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