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

3, 9, 18, 20, 25, 33, 35, 43, 57, 57, 66, 84, 111, 119, 139, 171, 176, 196, 231, 231, 249, 285, 339, 353, 388, 444, 452, 484, 540, 540, 567, 621, 702, 702, 702, 702, 703, 707, 714, 714, 720, 732, 750, 756, 771, 795, 799, 815, 843, 843, 858, 888, 933, 945, 975, 1023, 1030, 1058, 1107

Offset: 1

Author

Luca Onnis, Feb 09 2023

Keywords

• nonn
• base

Examples

			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.
		

Crossrefs

Cf. A061076, A061077, A061078.

Programs

Mathematica

Accumulate[Times @@@ IntegerDigits[Range[3, 999 , 3]]]

PARI

a(n)={sum(k=1, n, vecprod(digits(3*k)))} \\ Andrew Howroyd, Feb 09 2023

Python

from math import prod
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

Formula

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))).

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)).

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.