713, 90075, 1657009, 34867844, 826446281, 21794641505, 633095889817, 20088655029078, 691413758034721, 25657845139503479, 1021273028302258913, 43404581642184336392, 1961870762757168078553
Offset: 0

Views

Author

Vladeta Jovovic, Apr 17 2003

Keywords

Comments

From Mathew Englander, Oct 19 2020: (Start)
The sum of two adjacent terms of the sequence cannot be prime.
In base n, a(n) has n-1 digits, which are (beginning from the left): n-2, 2, n-4, 4, and so on, except that if n is even the rightmost digit is 1 instead of 0. In that case, the other digits form a palindrome with every even digit from 2 to n-2 appearing twice. For example, a(14) in base 14 is c2a486684a2c1. If n is odd, then all digits from 1 to n-1 occur exactly once. For example, a(15) in base 15 is d2b496785a3c1e.
For any positive integer k, any prime p, and any positive integer h such that h*p > 2, a(h*p^k - 2) == (-1)^h * (1 - 2^(h-1)) (mod p). For example, a(7*p^k - 2) == 63 (mod p); a(10*p^k - 2) == -511 (mod p).
Suppose k and m are positive integers. If k is even, then a(k*m) == 1, a(k*m+1) == 0, and a(k*m-1) == -1 (all mod m). If k is odd, then a(k*m) == (-1)^m and a(k*m+1) == ceiling(m/2) (both mod m), while a(k*m-1) == m/2 - 1 for m even, and a(k*m-1) == 1 for m odd (mod m).
For proofs of the above, see the Englander link. (End)

Links

Crossrefs

Cf. A081209, A110567, A076951, A273319, A193746, and A060073.

Programs

  • Maple
    seq((j^(j+1)+(-1)^j)/(j+1)^2, j=0..50); # Robert Israel, May 19 2016
  • Mathematica
    Array[(#^(# + 1) + (-1)^#)/(# + 1)^2 &, 19, 0] (* Michael De Vlieger, Nov 13 2020 *)
  • PARI
    a(n) = (n^(n+1)+(-1)^n)/(n+1)^2; \\ Michel Marcus, Oct 20 2020

Formula

a(n) = (-1)^n + Sum_{k=1..n} (-1)^(k+1)*(n+1)^(n-k)*C(n+1,n+2-k). - Gionata Neri, May 19 2016
E.g.f.: (Ei(1,x) - Ei(1,-LambertW(-x)))/x. - Robert Israel, May 19 2016
For n > 1, a(n) = Sum_{k=1..floor(n/2)} (n^(n-2*k) * (2*k/n + n - 2*k)). - Mathew Englander, Oct 19 2020
Showing 1-3 of 3 results.

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