cp's OEIS Frontend

A061078 Sum of the products of the digits of the first n positive even numbers.

%I A061078 #84 Mar 21 2022 11:48:21
%S A061078 2,6,12,20,20,22,26,32,40,40,44,52,64,80,80,86,98,116,140,140,148,164,
%T A061078 188,220,220,230,250,280,320,320,332,356,392,440,440,454,482,524,580,
%U A061078 580,596,628,676,740,740,758,794,848,920,920,920,920,920,920,920,922
%N A061078 Sum of the products of the digits of the first n positive even numbers.
%C A061078 For n = (10^r)/2, a(n) is the sum of the r terms of the geometric progression with first term 20 and common ratio 45.
%D A061078 Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.
%H A061078 Luca Onnis, <a href="/A061078/b061078.txt">Table of n, a(n) for n = 1..10000</a>
%H A061078 Luca Onnis, <a href="/A061078/a061078.txt">Mathematica prog which computes the first 10000 terms</a>
%H A061078 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 A061078 From _Luca Onnis_, Mar 13 2022: (Start)
%F A061078 a(5*10^n-1) = a(5*10^n) = (5/11)*(45^(n+1)-1).
%F A061078 a(n) <= (5/11)*(45^(log((n+1)/5)+1)-1) for all n.
%F A061078 a(n) ~ (4/5)*A061077(n) as n -> infinity.
%F A061078 Conjecture: let a >= 1, b >= 0, where p is not a multiple of 2 nor 5. Then:
%F A061078 a(5^a*2^b*p-1) = a(5^a*2^b*p) = ... = a(5^a*2^b*p + 55...5) where the number of fives is equal to b if a > b, and is equal to a-1 if 1 <= a <= b. (End)
%e A061078 a(5) = 2 + 4 + 6 + 8 + 1*0 = 20; (a(18)=116, not 114).
%e A061078 a(1199) = a(5^2*2^4*3 - 1) = ... = a(5^2*2^4*3 + 5) = a(1205). In fact, the number of "fives" is exactly equal to 1 = 2-1 (where 2 is the exponent of 5).
%t A061078 Accumulate[Times@@@IntegerDigits[Range[2,120,2]]] (* _Harvey P. Dale_, Jun 18 2021 *)
%o A061078 (PARI) pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
%o A061078 a(n) = sum(k=1, n, pd(2*k)); \\ _Michel Marcus_, Feb 01 2015
%o A061078 (PARI) a(n) = sum(k=1, n, vecprod(digits(2*k))); \\ _Michel Marcus_, Mar 13 2022
%o A061078 (PARI) a(n) = {t=digits(2*n); p=1; d=#t; for(i=1, #t, if(t[i]==0, d=i-1; break));
%o A061078 (5/11) * (45^(#t-1)-1) + (sum(i=1, #t-1, ((prod(j=1,#t-i-1,t[j])) * (t[#t-i]) * (t[#t-i]-1) * 2 * (5^(i))* (9^(i-1)))))+(prod(k=1,#t-1,t[k]))*((((t[#t])^2))/4+(t[#t])/2)} \\ _Luca Onnis_, Mar 17 2022
%o A061078 (Python)
%o A061078 from math import prod
%o A061078 from itertools import accumulate
%o A061078 def p(n): return prod(map(int, str(n)))
%o A061078 def a(n): return sum(p(2*i) for i in range(1, n+1))
%o A061078 def aupton(nn): return list(accumulate([pd(2*k) for k in range(1, nn+1)]))
%o A061078 print(aupton(56)) # _Michael S. Branicky_, Mar 13 2022
%Y A061078 Cf. A061076, A061077.
%K A061078 nonn,base
%O A061078 1,1
%A A061078 _Amarnath Murthy_, Apr 14 2001
%E A061078 Corrected and extended by _Matthew Conroy_, Apr 17 2001
%E A061078 Incorrect formula removed by _Luca Onnis_, Mar 13 2022
%E A061078 Name clarified by _Chai Wah Wu_, Mar 21 2022