A349279 -id:A349279 - cp's OEIS frontend

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

Michel Marcus, Nov 13 2021

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

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

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

Cf. A349194, A349279 (fixed points).

Cf. A349864, A349865.

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

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

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

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)

A349190 Numbers k such that k equals the product of the sum of its first i digits, with i going from 1 to the total number of digits of k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 48, 24192

Offset: 1

Malo David, Nov 09 2021

a(12) > 10^22 if it exists. - Max Alekseyev, Jan 19 2025

			24192 is a term since 24192 = 2*(2+4)*(2+4+1)*(2+4+1+9)*(2+4+1+9+2).

Cf. A055642, A349194, A349279.

Select[Range[10^5],Times@@Total/@Table[IntegerDigits[#][[;;k]],{k,IntegerLength@#}]==#&] (* Giorgos Kalogeropoulos, Nov 10 2021 *)

isok(k) = {my(d=digits(k)); prod(i=1, #d, sum(j=1, i, d[j])) == k;} \\ Michel Marcus, Nov 10 2021

def main(N): # prints all terms <= N
  for k in range(1,N+1):
    n1=str(k)
    n2 = 1
    for i in range(1,len(n1)+1):
      sum1 = 0
      for j in range(0,i):
        sum1 += int(n1[j])
      n2 = n2*sum1
    if n2 == k:
       print(k, end=", ")

from itertools import islice, accumulate, count
from math import prod
def A349190gen(): return filter(lambda n:prod(accumulate(int(d) for d in str(n))) == n,count(1)) # generator of terms
A349190_list = list(islice(A349190gen(),11)) # Chai Wah Wu, Dec 02 2021

cp's OEIS Frontend

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.

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A349190 Numbers k such that k equals the product of the sum of its first i digits, with i going from 1 to the total number of digits of k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Python

Python