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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.
6
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
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).
Sequence in context: A035930 A347470 A088117 * A195833 A257294 A330970
KEYWORD
nonn,base
AUTHOR
Michel Marcus, Nov 13 2021
STATUS
approved