A061076 -id:A061076 - cp's OEIS frontend

A007954 Product of decimal digits of n.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 0, 9, 18, 27, 36, 45, 54, 63, 72, 81, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

R. Muller

Moebius transform of A093811(n). a(n) = A093811(n) * A008683(n), where operation * denotes Dirichlet convolution, namely b(n) * c(n) = Sum_{d|n} b(d) * c(n/d). Simultaneously holds Dirichlet multiplication: a(n) * A000012(n) = A093811(n). - Jaroslav Krizek, Mar 22 2009
Apart from the 0's, all terms are in A002473. Further, for all m in A002473 there is some n such that a(n) = m, see A096867. - Charles R Greathouse IV, Sep 29 2013
a(n) = 0 asymptotically almost surely, namely for all n except for the set of numbers without digit '0'; this set is of density zero, since it is less and less probable to have no '0' as the number of digits of n grows. (See also A054054.) - M. F. Hasler, Oct 11 2015

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Rigoberto Flórez, Robinson A. Higuita and Antara Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Florentin Smarandache, Only Problems, Not Solutions!.
Index entries for Colombian or self numbers and related sequences

Cf. A031347 (different from A035930), A007953, A007602, A010888, A093811, A008683, A000012, A061076 (partial sums), A230099.

Cf. A051802 (ignoring zeros).

a007954 n | n < 10 = n
          | otherwise = m * a007954 n' where (n', m) = divMod n 10
-- Reinhard Zumkeller, Oct 26 2012, Mar 14 2011

[0] cat [&*Intseq(n): n in [1..110]]; // Vincenzo Librandi, Jan 03 2020

A007954 := proc(n::integer)
    if n = 0 then
        0;
    else
        mul( d,d=convert(n,base,10)) ;
    end if;
end proc: # R. J. Mathar, Oct 02 2019

Array[Times @@ IntegerDigits@ # &, 108, 0] (* Robert G. Wilson v, Mar 15 2011 *)

A007954(n)= { local(resul = n % 10); n \= 10; while( n > 0, resul *= n %10; n \= 10; ); return(resul); } \\ R. J. Mathar, May 23 2006, edited by M. F. Hasler, Apr 23 2015

A007954(n)=prod(i=1,#n=Vecsmall(Str(n)),n[i]-48) \\ (...eval(Vec(...)),n[i]) is about 50% slower; (...digits(n)...) about 6% slower. \\ M. F. Hasler, Dec 06 2009

a(n)=if(n,factorback(digits(n)),0) \\ Charles R Greathouse IV, Apr 14 2020

from math import prod
def a(n): return prod(map(int, str(n)))
print([a(n) for n in range(108)]) # Michael S. Branicky, Jan 16 2022

(0 to 99).map(.toString.toCharArray.map( - 48).scanRight(1)( * ).head) // Alonso del Arte, Apr 14 2020

A000035(a(A014261(n))) = 1. - Reinhard Zumkeller, Nov 30 2007
a(n) = abs(A187844(n)). - Reinhard Zumkeller, Mar 14 2011
a(n) > 0 if and only if A054054(n) > 0. a(n) = d in {1, ..., 9} if n = (10^k - 1)/9 + (d - 1)*10^m = A002275(k) + (d - 1)*A011557(m) for some k > m >= 0. The statement holds with "if and only if" for d in {1, 2, 3, 5, 7}. For d = 4, 6, 8 or 9, one has a(n) = d if n = (10^k - 1)/9 + (a - 1)*10^m + (b - 1)*10^p with integers k > m > p >= 0 and a, b > 0 such that d = a*b. - M. F. Hasler, Oct 11 2015
From Robert Israel, May 17 2016: (Start)
G.f.: 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 + A(x^10)). (End)
a(n) <= 9^(1 + log_10(n/9)). - Lucas A. Brown, Jun 22 2023

Error in term 25 corrected, Nov 15 1995

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

Original entry on oeis.org

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

Amarnath Murthy, Apr 14 2001

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

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

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

Cf. A061076, A061078.

