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

1,2

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

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(p_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

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A276381 A259728 A000741 * A355498 A049991 A368611

Adjacent sequences: A133202 A133203 A133204 * A133206 A133207 A133208

Keyword

mult,easy,nonn

Author

Jonathan Vos Post, Oct 10 2007

Status

approved