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A111080
Sum of numbers under a triangle on a spiral staircase of width 10.
2
11, 44, 110, 220, 385, 616, 924, 1320, 1815, 2420, 3146, 4004, 5005, 6160, 7480, 8976, 10659, 12540, 14630, 16940, 19481, 22264, 25300, 28600, 32175, 36036, 40194, 44660, 49445, 54560, 60016, 65824, 71995, 78540, 85470, 92796, 100529, 108680, 117260
OFFSET
0,1
COMMENTS
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).
FORMULA
a(n) = 11n(n^2 + 3n + 2)/6. - Charles R Greathouse IV, Jun 22 2013
From Colin Barker, Apr 03 2014: (Start)
a(n) = (11*(1+n)*(2+n)*(3+n))/6.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4).
G.f.: 11 / (x-1)^4. (End).
MAPLE
A111080:=n->(11*(1+n)*(2+n)*(3+n))/6; seq(A111080(n), n=0..50); # Wesley Ivan Hurt, Apr 03 2014
MATHEMATICA
Table[(11 (1 + n) (2 + n) (3 + n))/6, {n, 0, 50}] (* Wesley Ivan Hurt, Apr 03 2014 *)
LinearRecurrence[{4, -6, 4, -1}, {11, 44, 110, 220}, 40] (* Harvey P. Dale, Dec 22 2023 *)
PROG
(PARI) a(n)=11*n*(n^2+3*n+2)/6 \\ Charles R Greathouse IV, Jun 22 2013
(PARI) Vec(11/(x-1)^4 + O(x^100)) \\ Colin Barker, Apr 03 2014
CROSSREFS
Sequence in context: A248126 A253445 A172526 * A299288 A299286 A022816
KEYWORD
nonn,easy
AUTHOR
Maria Tkachenko (marytkachenko(AT)mail.com), Oct 12 2005
EXTENSIONS
Entry rewritten by Sophia Greathouse and Charles R Greathouse IV, Jun 22 2013
STATUS
approved