A101860 a(n) = (3+n)*(2 + 33*n + n^2)/6.

1, 24, 60, 110, 175, 256, 354, 470, 605, 760, 936, 1134, 1355, 1600, 1870, 2166, 2489, 2840, 3220, 3630, 4071, 4544, 5050, 5590, 6165, 6776, 7424, 8110, 8835, 9600, 10406, 11254, 12145, 13080, 14060, 15086, 16159, 17280, 18450, 19670, 20941

Offset: 0

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 18 2004

Keywords

• nonn
• easy

Comments

The 4th partial summation within series as series accumulate n times from an initial sequence of Euler Triangle's row 4: 1,11,11,1.

The partial sums of A101859 (plus a leading 1). 4th row in the array shown in the examples. The 2nd column is A101104, the 3rd column is A101103, the 4th column is A005914.

Examples

			Array with first column equal to the 4th row of A008292, and column k defined by partial sums of the preceding column k-1:
1  1 1 1  1  1  1  1  1  1  1
11 12 13 14 15 16 17 18 19 20 21
11 23 36 50 65 81 98 116 135 155 176
1  24 60 110 175 256 354 470 605 760 936   A101860
0  24 84 194 369 625 979 1449 2054 2814 3750   A101861
0  24 108 302 671 1296 2275 3724 5778 8592 12342  A101862
0  24 132 434 1105 2401 4676 8400 14178 22770 35112
0  24 156 590 1695 4096 8772 17172 31350 54120 89232
0  24 180 770 2465 6561 15333 32505 63855 117975 207207
... ... ... ... ... ... ... ... ... ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Programs

Magma

I:=[1, 24, 60, 110]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jun 26 2012

Mathematica

LinearRecurrence[{4,-6,4,-1},{1,24,60,110},50] (* or *) CoefficientList[Series[(1+20*x-30*x^2+10*x^3)/(x-1)^4,{x,0,50}],x] (* Vincenzo Librandi, Jun 26 2012 *)

PARI

a(n) = (n+3)*(n^2+33*n+2)/6; \\ Altug Alkan, Sep 23 2018

Formula

G.f.: ( 1 + 20*x - 30*x^2 + 10*x^3 ) / (x-1)^4 . - R. J. Mathar, Dec 06 2011

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 26 2012