A133205 Fully multiplicative with a(p) = p*(p+1)/2 for prime p.

1, 3, 6, 9, 15, 18, 28, 27, 36, 45, 66, 54, 91, 84, 90, 81, 153, 108, 190, 135, 168, 198, 276, 162, 225, 273, 216, 252, 435, 270, 496, 243, 396, 459, 420, 324, 703, 570, 546, 405, 861, 504, 946, 594, 540, 828, 1128, 486, 784, 675, 918, 819, 1431, 648, 990, 756

Offset: 1

Author

Jonathan Vos Post, Oct 10 2007

Keywords

• mult
• easy
• nonn

Comments

There are analogs with the triangular numbers replaced by some other sequence, but this was chosen because of the parity coincidences of A034953.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A000217, A003958, A003959, A003960, A003961, A003964, A034953, A061142, A167338.

Programs

Mathematica

f[p_, e_] := (p*(p + 1)/2)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 60] (* Amiram Eldar, Dec 24 2022 *)

PARI

a(n)=my(f=factor(n));prod(i=1,#f[,1],binomial(f[i,1]+1,2)^f[i,2]) /* Charles R Greathouse IV, Sep 09 2010 */

PARI

for(n=1, 100, print1(direuler(p=2, n, 1 + (p^2 + p) / (2/X - p^2 - p))[n], ", ")) \\ Vaclav Kotesovec, Apr 05 2023

Formula

a((p_1)^(e_1)*(p_2)^(e_2)*...*(p_k)^(e_k)) = T(p_1)^(e_1)*T(p_2)^(e_2)*...*T(p_k)^(e_k), where T(i) = A000217(i). a(prime(i)) = A034953(i).

Sum_{n>=1} 1/a(n) = Product_{p prime} (1 - 2/(p*(p+1)))^(-1) = 2.12007865309570462566... . - Amiram Eldar, Dec 24 2022

Dirichlet g.f.: Product_{p prime} (1 + (p^2 + p) / (2*p^s - p^2 - p)). - Vaclav Kotesovec, Apr 05 2023

a(n) = A167338(n)/A061142(n). - Vaclav Kotesovec, Jan 28 2025

Conjecture: Sum_{k=1..n} a(k) = O(n^3/log(n)). - Vaclav Kotesovec, Jan 28 2025