A260955 - OEIS

%I #52 Jan 12 2025 16:38:39

%S 6,54,150,294,486,726,1014,1350,1734,2166,2646,3174,3750,4374,5046,

%T 5766,6534,7350,8214,9126,10086,11094,12150,13254,14406,15606,16854,

%U 18150,19494,20886,22326,23814,25350,26934,28566,30246,31974,33750,35574,37446,39366,41334

%N Differences of the increasing arithmetic progression a^2+a, b^2+b, c^2+c, where b = 5*a+2, c = 7*a+3 and a >= 0.

%H Colin Barker, <a href="/A260955/b260955.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 24*n^2 + 24*n + 6.

%F From _Colin Barker_, Aug 05 2015: (Start)

%F G.f.: 6*(1 + 6*x + x^2)/(1 - x)^3.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)

%F From _Bruno Berselli_, Aug 05 2015: (Start)

%F a(n) = A032528(4*n+2).

%F a(n)*(2*h-1)^2 = a((2*h-1)*n+h-1). For h=0, a(n) = a(-n-1); for h=7, 169*a(n) = a(13*n+6). (End)

%F From _Elmo R. Oliveira_, Dec 28 2024: (Start)

%F E.g.f.: 6*exp(x)*(1 + 8*x + 4*x^2).

%F a(n) = 6*A016754(n). (End)

%e By the definition, given a = 7 and b = 5*7+2 = 37, c = 7*7+3 = 52, it follows that a^2+a = 56, b^2+b = 1406, c^2+c = 2756, where 56, 1406, 2756 are in arithmetic progression. Therefore, 2756-1406 = 1406-56 = 1350 and 1350 is in the sequence (8th term).

%t Table[24 n^2 + 24 n + 6, {n, 0, 40}] (* _Bruno Berselli_, Aug 05 2015 *)

%t LinearRecurrence[{3, -3, 1}, {6, 54, 150}, 50] (* _Vincenzo Librandi_, Aug 05 2015 *)

%o (PARI) Vec(6*(1+6*x+x^2)/(1-x)^3 + O(x^100)) \\ _Colin Barker_, Aug 05 2015

%o (Magma) [24*n^2+24*n+6: n in [0..40]]; // _Vincenzo Librandi_, Aug 05 2015

%Y Cf. A016754, A032528.

%Y Second bisection of A033581 (the first bisection is A195824).

%K nonn,easy

%O 0,1

%A _Marco Ripà_, Aug 05 2015