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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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%I A061076 #91 Oct 20 2022 12:43:55
%S A061076 1,3,6,10,15,21,28,36,45,45,46,48,51,55,60,66,73,81,90,90,92,96,102,
%T A061076 110,120,132,146,162,180,180,183,189,198,210,225,243,264,288,315,315,
%U A061076 319,327,339,355,375,399,427,459,495,495,500,510,525,545,570,600,635
%N A061076 a(n) is the sum of the products of the digits of all the numbers from 1 to n.
%C A061076 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
%C A061076 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
%C A061076 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
%C A061076 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
%D A061076 Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.
%H A061076 Daria Micovic, <a href="/A061076/b061076.txt">Table of n, a(n) for n = 1..10000</a>
%H A061076 Amarnath Murthy, <a href="https://vixra.org/abs/1403.0845">Smarandache friendly numbers and a few more sequences</a>, vixra, 2014.
%H A061076 Luca Onnis, <a href="https://arxiv.org/abs/2203.07227">On the general Smarandache's sigma product of digits</a>, arXiv:2203.07227 [math.GM], 2022.
%F A061076 a(n) = Sum_{k = 1..n} (product of the digits of k).
%F A061076 a(10^k-1) = (45/44)*(45^k-1). - _Giovanni Resta_, Oct 18 2012
%F A061076 From _Robert Israel_, May 17 2016: (Start)
%F A061076 Partial sums of A007954.
%F A061076 G.f.: (1-x)^(-1) * Sum_{n>=0} Product_{j=0..n} Sum_{k=1..9} k * x^(k*10^j).
%F A061076 G.f. satisfies A(x) = (x + 2*x^2 + ... + 9*x^9)*(1+(1-x^10)*A(x^10))/(1-x).
%F A061076 (End)
%F A061076 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
%e A061076 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.
%p A061076 A007954:= n -> convert(convert(n,base,10),`*`):
%p A061076 ListTools:-PartialSums(map(A007954,[$1..100])); # _Robert Israel_, May 17 2016
%t A061076 Accumulate[Times@@IntegerDigits[#]& /@ Range[100]]
%o A061076 (PARI) pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
%o A061076 a(n) = sum(k=1, n, pd(k)); \\ _Michel Marcus_, Feb 01 2015
%o A061076 (PARI) a(n) = {n=digits(n); p=1; d=#n; for(i=1,#n,if(n[i]==0,d=i-1;break));
%o A061076 (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
%o A061076 (Sage)
%o A061076 def A061076(n):
%o A061076     p = 0
%o A061076     i = 0
%o A061076     while i < n + 1:
%o A061076         p += prod(int(digit) for digit in str(i))
%o A061076         i += 1
%o A061076     return p # _Daria Micovic_, Apr 13 2016
%o A061076 (Python)
%o A061076 from math import prod
%o A061076 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
%Y A061076 Cf. A002275, A007954, A037123, A047855.
%K A061076 nonn,base,easy,look
%O A061076 1,2
%A A061076 _Amarnath Murthy_, Apr 14 2001
%E A061076 Corrected and extended by _Matthew Conroy_, Apr 16 2001