A130827 -id:A130827 - cp's OEIS frontend

A130824 a(n) = 2*A004273(n).

0, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, 186, 190, 194, 198, 202, 206, 210, 214, 218, 222, 226, 230

Offset: 0

Paul Curtz, Jul 17 2007

Equals A111284 from the 2nd term on. - R. J. Mathar, Jun 13 2008
Besides the first term, this sequence gives the denominators of the alternating series Pi/8 = 1/2 - 1/6 + 1/10 - 1/14 + 1/18 - 1/22 + .... - Mohammad K. Azarian, Oct 14 2011 [edited by Jon E. Schoenfield, Mar 07 2015]
Numbers that cannot be a side of a primitive Pythagorean triangle. - Torlach Rush, Nov 07 2019
Simple continued fraction expansion of tanh(1/2) = (e - 1)/(e + 1) = 1/(2 + 1/(6 + 1/(10 + 1/(14 + ...)))). - Peter Bala, Oct 01 2023

Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A016825, A111284, A160327.

Concatenation([0], List([1..60], n-> 4*n-2 )); # G. C. Greubel, Dec 30 2019

[4*n-2*Floor((n+2) mod (n+1)):n in [0..60]]; // Vincenzo Librandi, Sep 22 2011

A130827 := proc(n) if n =0 then 0 ; else 4*n-2 ; fi ; end: seq(A130827(n),n=0..120) ; # R. J. Mathar, Oct 28 2007

2 Join[{0}, Range[1, 200, 2]] (* Michael De Vlieger, Mar 07 2015 *)

vector(61, n, if(n==1, 0, 4*(n-1) -2) ) \\ G. C. Greubel, Dec 30 2019

[0]+[4*n-2 for n in (1..60)] # G. C. Greubel, Dec 30 2019

From Stefano Spezia, Dec 09 2019: (Start)
G.f.: 2*x*(1+x)/(1-x)^2.
a(n) = 2*a(n-1) - a(n-2) for n > 0.
a(n) = 4*n - 1 - (-1)^(2^n). (End)
E.g.f: 2*(1 - (1-2*x)*exp(x)). - G. C. Greubel, Dec 30 2019

More terms from R. J. Mathar, Oct 28 2007

A361803 Least k > 1 such that k^n - n > 1 is semiprime, or 0 if no such k exists.

Original entry on oeis.org

5, 4, 5, 3, 6, 2, 2, 5, 8, 3, 4, 11, 15, 5, 2, 0, 4, 2, 14, 7, 48, 42, 6, 35, 2, 7, 602, 3, 16, 13, 2, 3, 2, 6, 37, 3185, 6, 9, 2, 33, 28, 2, 20, 9, 2, 135, 6, 5, 2, 49, 100, 5, 166, 5, 4, 9, 98, 15, 4, 27, 24, 2, 4, 17343, 34, 19, 24, 15, 56, 6, 90, 5, 2, 85

Offset: 1

Kevin P. Thompson, Jun 12 2023

nonn

For n = 16, k^16 - 16 = (k^8 - 4)(k^8 + 4) = (k^4 - 2)(k^4 + 2)(k^8 + 4) always has at least three factors, so a(16) = 0. Similarly for any n of the form (2m)^4, so a(A016744(n)) = 0.

			For n = 3:
k = 1: 1^3 - 3 = -2 < 0 so ignore.
k = 2: 2^3 - 3 = 5 which is not semiprime.
k = 3: 3^3 - 3 = 24 = 2 * 2 * 2 * 3 which is not semiprime.
k = 4: 4^3 - 3 = 61 which is not semiprime.
k = 5: 5^3 - 3 = 122 = 2 * 61 which is semiprime.
Therefore, a(3) = 5 since k = 5 is the first value for which k^3 - 3 is semiprime.

Kevin P. Thompson, Table of n, a(n) for n = 1..115
Kevin P. Thompson, Table of n, a(n) for n = 1..154 with uncertain values

Cf. A001358, A016744, A130827.

cp's OEIS Frontend

A130824 a(n) = 2*A004273(n).

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