A061076 a(n) is the sum of the products of the digits of all the numbers from 1 to n.

1, 3, 6, 10, 15, 21, 28, 36, 45, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 90, 92, 96, 102, 110, 120, 132, 146, 162, 180, 180, 183, 189, 198, 210, 225, 243, 264, 288, 315, 315, 319, 327, 339, 355, 375, 399, 427, 459, 495, 495, 500, 510, 525, 545, 570, 600, 635

Offset: 1

Author

Amarnath Murthy, Apr 14 2001

Keywords

• nonn
• base
• easy
• look

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

Examples

			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.
		

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.

Crossrefs

Cf. A002275, A007954, A037123, A047855.

Programs

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]]

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

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

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

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

Extensions

Corrected and extended by Matthew Conroy, Apr 16 2001