A279495 Number of tetrahedral numbers dividing n.
1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5
Offset: 1
Examples
a(10) = 2 because 10 has 4 divisors {1,2,5,10} among which 2 divisors {1,10} are tetrahedral numbers.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
- Eric Weisstein's World of Mathematics, Tetrahedral Number.
- Index to sequences related to pyramidal numbers.
Programs
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Mathematica
Table[SeriesCoefficient[Sum[x^(k (k + 1) (k + 2)/6)/(1 - x^(k (k + 1) (k + 2)/6)), {k, 1, n}], {x, 0, n}], {n, 1, 120}]
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PARI
a(n)=sum(k=1,sqrtnint(6*n,3),n%(k*(k+1)*(k+2)/6)==0) \\ Charles R Greathouse IV, Dec 13 2016
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PARI
isA000292(n)=my(k=sqrtnint(6*n,3)); k*(k+1)*(k+2)==6*n a(n)=sumdiv(n,d,isA000292(d)) \\ Charles R Greathouse IV, Dec 13 2016
Formula
G.f.: Sum_{k>=1} x^(k*(k+1)*(k+2)/6)/(1 - x^(k*(k+1)*(k+2)/6)).
a(n) = Sum_{d|n} A023533(d). - Antti Karttunen, Oct 01 2018
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3/2. - Amiram Eldar, Jan 02 2024
Comments