A061077 a(n) is the sum of the products of the digits of the first n odd numbers.

1, 4, 9, 16, 25, 26, 29, 34, 41, 50, 52, 58, 68, 82, 100, 103, 112, 127, 148, 175, 179, 191, 211, 239, 275, 280, 295, 320, 355, 400, 406, 424, 454, 496, 550, 557, 578, 613, 662, 725, 733, 757, 797, 853, 925, 934, 961, 1006, 1069, 1150, 1150, 1150, 1150

Offset: 1

Author

Amarnath Murthy, Apr 14 2001

Keywords

• nonn
• base
• easy

Examples

			a(7) = 1 + 3 + 5 + 7 + 9 + 1*1 + 1*3 = 29.
		

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, On the general Smarandache's sigma product of digits, arXiv:2203.07227 [math.GM], 2022.

Crossrefs

Cf. A061076, A061078.

Programs

Mathematica

Accumulate[Times @@@ IntegerDigits[Range[1, 99, 2]]] (* Luca Onnis, Mar 20 2022 *)

PARI

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

PARI

a(n) = {m=digits(2*n - 1); p=1; d=#m; for(i=1, #m, if(m[i]==0, d=i-1; break));
(25/44) * (45^(#m-1)-1) + sum(i=1, #m-1, (prod(j=1,#m-i-1,m[j])) * (m[#m-i]) * (m[#m-i]-1) * (5^(i + 1) * 9^(i-1)) / 2)+prod(k=1,#m-1,m[k])*(((m[#m]+1)^2)/4)} \\ Luca Onnis, Mar 20 2022

Python

from math import prod
def A061077(n): return sum(prod(int(d) for d in str(2*i+1)) for i in range(n)) # Chai Wah Wu, Mar 21 2022

Formula

a(n) = Sum_{k = 1..n} (product of the digits of 2k-1).

From Luca Onnis, Mar 20 2022: (Start)

a(5*10^n) = (25/44)*(45^(n+1)-1).

a(n) <= (25/44)*(45^(log(n/5)+1)-1) for all n.

a(n) ~ (5/4)*A061078(n) as n -> infinity. (End)

Extensions

More terms from Matthew Conroy, Apr 16 2001

Offset corrected by Charles R Greathouse IV, Feb 01 2015

Incorrect formula removed by Luca Onnis, Mar 20 2022