A193576 a(n) = T(n)^3 + n^3 where T(n) is a triangular number.

2, 35, 243, 1064, 3500, 9477, 22295, 47168, 91854, 167375, 288827, 476280, 755768, 1160369, 1731375, 2519552, 3586490, 5006043, 6865859, 9269000, 12335652, 16204925, 21036743, 27013824, 34343750, 43261127, 54029835, 66945368, 82337264, 100571625, 122053727

Offset

1,1

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Formula

a(n) = (n^3*(n^3+3*n^2+3*n+9)/8) = (1/8)*(n+3)*(n^2+3)*n^3.

From Chai Wah Wu, Jun 12 2025: (Start)

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n > 7.

G.f.: x*(x^5 - 28*x^3 - 40*x^2 - 21*x - 2)/(x - 1)^7. (End)

a(n) = A000578(n) + A059827(n). - Alois P. Heinz, Jun 12 2025

E.g.f.: exp(x)*x*(16 + 124*x + 192*x^2 + 98*x^3 + 18*x^4 + x^5)/8. - Stefano Spezia, Jun 13 2025

Prog

(Magma) [(n^3*(n^3+3*n^2+3*n+9)/8): n in [1..40]];
(Python)
def A193576(n): return n**3*(n*(n*(n+3)+3)+9)>>3 # Chai Wah Wu, Jun 12 2025

Crossrefs

Cf. A000217, A000578, A059827.

Sequence in context: A257601 A282727 A042353 * A381146 A384726 A297538

Adjacent sequences: A193573 A193574 A193575 * A193577 A193578 A193579

Keyword

nonn,easy

Author

Vincenzo Librandi, Sep 08 2011

Status

approved