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A133205
Fully multiplicative with a(p) = p*(p+1)/2 for prime p.
2
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
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
KEYWORD
mult,easy,nonn
AUTHOR
Jonathan Vos Post, Oct 10 2007
STATUS
approved