OFFSET
1,4
COMMENTS
Inverse Möbius transform of A023533. - Antti Karttunen, Oct 01 2018
Records are a(1) = 1, a(4) = 2, a(20) = 4, a(120) = 5, a(280) = 6, a(560) = 7, a(840) = 8, a(1680) = 9, a(9240) = 11, a(18480) = 12, a(55440) = 13, a(120120) = 14, a(240240) = 15, a(314160) = 16, a(628320) = 17, a(1441440) = 18, a(2282280) = 19, a(4564560) = 21, a(9129120) = 22, a(13693680) = 23, a(27387360) = 24, a(54774720) = 25, a(68468400) = 26, a(77597520) = 27, a(136936800) = 28, a(155195040) = 29, a(310390080) = 30, a(465585120) = 31, a(775975200) = 32, a(1163962800) = 33, a(2327925600) = 36, a(4655851200) = 37, a(13967553600) = 38, a(16295479200) = 40. - Charles R Greathouse IV, Dec 19 2016
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..65537
Eric Weisstein's World of Mathematics, Tetrahedral Number.
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
EXAMPLE
a(10) = 2 because 10 has 4 divisors {1,2,5,10} among which 2 divisors {1,10} are tetrahedral numbers.
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}]
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ilya Gutkovskiy, Dec 13 2016
STATUS
approved