A051937 Truncated triangular pyramid numbers: a(n) = Sum_{k=4..n} (k*(k+1)/2 - 9).
1, 7, 19, 38, 65, 101, 147, 204, 273, 355, 451, 562, 689, 833, 995, 1176, 1377, 1599, 1843, 2110, 2401, 2717, 3059, 3428, 3825, 4251, 4707, 5194, 5713, 6265, 6851, 7472, 8129, 8823, 9555, 10326, 11137, 11989, 12883, 13820, 14801, 15827, 16899, 18018
Offset: 4
Links
- Vincenzo Librandi, Table of n, a(n) for n = 4..5000
- Milan Janjic, Binomial Coefficients and Enumeration of Restricted Words, Journal of Integer Sequences, 2016, Vol 19, #16.7.3.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
Cf. A000292.
Programs
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Magma
I:=[1, 7, 19, 38]; [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, Apr 28 2012
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Mathematica
CoefficientList[Series[(1+3*x-3*x^2)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Apr 28 2012 *) LinearRecurrence[{4,-6,4,-1},{1,7,19,38},50] (* Harvey P. Dale, Aug 12 2012 *)
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PARI
a(n)=(n-3)*(n^2+6*n-34)/6 \\ Charles R Greathouse IV, Nov 10 2015
Formula
a(n) = (1/6)*(n-3)*(n^2+6*n-34).
G.f.: x^4*(1+3*x-3*x^2)/(1-x)^4. - Colin Barker, Mar 19 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Apr 28 2012
Comments