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a(n) = n*(7*n + 11)/2 + 1.
11

%I #39 Dec 25 2024 11:03:15

%S 1,10,26,49,79,116,160,211,269,334,406,485,571,664,764,871,985,1106,

%T 1234,1369,1511,1660,1816,1979,2149,2326,2510,2701,2899,3104,3316,

%U 3535,3761,3994,4234,4481,4735,4996,5264,5539,5821,6110,6406,6709,7019,7336,7660,7991

%N a(n) = n*(7*n + 11)/2 + 1.

%C First bisection of A193053 (see also the numerical spiral illustrated in the Links section).

%C The inverse binomial transform yields 1, 9, 7, 0, 0 (0 continued).

%H Bruno Berselli, <a href="/A198017/b198017.txt">Table of n, a(n) for n = 0..1000</a>

%H Bruno Berselli, <a href="http://www.base5forum.it/upload/a198017.jpg">Illustration of initial terms</a>.

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

%F G.f.: (1 + 7*x - x^2)/(1-x)^3.

%F a(n) = A195020(2*n) + 2*n + 1.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) = 2*a(n-1) - a(n-2) + 7.

%F From _Elmo R. Oliveira_, Dec 24 2024: (Start)

%F E.g.f.: exp(x)*(2 + 18*x + 7*x^2)/2.

%F a(n) = n + A001106(n+1). (End)

%t Table[(n(7n+11))/2+1,{n,0,60}] (* or *) LinearRecurrence[{3,-3,1},{1,10,26},60] (* _Harvey P. Dale_, Mar 03 2013 *)

%o (PARI) for(n=0, 47, print1(n*(7*n+11)/2+1", "));

%o (Magma) [n*(7*n+11)/2+1: n in [0..47]];

%Y Cf. A195020 (vertices of the numerical spiral in link).

%Y Cf. A017005 (first differences).

%Y Cf. A069099, A001106.

%Y Cf. A001106, A022264, A033572, A144555, A152760, A158482, A158485, A185019, A193053, A195021, A195023-A195030, A195320 [incomplete list].

%K nonn,easy

%O 0,2

%A _Bruno Berselli_, Oct 21 2011 - based on remarks and sequences by _Omar E. Pol_