A038366 -id:A038366 - cp's OEIS frontend

A061016 Duplicate of A038366.

10, 19, 20, 29, 30, 39, 40, 42, 49, 50, 59, 60, 69, 70, 79, 80, 89, 90, 99, 100, 102, 108

Offset: 1

dead

A061762 a(n) = (sum of digits of n) + (product of digits of n).

0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 6, 13, 20, 27, 34, 41, 48, 55, 62, 69, 7, 15, 23, 31, 39, 47

Offset: 0

Amarnath Murthy, May 20 2001

Fixed points a(m) = m are m = {0, 19, 29, 39, 49, 59, 69, 79, 89, 99}. Is this list complete? - Zak Seidov, Aug 22 2007
The above list of fixed points is complete. If a(m) = m, then m < 10^21 and there are no other fixed points below 10^21. - Chai Wah Wu, Aug 14 2017
All numbers are in this sequence. Proof: One can create a number m whose digital sum is any number p and one can create a number k by concatenating digit "0" to m. Then this number k will be a term. - Metin Sariyar, Oct 29 2019

			a(14) = 1+4 + 1*4 = 9.

S. Parmeswaran, S+P numbers, Mathematics Informatics Quarterly, Vol. 9, No. 3 (Sep 1999), Bulgaria.

Harry J. Smith, Table of n, a(n) for n = 0..1000

Cf. A007953, A007954, A061763, A038366, A074871.

See A130858 for the smallest inverse.

[0] cat [&+Intseq(n)+&*Intseq(n): n in [1..80]];// Vincenzo Librandi, Jan 03 2020

read("transforms") :
A061762 := proc(n)
    digsum(n)+A007954(n) ;
end proc: # R. J. Mathar, Aug 13 2012

Table[Plus @@ IntegerDigits[n] + Times @@ IntegerDigits[n], {n, 0, 75}] (* Jayanta Basu, Apr 05 2013 *)

a(n) = if (n==0, 0, my(d=digits(n)); vecsum(d) + vecprod(d)); \\ Michel Marcus, Oct 29 2019, Jan 03 2020

from operator import mul
from functools import reduce
def A061762(n):
    a = [int(d) for d in str(n)]
    return sum(a)+reduce(mul,a) # Chai Wah Wu, Aug 14 2017

a(n) = A007953(n) + A007954(n).

Corrected and extended by Larry Reeves (larryr(AT)acm.org) and Matthew Conroy, May 23 2001

A061763 Numbers k such that k is divisible by A061762(k) and the product of digits of k (A007954(k)) is not zero.

Original entry on oeis.org

19, 29, 39, 42, 49, 59, 69, 79, 89, 99, 126, 132, 285, 312, 522, 594, 1134, 1144, 1159, 1211, 1275, 1323, 1365, 1573, 1632, 1634, 1674, 1715, 1813, 1815, 1911, 1919, 1932, 1944, 2133, 2139, 2516, 2793, 3132, 3135, 3161, 3211, 3213, 3216, 3321, 3363, 3393

Offset: 1

Amarnath Murthy, May 20 2001

Intersection of A038366 and A052382 (zeroless numbers). - Michel Marcus, Oct 29 2019

			42 is a term as 4+2 + 2*4 = 14 and 42 = 14*3.

S. Parmeswaran, S+P numbers, Mathematics Informatics Quarterly, Vol. 9, No. 3 Sept. 1999, Bulgaria.

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 1001 terms from Harry J. Smith)

Cf. A061762, A038366, A052382.

Select[Range[3400], (y = Times @@ (x = IntegerDigits[#])) != 0 && Divisible[#, Plus @@ x + y] &] (* Jayanta Basu, Jul 14 2013 *)

isok(k) = my(d=digits(k)); vecmin(d) && ((k % (vecprod(d) + vecsum(d))) == 0);

Corrected and extended by Larry Reeves (larryr(AT)acm.org), May 23 2001

Offset corrected by Giovanni Resta, Oct 29 2019

A328864 For any three-digit number k = hdu, f(k) = (h+d+u) + (hd+du+uh) + (hd*u). This sequence consists of the numbers k for which the ratio k/f(k) is an integer.

