A185505 a(n) = (7*n^4 + 5*n^2)/12.

1, 11, 51, 156, 375, 771, 1421, 2416, 3861, 5875, 8591, 12156, 16731, 22491, 29625, 38336, 48841, 61371, 76171, 93500, 113631, 136851, 163461, 193776, 228125, 266851, 310311, 358876, 412931, 472875, 539121, 612096, 692241, 780011, 875875, 980316, 1093831, 1216931, 1350141, 1494000, 1649061, 1815891, 1995071

Offset

1,2

Comments

a(n) is the sum of terms in the square [1,n]x[1,n] of the natural number array A000027; e.g., the [1,3]x[1,3] square is

1..2..4

3..5..8

6..9..13,

so that a(1) = 1, a(2) = 1+2+3+5 = 11, a(3) = 1+2+3+4+5+6+8+9+13 = 51.

Partial sums of A063490. - Omar E. Pol, Oct 23 2019

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = (7*n^4 + 5*n^2)/12.

From Chai Wah Wu, Sep 05 2016: (Start)

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.

G.f.: x*(1 + x)*(1 + 5*x + x^2)/(1 - x)^5. (End)

E.g.f.: (1/12)*x*(12 + 54*x + 42*x^2 + 7*x^3)*exp(x). - G. C. Greubel, Jul 07 2017

Example

a(1)=(7+5)/12, a(2)=(7*16+5*4)/12.

Mathematica

Table[(7*n^4+5*n^2)/12, {n, 1, 60}]
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 11, 51, 156, 375}, 50] (* Harvey P. Dale, Jan 26 2024 *)

Prog

(PARI) a(n)=(7*n^4+5*n^2)/12 \\ Charles R Greathouse IV, Sep 05 2016

Crossrefs

Cf. A000027, A063490.

Sequence in context: A067983 A175360 A226451 * A051843 A107464 A027942

Adjacent sequences: A185502 A185503 A185504 * A185506 A185507 A185508

Keyword

nonn,easy

Author

Clark Kimberling, Jan 29 2011

Status

approved