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

1,2

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

Crossrefs

Cf. A349194, A349279 (fixed points).

Cf. A349864, A349865.

Sequence in context: A035930 A347470 A088117 * A195833 A257294 A330970

Adjacent sequences: A349275 A349276 A349277 * A349279 A349280 A349281

Keyword

nonn,base

Author

Michel Marcus, Nov 13 2021

Status

approved