A349278 - OEIS

A349278

a(n) is the product of the sum of the last i digits of n, with i going from 1 to the total number of digits of n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 0, 4, 10, 18, 28, 40, 54, 70, 88, 108, 0, 5, 12, 21, 32, 45, 60, 77, 96, 117, 0, 6, 14, 24, 36, 50, 66, 84, 104, 126, 0, 7, 16, 27, 40, 55, 72, 91, 112, 135, 0

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OFFSET

1,2

COMMENTS

This is similar to A349194 but with digits taken in reversed order.

The only primes in the sequence are 2, 3, 5 and 7. - Bernard Schott, Dec 04 2021

The positive terms form a subsequence of A349194. - Bernard Schott, Dec 19 2021

LINKS

Michel Marcus, Table of n, a(n) for n = 1..10000

FORMULA

From Bernard Schott, Dec 04 2021: (Start)

a(n) = 0 iff n is a multiple of 10 (A008592).

a(n) = 1 iff n = 1.

a(n) = 2 (resp. 3, 4, 5, 7, 9) iff n = 10^k+1 (A000533) (resp. 2*10^k+1 (A199682), 3*10^k+1 (A199683), 4*10^k+1 (A199684), 6*10^k+1 (A199686), 8*10^k+1 (A199689)).

a(R_n) = n! where R_n = A002275(n) is repunit > 0, and n! = A000142(n).

a(n) = A349194(n) if n is palindrome (A002113). (End)

EXAMPLE

For n=256, a(256) = 6*(6+5)*(6+5+2) = 858.

MATHEMATICA

a[n_] := Times @@ Accumulate @ Reverse @ IntegerDigits[n]; Array[a, 70] (* Amiram Eldar, Nov 13 2021 *)

PROG

(PARI) a(n) = my(d=Vecrev(digits(n))); prod(i=1, #d, sum(j=1, i, d[j]));

(Python)

from math import prod

from itertools import accumulate

def a(n): return 0 if n%10==0 else prod(accumulate(map(int, str(n)[::-1])))