A076921 - OEIS

%I #31 Dec 11 2025 03:35:25

%S 1,1,2,4,6,9,12,16,20,25,30,36,42,49,56,64,72,81,90,100,110,121,132,

%T 144,156,169,182,196,210,225,240,256,272,289,306,324,342,361,380,400,

%U 420,441,462,484,506,529,552,576,600,625,650,676,702,729,756,784

%N Smallest number such that the highest common factor of pair of successive terms follows the pattern 1, 1, 2, 2, 3, 3, 4, 4, ..., i.e., b(2n-1) = b(2n) = n given by A004526.

%C (1) a(2n) = n^2, a(2n-1) = n(n+1) = twice the n-th triangular number.

%C (2) Geometric mean of the successive squares interleaved between them.

%C Essentially the same as A002620.

%H G. C. Greubel, <a href="/A076921/b076921.txt">Table of n, a(n) for n = 1..5000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).

%F a(n+1) + a(n) = binomial(n+1,2), a(1) = a(2) = 1. - _G. C. Greubel_, Oct 29 2017

%F From _Stefano Spezia_, Nov 16 2024: (Start)

%F a(n) = (2*n^2 + (-1)^n - 1)/8 for n > 1.

%F G.f.: x*(1 - x + 2*x^2 - x^4)/((1 - x)^3*(1 + x)).

%F E.g.f.: (4*x + x*(1 + x)*cosh(x) - (1 - x - x^2)*sinh(x))/4. (End)

%F From _Amiram Eldar_, Dec 11 2025: (Start)

%F Sum_{n>=1} 1/a(n) = Pi^2/6 + 2 = A013661 + 2.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 2 - Pi^2/6 (A152416). (End)

%t Join[{1}, LinearRecurrence[{2, 0, -2, 1}, {1, 2, 4, 6}, 50]] (* _G. C. Greubel_, Oct 29 2017 *)

%Y Cf. A002620, A004526, A013661, A152416.

%K nonn,easy

%O 1,3

%A _Amarnath Murthy_, Oct 17 2002

%E More terms from _Philippe Deléham_, Jun 20 2005