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

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

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

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

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)

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

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

Original entry on oeis.org

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

Amarnath Murthy, Apr 14 2001

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.

			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).

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

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.

Cf. A061076, A061077.

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

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

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

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

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

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)

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

A187845 Partial sums of A187844.

Original entry on oeis.org

0, -1, -3, -6, -10, -15, -21, -28, -36, -45, -45, -44, -42, -39, -35, -30, -24, -17, -9, 0, 0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 90, 93, 99, 108, 120, 135, 153, 174, 198, 225, 225, 229, 237, 249, 265, 285, 309, 337, 369, 405, 405, 410, 420, 435, 455, 480, 510, 545, 585, 630, 630, 636, 648, 666

Offset: 0

Reinhard Zumkeller, Mar 14 2011

Reinhard Zumkeller, Table of n, a(n) for n = 0..25000

Cf. A061076.

a187845 n = a187845_list !! (n-1)
a187845_list = scanl1 (+) $ map a187844 [0..]
-- Reinhard Zumkeller, Jan 29 2014

A217627 a(n) is the sum of the products of the nonzero digits of the numbers from 1 to n.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 28, 36, 45, 46, 47, 49, 52, 56, 61, 67, 74, 82, 91, 93, 95, 99, 105, 113, 123, 135, 149, 165, 183, 186, 189, 195, 204, 216, 231, 249, 270, 294, 321, 325, 329, 337, 349, 365, 385, 409, 437, 469, 505, 510, 515, 525, 540, 560, 585, 615, 650

Offset: 1

Giovanni Resta, Oct 18 2012

The formula a(10^k) = 46^k can be easily derived from the Multinomial Theorem, inspecting the expansion of (1+1+2+3+...+9)^k, where the second '1's takes the place of '0' (since we are neglecting the zeros in the products). This formula can be generalized as follows:

Let B>1 be the base used for representation. Let D be a subset of {1,2,...,B-1}. Using base B, let A(n) be the sum of the products of the digits in D of the numbers up to n. Then, A(B^k)=(B+S-|D|)^k, where |D| is the cardinality of D and S is the sum of the elements of D. For example, in base 10, with D={1,3,5,7,9}, (i.e., A(n)= sum of the products of the odd digits of the numbers up to n) we have A(k)=(10+(1+3+5+7+9)-5)^k = 30^k.

			a(10) = 1+2+3+4+5+6+7+8+9+1 = 46

Giovanni Resta, Table of n, a(n) for n = 1..10000
Wikipedia, Multinomial theorem

Cf. A061076 (the same sum, when zeros are taken into account).

pp[n_]:=Times@@Select[IntegerDigits[n],#>0 &]; Accumulate[pp /@ Range[100]]

a(10^k) = 46^k.

A352346 Common terms between A061078 and A061077.

Original entry on oeis.org

26, 52, 148, 280, 320, 454, 1150, 1480, 8000, 41650, 80300, 165656, 166088, 614900, 2353700, 2859460, 28233200, 66130400, 68941640, 85717240, 107300320, 131507080, 155478800, 207666520, 1426680920, 1824596800, 2468014900, 2475648820, 5342351060, 5355218900, 5857281500, 8550475900, 36025361120

Offset: 1

Luca Onnis, Mar 12 2022

Smarandache's conjecture: there are infinitely many terms.

This is a subsequence of A061076.

			26 is a term of this sequence, in fact:
26 = 1+3+5+7+9+1*1 (A061077(6)=26);
26 = 2+4+6+8+1*0+1*2+1*4 (A061078(7)=26).

A. Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001. Page 267

Chai Wah Wu, Table of n, a(n) for n = 1..205

Cf. A061076, A061077, A061078.

