

A008865


a(n) = n^2  2.


60



1, 2, 7, 14, 23, 34, 47, 62, 79, 98, 119, 142, 167, 194, 223, 254, 287, 322, 359, 398, 439, 482, 527, 574, 623, 674, 727, 782, 839, 898, 959, 1022, 1087, 1154, 1223, 1294, 1367, 1442, 1519, 1598, 1679, 1762, 1847, 1934, 2023, 2114, 2207, 2302, 2399, 2498
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OFFSET

1,2


COMMENTS

For n >= 2, least m >= 1 such that f(m, n) = 0 where f(m,n) = Sum_{i=0..m} Sum_{k= 0..i} (1)^k*(floor(i/n^k)  n*floor(i/n^(k+1))).  Benoit Cloitre, May 02 2004
For n >= 3, the a(n)th row of Pascal's triangle always contains a triple forming an arithmetic progression.  Lekraj Beedassy, Jun 03 2004
Let C = 1 + sqrt(2) = 2.414213...; and 1/C = 0.414213... Then a(n) = (n + 1 + 1/C) * (n + 1  C). Example: a(6) = 34 = (7 + 0.414...) * (7  2.414...).  Gary W. Adamson, Jul 29 2009
The sequence (n4)^22, n = 7, 8, ... enumerates the number of nonisomorphic sequences of length n, with entries from {1, 2, 3} and no two adjacent entries the same, that minimally contain each of the thirteen rankings of three players (111, 121, 112, 211, 122, 212, 221, 123, 132, 213, 231, 312, 321) as embedded order isomorphic subsequences. By "minimally", we mean that the nth symbol is necessary for complete inclusion of all thirteen words. See the arXiv paper below for proof. If n = 7, these sequences are 1213121, 1213212, 1231213, 1231231, 1231321, 1232123, and 1232132, and for each case, there are 3! = 6 isomorphs.  Anant Godbole, Feb 20 2013
a(n), n >= 0, with a(0) = 2, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 8 for b = 2*n. In general D = b^2  4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2.  Wolfdieter Lang, Aug 15 2013
With a different offset, this is 2*n^2  (n + 1)^2, which arises in one explanation of why Bertrand's postulate does not automatically prove Legendre's conjecture: as n gets larger, so does the range of numbers that can have primes that satisfy Bertrand's postulate yet do nothing for Legendre's conjecture.  Alonso del Arte, Nov 06 2013
x*(x + r*y)^2 + y*(y + r*x)^2 can be written as (x + y)*(x^2 + s*x*y + y^2). For r >= 0, the sequence gives the values of s: in fact, s = (r + 1)^2  2.  Bruno Berselli, Feb 20 2019
For n >= 2, the continued fraction expansion of sqrt(a(n)) is [n1; {1, n2, 1, 2n2}]. For n=2, this collapses to [1; {2}].  Magus K. Chu, Sep 06 2022


LINKS



FORMULA

For n > 1, a(n) = floor(n^5/(n^3 + n + 1)).  Gary Detlefs, Feb 10 2010
Sum_{n>=1} 1/a(n) = (1  sqrt(2)*Pi*cot(sqrt(2)*Pi))/4.
Sum_{n>=1} (1)^n/a(n) = (1  sqrt(2)*Pi*cosec(sqrt(2)*Pi))/4. (End)
Assume offset 0. Then a(n) = 2*LaguerreL(2, 1  n).  Peter Luschny, May 09 2021


EXAMPLE

G.f. = x + 2*x^2 + 7*x^3 + 14*x^4 + 23*x^5 + 34*x^6 + 47*x^7 + 62*x^8 + 79*x^9 + ...


MATHEMATICA



PROG

(Haskell)
a008865 = (subtract 2) . (^ 2) :: Integral t => t > t
a008865_list = scanl (+) (1) [3, 5 ..]


CROSSREFS



KEYWORD

sign,easy


AUTHOR



STATUS

approved



