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A051895
Partial sums of second pentagonal numbers with even index (A049453).
5
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
0,2
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.
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
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 *)
PROG
(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
(PARI) a(n) = n*(n+1)*(4*n+3)/2; \\ Altug Alkan, Apr 20 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Barry E. Williams, Dec 17 1999
STATUS
approved