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A360365 a(n) = sum of the products of the digits of the first n positive multiples of 3.

%I A360365 #37 Feb 28 2023 12:34:43
%S A360365 3,9,18,20,25,33,35,43,57,57,66,84,111,119,139,171,176,196,231,231,
%T A360365 249,285,339,353,388,444,452,484,540,540,567,621,702,702,702,702,703,
%U A360365 707,714,714,720,732,750,756,771,795,799,815,843,843,858,888,933,945,975,1023,1030,1058,1107
%N A360365 a(n) = sum of the products of the digits of the first n positive multiples of 3.
%F A360365 a((10^n-1)/3) = (1/836)*(15*(19*45^n-63) + 44*3^((3*n)/2)*(sqrt(3)*sin((Pi*n)/6) + 15*cos((Pi*n)/6))).
%F A360365 G.f. of the subsequence a((10^n-1)/3): 9*(2 - 32*x + 135*x^2)/((1 - x)*(1 - 45*x)*(1 - 9*x + 27*x^2)).
%F A360365 a((10^n-1)/3) = 55*a((10^(n-1)-1)/3) - 486*a((10^(n-2)-1)/3) + 1647*a((10^(n-3)-1)/3) - 1215*a((10^(n-4)-1)/3) for n > 4.
%e A360365 a(4) = 20 since the first 4 positive multiples of 3 are 3,6,9 and 12 and the sum of the product of their digits is 3 + 6 + 9 + 1*2 = 20.
%t A360365 Accumulate[Times @@@ IntegerDigits[Range[3, 999 , 3]]]
%o A360365 (PARI) a(n)={sum(k=1, n, vecprod(digits(3*k)))} \\ _Andrew Howroyd_, Feb 09 2023
%o A360365 (Python)
%o A360365 from math import prod
%o A360365 def A360365(n): return sum(prod(int(d) for d in str(m)) for m in range(3,3*n+1,3)) # _Chai Wah Wu_, Feb 28 2023
%Y A360365 Cf. A061076, A061077, A061078.
%K A360365 nonn,base
%O A360365 1,1
%A A360365 _Luca Onnis_, Feb 09 2023