A349194 a(n) is the product of the sum of the first i digits of n, as i goes from 1 to the total number of digits of n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 49, 56, 63, 70, 77

Offset: 1

Author

Malo David, Nov 10 2021

Keywords

• nonn
• base
• look

Comments

The only primes in the sequence are 2, 3, 5 and 7. - Bernard Schott, Nov 23 2021

Examples

			For n=256, a(256) = 2*(2+5)*(2+5+6) = 182.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Malo David, About the Product Sum Digit Sequence

Crossrefs

Cf. A055642, A284001 (binary analog), A349190 (fixed points).

Cf. A007953 (sum of digits), A059995 (floor(n/10)).

Cf. A349278 (similar, with the last digits).

Cf. A349898, A349899.

Programs

Magma

f:=func; [f(n):n in [1..100]]; // Marius A. Burtea, Nov 23 2021

Mathematica

Table[Product[Sum[Part[IntegerDigits[n],j],{j,i}],{i,Length[IntegerDigits[n]]}],{n,74}] (* Stefano Spezia, Nov 10 2021 *)

PARI

a(n) = my(d=digits(n)); prod(i=1, #d, sum(j=1, i, d[j])); \\ Michel Marcus, Nov 10 2021

PARI

first(n)=if(n<9,return([1..n])); my(v=vector(n)); for(i=1,9,v[i]=i); for(i=10,n, v[i]=sumdigits(i)*v[i\10]); v \\ Charles R Greathouse IV, Dec 04 2021

Python

from math import prod
from itertools import accumulate
def a(n): return prod(accumulate(map(int, str(n))))
print([a(n) for n in range(1, 100)]) # Michael S. Branicky, Nov 10 2021

Formula

For n>10: a(n) = a(A059995(n))*A007953(n) where A059995(n) = floor(n/10).

In particular, for n<100: a(n) = floor(n/10)*A007953(n)

From Bernard Schott, Nov 23 2021: (Start)

a(n) = 1 iff n = 10^k, k >= 0 (A011557).

a(n) = 2 iff n = 10^k + 1, k >= 0 (A000533 \ {1}).

a(n) = 3 iff n = 10^k + 2, k >= 0 (A133384).

a(n) = 5 iff n = 10^k + 4, k >= 0.

a(n) = 7 iff n = 10^k + 6, k >= 0. (End)

From Marius A. Burtea, Nov 23 2021: (Start)

a(A002275(n)) = n! = A000142(n), n >= 1.

a(A090843(n - 1)) = (2*n - 1)!! = A001147(n), n >= 1.

a(A097166(n)) = (3*n - 2)!!! = A007559(n).

a(A093136(n)) = 2^n = A000079(n).

a(A093138(n)) = 3^n = A000244(n). (End)