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A111080 Sum of numbers under a triangle on a spiral staircase of width 10.

%I A111080 #26 Dec 22 2023 18:30:54
%S A111080 11,44,110,220,385,616,924,1320,1815,2420,3146,4004,5005,6160,7480,
%T A111080 8976,10659,12540,14630,16940,19481,22264,25300,28600,32175,36036,
%U A111080 40194,44660,49445,54560,60016,65824,71995,78540,85470,92796,100529,108680,117260
%N A111080 Sum of numbers under a triangle on a spiral staircase of width 10.
%C A111080 Write the numbers starting from 0 on a spiral of width 10, so 10 is above 0, 11 is above 1, and so on. Draw a triangle n+1 units high and n+1 units across on the spiral, overlapping if n > 9. The sequence gives the sum of the covered numbers (with multiplicity).
%H A111080 Harvey P. Dale, <a href="/A111080/b111080.txt">Table of n, a(n) for n = 0..1000</a>
%H A111080 Duncan Keith, <a href="http://www.subtangent.com/maths/ig-numberstairs.php">Number stairs</a>
%H A111080 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A111080 a(n) = 11n(n^2 + 3n + 2)/6. - _Charles R Greathouse IV_, Jun 22 2013
%F A111080 From _Colin Barker_, Apr 03 2014: (Start)
%F A111080 a(n) = (11*(1+n)*(2+n)*(3+n))/6.
%F A111080 a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4).
%F A111080 G.f.: 11 / (x-1)^4. (End).
%p A111080 A111080:=n->(11*(1+n)*(2+n)*(3+n))/6; seq(A111080(n), n=0..50); # _Wesley Ivan Hurt_, Apr 03 2014
%t A111080 Table[(11 (1 + n) (2 + n) (3 + n))/6, {n, 0, 50}] (* _Wesley Ivan Hurt_, Apr 03 2014 *)
%t A111080 LinearRecurrence[{4,-6,4,-1},{11,44,110,220},40] (* _Harvey P. Dale_, Dec 22 2023 *)
%o A111080 (PARI) a(n)=11*n*(n^2+3*n+2)/6 \\ _Charles R Greathouse IV_, Jun 22 2013
%o A111080 (PARI) Vec(11/(x-1)^4 + O(x^100)) \\ _Colin Barker_, Apr 03 2014
%K A111080 nonn,easy
%O A111080 0,1
%A A111080 Maria Tkachenko (marytkachenko(AT)mail.com), Oct 12 2005
%E A111080 Entry rewritten by Sophia Greathouse and _Charles R Greathouse IV_, Jun 22 2013