A349682 a(n) = A000292(6*n + 1) where A000292 are the tetrahedral numbers.
1, 84, 455, 1330, 2925, 5456, 9139, 14190, 20825, 29260, 39711, 52394, 67525, 85320, 105995, 129766, 156849, 187460, 221815, 260130, 302621, 349504, 400995, 457310, 518665, 585276, 657359, 735130, 818805, 908600, 1004731, 1107414, 1216865, 1333300, 1456935, 1587986
Offset: 0
Links
- Euclid of Alexandria, Elements, VII Def. 17, p. 194, 300 BCE.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Programs
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Mathematica
nterms=50;Table[36n^3+36n^2+11n+1,{n,0,nterms-1}] (* Paolo Xausa, Nov 25 2021 *) -
PARI
a(n) = subst(m*(m+1)*(m+2)/6, 'm, 6*n+1); \\ Michel Marcus, Dec 16 2021
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Python
def A349682(n): return n*(n*(36*n + 36) + 11) + 1 # Chai Wah Wu, Dec 27 2021
Formula
a(n) = 1 + 11*n + 36*n^2 + 36*n^3 = (1 + 2*n)*(1 + 3*n)*(1 + 6*n).
G.f.: (1 + 80*x + 125*x^2 + 10*x^3)/(1 - x)^4. - Stefano Spezia, Nov 29 2021
From Elmo R. Oliveira, Aug 22 2025: (Start)
E.g.f.: exp(x)*(1 + 83*x + 144*x^2 + 36*x^3).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
From Amiram Eldar, Aug 31 2025: (Start)
Sum_{n>=0} 1/a(n) = Pi/(4*sqrt(3)) + 2*log(2) - 3*log(3)/4.
Sum_{n>=0} (-1)^n/a(n) = (3/4 - 1/sqrt(3))*Pi + sqrt(3)*log(2 + sqrt(3))/2 - log(2). (End)