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

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

Author

Maria Tkachenko (marytkachenko(AT)mail.com), Oct 12 2005

Keywords

• nonn
• easy

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).

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Duncan Keith, Number stairs
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Programs

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 *)

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

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).

Extensions

Entry rewritten by Sophia Greathouse and Charles R Greathouse IV, Jun 22 2013