Intersection[Accumulate[Times @@@ IntegerDigits[Range[2, 10000000, 2]]],
Accumulate[Times @@@ IntegerDigits[Range[1, 10000000, 2]]]]

from math import prod
from itertools import islice
def A352346_gen(): # generator of terms
    n1, m1, n2, m2 = 1, 1, 2, 2
    while True:
        if m1 == m2:
            yield m1
        k = 0
        while k == 0:
            n1 += 2
            m1 += (k := prod(int(d) for d in str(n1)))
        while m2 < m1:
            n2 += 2
            m2 += prod(int(d) for d in str(n2))
A352346_list = list(islice(A352346_gen(),20)) # Chai Wah Wu, Mar 21 2022

A360365 a(n) = sum of the products of the digits of the first n positive multiples of 3.

Original entry on oeis.org

3, 9, 18, 20, 25, 33, 35, 43, 57, 57, 66, 84, 111, 119, 139, 171, 176, 196, 231, 231, 249, 285, 339, 353, 388, 444, 452, 484, 540, 540, 567, 621, 702, 702, 702, 702, 703, 707, 714, 714, 720, 732, 750, 756, 771, 795, 799, 815, 843, 843, 858, 888, 933, 945, 975, 1023, 1030, 1058, 1107

Offset: 1

Luca Onnis, Feb 09 2023

			a(4) = 20 since the first 4 positive multiples of 3 are 3,6,9 and 12 and the sum of the product of their digits is 3 + 6 + 9 + 1*2 = 20.

Cf. A061076, A061077, A061078.

Accumulate[Times @@@ IntegerDigits[Range[3, 999 , 3]]]

a(n)={sum(k=1, n, vecprod(digits(3*k)))} \\ Andrew Howroyd, Feb 09 2023

from math import prod
def A360365(n): return sum(prod(int(d) for d in str(m)) for m in range(3,3*n+1,3)) # Chai Wah Wu, Feb 28 2023

a((10^n-1)/3) = (1/836)*(15*(19*45^n-63) + 44*3^((3*n)/2)*(sqrt(3)*sin((Pi*n)/6) + 15*cos((Pi*n)/6))).
G.f. of the subsequence a((10^n-1)/3): 9*(2 - 32*x + 135*x^2)/((1 - x)*(1 - 45*x)*(1 - 9*x + 27*x^2)).
a((10^n-1)/3) = 55*a((10^(n-1)-1)/3) - 486*a((10^(n-2)-1)/3) + 1647*a((10^(n-3)-1)/3) - 1215*a((10^(n-4)-1)/3) for n > 4.

A350744 Numbers m such that A061078(m)/A061077(m) = 4/5.

Original entry on oeis.org

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 51, 52, 53, 54, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 101, 102, 103, 104, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 151, 152, 153, 154, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 201, 202, 203, 204, 205, 210, 215, 220, 225

Offset: 1

Luca Onnis, Mar 20 2022

nonn

All positive multiples of 5 are terms of the sequence.

			30 is a term, in fact A061078(30)=320, A061077(30)=400 and a(n) = 320/400 = is 4/5.
500, 501, 502, ..., 554, 555 are all terms. In fact 500=5*10^2 and for the formula above also 501, ..., 500+(5/9)*(10^2-1) = 555 are all terms of the sequence.

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

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

Cf. A061076, A061077, A061078.

Flatten[Position[(Accumulate[Times @@@ IntegerDigits[Range[2, 10000, 2]]]/
    Accumulate[Times @@@ IntegerDigits[Range[1, 9999, 2]]]), 4/5]]

pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
isok(k) = sum(i=1, k, pd(2*i))/sum(i=1, k, pd(2*i-1)) == 4/5; \\ Michel Marcus, Mar 21 2022

Let k be a positive integer not divisible by 5 and j >= 0; then 5*k*10^j, 5*k*10^j+1, ..., 5*k*10^j+(5/9)*(10^j-1) are all terms of the sequence.

Limit_{n->oo} A061078(n)/A061077(n) = 4/5.

cp's OEIS Frontend

A007954 Product of decimal digits of n.

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

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

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A187845 Partial sums of A187844.

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A217627 a(n) is the sum of the products of the nonzero digits of the numbers from 1 to n.

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A352346 Common terms between A061078 and A061077.

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