Original entry on oeis.org

100, 114, 115, 120, 121, 190, 199, 200, 207, 208, 210, 221, 260, 290, 299, 300, 301, 304, 330, 390, 399, 400, 420, 441, 448, 490, 499, 500, 572, 573, 590, 599, 600, 620, 624, 625, 690, 699, 700, 705, 790, 799, 800, 806, 880, 890, 899, 900, 990, 999

Offset: 1

Bernard Schott, Oct 29 2019

The idea of this sequence comes from the 1st problem of the 30th British Mathematical Olympiad in 1994 [see link BMO].
This sequence is finite with 50 terms.
The values of k/f(k) obtained are 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 13, 15, 16, 22, 24, 30, 31, 42, 43, 100.
Three particular subsequences:
k/f(k) = 1 for k = 199, 299, 399, 499, 599, 699, 799, 899, 999 (answer to part (ii) of the BMO problem).
k/f(k) = 10 for k = 190, 290, 390, 490, 590, 690, 790, 890, 990.
k/f(k) = 100 for k = 100, 200, 300, 400, 500, 600, 700, 800, 900.
Other definition: three-digit numbers k = hdu such as k/(e_1(h,d,u) + e_2(h,d,u) + e_3(h,d,u)) is an integer, where e_1, e_2, e_3 are the elementary symmetric polynomials in 3 variables.
Remark: When k has two digits du, the numbers that are divisible by (e_1(d,u) + e_2(d,u)) = (d+u) + (d*u) are the first 19 terms of A038366.

			For k = 625, f(k) = 6+2+5 + 6*2+2*5+6*5 + 6*2*5 = 13 + 52 + 60 =  125 and 625/125 = 5, hence, 625 is a term, and 5 is the solution to part (i) of the BMO problem.

A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 1 pp. 55 and 99-100 (1994)

British Mathematical Olympiad, 1994 - Problem 1.
Eric Weisstein's World of Mathematics, Symmetric polynomial
Wikipedia, Elementary symmetric polynomial.

Cf. A038366 (similar, with 2 digits, the first 19 terms).

Cf. A005349 (Niven numbers), A007602 (Zuckerman numbers).

for i from 1 to 9 do
for j from 0 to 9 do
for k from 0 to 9 do
     n := 100*i + 10*j + k ;
     m := i + j + k + i*j + j*k + k*i + i*j*k ;
     if n/m = floor(n/m) then print(n,m,n/m) ; end if ;
end do ;
end do ;
end do ;

Select[Range[100, 999], ({h,d,u} = IntegerDigits@ #; IntegerQ[# / (d + u + d u + (1 + d) h (1 + u))]) &] (* Giovanni Resta, Oct 29 2019 *)

A343131 For m >= 1, the m-digit number k = d_m||...||d_2||d_1 is a term if it is divisible by f_m(k) that is the sum of the m elementary symmetric polynomials in m variables e_i(k): f_m(k) = Sum_{i=1..m} e_i(d_1, ..., d_m).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 19, 20, 29, 30, 39, 40, 42, 49, 50, 59, 60, 69, 70, 79, 80, 89, 90, 99, 100, 114, 115, 120, 121, 190, 199, 200, 207, 208, 210, 221, 260, 290, 299, 300, 301, 304, 330, 390, 399, 400, 420, 441, 448, 490, 499, 500, 572, 573, 590, 599, 600, 620

Offset: 1

Bernard Schott, Apr 06 2021

Equivalently, integers k that are divisible by A061486(k); the quotients obtained when these terms k are divided by A061486(k) are in A343132.
The idea of this sequence comes from A328864 (1st problem of the 30th British Mathematical Olympiad in 1994) and also from the elementary symmetric polynomials in m variables (see links) that allow the generalization of this Olympiad problem to m-digit numbers.
-> When k = u has only one digit, then f_1(k) = e_1(u) = u and the numbers u from 1 to 9 satisfy u/e_1(u) = 1, so those are the first nine terms of this sequence.
-> When k = du has two digits, the numbers that are divisible by f_2(k) = (e_1(d,u) + e_2(d,u)) = (d+u) + (d*u) are the first 19 terms of A038366, from a(10) = 10 to a(28) = 99.
-> when k = hdu is a 3-digit number, then e_1(h,d,u) = h+d+u, e_2(h,d,u) = h*d+d*u+u*h and e_3(h,d,u) = h*d*u so f_3(k) = (h+d+u) + (hd+du+uh) + (hdu). This subsequence with 3-digit numbers A328864 has 50 terms that are here from a(29) = 100 to a(78) = 999.
-> When k = thdu is a 4-digit number, then e_1(t,h,d,u) = e_1(k) = t+h+d+u, e_2(t,h,d,u)= t*h+t*d+t*u+h*d+h*u+d*u, e_3(t,h,d,u) = t*h*d+t*h*u+t*d*u+h*d*u and e_4(t,h,d,u) = t*h*d*u with f_4(k) = e_1(t,h,d,u) + e_2(t,h,d,u) + e_3(t,h,d,u) + e_4(t,h,d,u). Numbers k such that k/f(k) is an integer form another subsequence with 87 terms. This subsequence with 4-digit numbers goes from a(79) = 1000 to a(165) = 9999.

