OFFSET
1,2
COMMENTS
What is the asymptotic behavior of this sequence? a(n) = a(n+1) for almost all n. A weak upper bound: a(n) << n^1.91. - Charles R Greathouse IV, Jan 13 2012
A check was done for k in {i^j | 1 <= i <= 10 AND 1 <= j <= 100}. For all these values, a(k) < k^1.733. Another check for k in {i^j | 101 <= i <= 110 AND 101 <= j <= 200} gave a(k) < k^1.65324. For k in {i | 10^6 <= i <= 10^7}, a(k) < k^1.6534. So I ask: is it true that a(n) < n^1.733 and a(n) -> n^(1.65323 + o(1)), or about n^(log(45)/log(10) + o(1))? - David A. Corneth, May 17 2016
For n = 10^(k-1), the closed-form formula from Mihai Teodor (see Formula section) gives a(n) = (45^k - 45)/44, so lim_{n->oo} log(a(n))/log_10(n) = log(45) = 3.80666248977.... - Jon E. Schoenfield, Apr 10 2022
For k >= 1, a(10^k-1) = a(10^k) = ... = a(10*R_k) where R = A002275; so there is a run of 10*R_{k-1} + 2 = A047855(k) consecutive terms equal to (45/44)*(45^k-1) when n runs from 10^k-1 up to 10*R_k, this is because those numbers have one or more 0's. Example: first runs with 2, 12, 112, 1112, ... consecutive terms equal to 45, 2070, 93195, 4193820, ... start at 9, 99, 999, 9999, ... and end at 10, 110, 1110, 11110, ... - Bernard Schott, Oct 18 2022
REFERENCES
Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.
LINKS
Daria Micovic, Table of n, a(n) for n = 1..10000
Amarnath Murthy, Smarandache friendly numbers and a few more sequences, vixra, 2014.
Luca Onnis, On the general Smarandache's sigma product of digits, arXiv:2203.07227 [math.GM], 2022.
FORMULA
a(n) = Sum_{k = 1..n} (product of the digits of k).
a(10^k-1) = (45/44)*(45^k-1). - Giovanni Resta, Oct 18 2012
From Robert Israel, May 17 2016: (Start)
Partial sums of A007954.
G.f.: (1-x)^(-1) * Sum_{n>=0} Product_{j=0..n} Sum_{k=1..9} k * x^(k*10^j).
G.f. satisfies A(x) = (x + 2*x^2 + ... + 9*x^9)*(1+(1-x^10)*A(x^10))/(1-x).
(End)
Let b(1), b(2), ..., b(k) be the digits of the base-10 expansion of n: n = b(1)*10^(k-1) + b(2)*10^(k-2) + ... + b(k). Then a(n) = b(1)*b(2)*...*b(k) + (45^k-45)/44 + (1/2)*Sum_{i=1..k} b(1)*b(2)*...*b(i)*(b(i)-1)*45^(k-i). - Mihai Teodor, Apr 09 2022
EXAMPLE
a(9) = a(10) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1*0 = 1+2+3+4+5+6+7+8+9 = 45.
MAPLE
A007954:= n -> convert(convert(n, base, 10), `*`):
ListTools:-PartialSums(map(A007954, [$1..100])); # Robert Israel, May 17 2016
MATHEMATICA
Accumulate[Times@@IntegerDigits[#]& /@ Range[100]]
PROG
(PARI) pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
a(n) = sum(k=1, n, pd(k)); \\ Michel Marcus, Feb 01 2015
(PARI) a(n) = {n=digits(n); p=1; d=#n; for(i=1, #n, if(n[i]==0, d=i-1; break));
(45/44) * (45^(#n-1)-1) + sum(i=1, d, p*=n[i]; p * (n[i]-1) * (45/44) * (45^(#n -i) - 45^(#n-i-1)) / 2)+p*(d==#n)} \\ David A. Corneth, May 17 2016
(Sage)
def A061076(n):
p = 0
i = 0
while i < n + 1:
p += prod(int(digit) for digit in str(i))
i += 1
return p # Daria Micovic, Apr 13 2016
(Python)
from math import prod
def A061076(n): return sum(prod(int(d) for d in str(i)) for i in range(1, n+1)) # Chai Wah Wu, Mar 21 2022
CROSSREFS
KEYWORD
AUTHOR
Amarnath Murthy, Apr 14 2001
EXTENSIONS
Corrected and extended by Matthew Conroy, Apr 16 2001
STATUS
approved