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A260810 a(n) = n^2*(3*n^2 - 1)/2. 6

%I #34 Aug 25 2022 04:35:43

%S 0,1,22,117,376,925,1926,3577,6112,9801,14950,21901,31032,42757,57526,

%T 75825,98176,125137,157302,195301,239800,291501,351142,419497,497376,

%U 585625,685126,796797,921592,1060501,1214550,1384801,1572352,1778337,2003926,2250325,2518776

%N a(n) = n^2*(3*n^2 - 1)/2.

%C Pentagonal numbers with square indices.

%C After 0, a(k) is a square if k is in A072256.

%H Bruno Berselli, <a href="/A260810/b260810.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: x*(1 + x)*(1 + 16*x + x^2)/(1 - x)^5.

%F a(n) = a(-n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

%F a(n) = A245288(2*n^2).

%F a(n) = A001318(2*n^2-1) with A001318(-1) = 0.

%F From _Amiram Eldar_, Aug 25 2022: (Start)

%F Sum_{n>=1} 1/a(n) = 3 - Pi^2/3 - sqrt(3)*Pi*cot(Pi/sqrt(3)).

%F Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi*cosec(Pi/sqrt(3)) - Pi^2/6 - 3. (End)

%p A260810:=n->n^2*(3*n^2 - 1)/2: seq(A260810(n), n=0..50); # _Wesley Ivan Hurt_, Apr 25 2017

%t Table[n^2 (3 n^2 - 1)/2, {n, 0, 40}]

%t LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 22, 117, 376}, 40] (* _Vincenzo Librandi_, Aug 23 2015 *)

%o (PARI) vector(40, n, n--; n^2*(3*n^2-1)/2)

%o (Sage) [n^2*(3*n^2-1)/2 for n in (0..40)]

%o (Magma) [n^2*(3*n^2-1)/2: n in [0..40]];

%o (Magma) I:=[0,1,22,117,376]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; // _Vincenzo Librandi_, Aug 23 2015

%Y Subsequence of A001318 and A245288 (see Formula field).

%Y Cf. A000326, A193218 (first differences).

%Y Cf. A000583 (squares with square indices), A002593 (hexagonal numbers with square indices).

%Y Cf. A232713 (pentagonal numbers with pentagonal indices), A236770 (pentagonal numbers with triangular indices).

%K nonn,easy

%O 0,3

%A _Bruno Berselli_, Jul 31 2015

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Last modified March 28 18:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)