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

Offset: 1

Author

Michel Marcus, Nov 13 2021

Keywords

• nonn
• base

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

Examples

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

Links

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

Crossrefs

Cf. A349194, A349279 (fixed points).

Cf. A349864, A349865.

Programs

Mathematica

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

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])))
print([a(n) for n in range(1, 71)]) # Michael S. Branicky, Nov 13 2021

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)