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

2, 6, 12, 20, 20, 22, 26, 32, 40, 40, 44, 52, 64, 80, 80, 86, 98, 116, 140, 140, 148, 164, 188, 220, 220, 230, 250, 280, 320, 320, 332, 356, 392, 440, 440, 454, 482, 524, 580, 580, 596, 628, 676, 740, 740, 758, 794, 848, 920, 920, 920, 920, 920, 920, 920, 922

Offset: 1

Author

Amarnath Murthy, Apr 14 2001

Keywords

• nonn
• base

Comments

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.

Examples

			a(5) = 2 + 4 + 6 + 8 + 1*0 = 20; (a(18)=116, not 114).
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).
		

References

• Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Links

• Luca Onnis, Table of n, a(n) for n = 1..10000
• Luca Onnis, Mathematica prog which computes the first 10000 terms
• Luca Onnis On the general Smarandache's sigma product of digits, arXiv:2203.07227 [math.GM], 2022.

Crossrefs

Cf. A061076, A061077.

Programs

Mathematica

Accumulate[Times@@@IntegerDigits[Range[2,120,2]]] (* Harvey P. Dale, Jun 18 2021 *)

PARI

pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
a(n) = sum(k=1, n, pd(2*k)); \\ Michel Marcus, Feb 01 2015

PARI

a(n) = sum(k=1, n, vecprod(digits(2*k))); \\ Michel Marcus, Mar 13 2022

PARI

a(n) = {t=digits(2*n); p=1; d=#t; for(i=1, #t, if(t[i]==0, d=i-1; break));
(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

Python

from math import prod
from itertools import accumulate
def p(n): return prod(map(int, str(n)))
def a(n): return sum(p(2*i) for i in range(1, n+1))
def aupton(nn): return list(accumulate([pd(2*k) for k in range(1, nn+1)]))
print(aupton(56)) # Michael S. Branicky, Mar 13 2022

Formula

From Luca Onnis, Mar 13 2022: (Start)

a(5*10^n-1) = a(5*10^n) = (5/11)*(45^(n+1)-1).

a(n) <= (5/11)*(45^(log((n+1)/5)+1)-1) for all n.

a(n) ~ (4/5)*A061077(n) as n -> infinity.

Conjecture: let a >= 1, b >= 0, where p is not a multiple of 2 nor 5. Then:

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)

Extensions

Corrected and extended by Matthew Conroy, Apr 17 2001

Incorrect formula removed by Luca Onnis, Mar 13 2022

Name clarified by Chai Wah Wu, Mar 21 2022