A051895 Partial sums of second pentagonal numbers with even index (A049453).

0, 7, 33, 90, 190, 345, 567, 868, 1260, 1755, 2365, 3102, 3978, 5005, 6195, 7560, 9112, 10863, 12825, 15010, 17430, 20097, 23023, 26220, 29700, 33475, 37557, 41958, 46690, 51765, 57195, 62992, 69168, 75735, 82705, 90090, 97902, 106153, 114855, 124020, 133660

Offset: 0

Author

Barry E. Williams, Dec 17 1999

Keywords

• nonn
• easy

Comments

For A049453(n+1), the corresponding formula would be a(n)=(n+1)*(6*n+7) and its partial sums would be given by a(n)=(n+1)*(n+2)*(4*n+7)/2.

References

• A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A002492, A016061, A017605, A049453.

Programs

Magma

I:=[0, 7, 33, 90]; [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..50]]; // Vincenzo Librandi, Apr 27 2012

Mathematica

Table[(n(4n-1)(n-1))/2,{n,40}]  (* Harvey P. Dale, Mar 11 2011 *)
CoefficientList[Series[x*(7+5*x)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Apr 27 2012 *)

PARI

a(n) = n*(n+1)*(4*n+3)/2; \\ Altug Alkan, Apr 20 2018

Formula

a(n) = n*(n+1)*(4*n+3)/2.

G.f.: x*(7+5*x)/(1-x)^4. - Colin Barker, Jan 12 2012

a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). - Vincenzo Librandi, Apr 27 2012

a(n) = A002492(n) + A016061(n). - J. M. Bergot, Apr 20 2018