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a(n) = 6^n + 1.
50

%I #33 Mar 13 2023 03:06:03

%S 2,7,37,217,1297,7777,46657,279937,1679617,10077697,60466177,

%T 362797057,2176782337,13060694017,78364164097,470184984577,

%U 2821109907457,16926659444737,101559956668417,609359740010497,3656158440062977

%N a(n) = 6^n + 1.

%H Vincenzo Librandi, <a href="/A062394/b062394.txt">Table of n, a(n) for n = 0..150</a>

%H Amelia Carolina Sparavigna, <a href="https://doi.org/10.18483/ijSci.2188">Some Groupoids and their Representations by Means of Integer Sequences</a>, International Journal of Sciences (2019) Vol. 8, No. 10.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,-6).

%F a(n) = 6*a(n-1) - 5.

%F a(n) = A000400(n) + 1.

%F a(n) = 7*a(n-1) - 6*a(n-2).

%F From _Mohammad K. Azarian_, Jan 02 2009: (Start)

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

%F E.g.f.: exp(x) + exp(6*x). (End)

%t 6^Range[0,30] +1

%t LinearRecurrence[{7,-6},{2,7},30] (* _Harvey P. Dale_, Aug 11 2015 *)

%o (Magma) [6^n + 1: n in [0..30] ]; // _Vincenzo Librandi_, Apr 30 2011

%o (PARI) vector(20, n, n--; 6^n + 1) \\ _Michel Marcus_, Jun 11 2015

%o (SageMath) [6^n+1 for n in range(31)] # _G. C. Greubel_, Mar 11 2023

%Y Sequences of the form m^n + 1: A000012 (m=0), A007395 (m=1), A000051 (m=2), A034472 (m=3), A052539 (m=4), A034474 (m=5), this sequence (m=6), A034491 (m=7), A062395 (m=8), A062396 (m=9), A062397 (m=10), A034524 (m=11), A178248 (m=12), A141012 (m=13), A228081 (m=64).

%Y Cf. A000400.

%K easy,nonn

%O 0,1

%A _Henry Bottomley_, Jun 22 2001