			For k = 7, f_1(7) = 7, and 7/7 = 1, hence 7 is a term.
For k = 42, f_2(4,2) = 4+2 + 4*2 = 14 and 42/14 = 3, hence 42 is a term.
For k = 572, f_3(5,7,2) = 5+7+2 + 5*7+7*2+2*5 + 5*7*2 = 14 + 59 + 70 = 143 and 572/143 = 4, hence 572 is a term.
For k = 3225, f_4(3,2,2,5) = 3+2+2+5 + 3*2+3*2+3*5+2*2+2*5+2*5 + 3*2*2+3*2*5+3*2*5+2*2*5 + 3*2*2*5 = 215 and 3225/215 = 15, hence 3225 is a term.

Eric Weisstein's World of Mathematics, Symmetric polynomial.
Wikipedia, Elementary symmetric polynomial.

Cf. A005349, A007602, A038366, A061486, A328864, A343132.

sympol(X, n)=my(s=0); forvec(i=vector(n, j, [1, #X]), s+=prod(k=1, n, X[i[k]]), 2); s ;
f(n) = my(d=digits(n)); sum(k=1, #d, sympol(d, k)); \\ A061486
isok(m) = (m % f(m)) == 0; \\ Michel Marcus, Apr 06 2021

A343132 a(n) is the quotient obtained when integer A343131(n) = k is divided by A061486(k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 10, 1, 10, 1, 10, 3, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 100, 6, 5, 24, 11, 10, 1, 100, 9, 8, 42, 13, 13, 10, 1, 100, 43, 16, 22, 10, 1, 100, 30, 9, 2, 10, 1, 100, 4, 3, 10, 1, 100, 31, 6, 5, 10, 1, 100, 15, 10, 1, 100, 13, 11, 10, 1, 100, 10, 1, 1000

Offset: 1

Bernard Schott, Apr 07 2021

The first 9 terms corresponding to the 1-digit numbers k = u are the quotients u/u = 1.
The next 19 terms from a(10) = 10 to a(28) = 1 corresponding to 2-digit numbers k = du are the quotients du/(d+u + d*u).
The next 50 terms from a(29) = 100 to a(78) = 1 corresponding to 3-digit numbers k = hdu (in A328864) are the quotients hdu/f_3(h,d,u) where f_3(h,d,u) = (h+d+u) + (h*d+d*u+u*h) + (h*d*u).
The next 87 terms, from a(79) = 1000 to a(165) = 1, corresponding to 4-digit numbers k = thdu are the quotients thdu/f_4(t,h,d,u) where f_4(t,h,d,u) = (t+h+d+u) + (t*h+t*d+t*u+h*d+h*u+d*u) + (t*h*d+t*h*u+t*d*u+h*d*u) + (t*h*d*u).
When A343131(n) = z*10^q = A037124(r) is a number that contains only one nonzero digit z, then A061486(A037124(r)) = this nonzero digit z and a(n) = 10^q.

			For A343131(7) = 7, A061486(7) = 7 and a(7) = 7/7 = 1.
For A343131(17) = 42, A061486(42) = 4+2 + 4*2 = 14 and a(17) = 42/14 = 3.
For A343131(58) = 573, A061486(573) = 5+7+3 + 5*7+7*3+3*5 + 5*7*3 = 191 and a(58) = 573/191 = 3.

Eric Weisstein's World of Mathematics, Symmetric polynomial.
Wikipedia, Elementary symmetric polynomial.

Cf. A005349, A007602, A038366, A061486, A328864, A343131.

sympol(X, n) = my(s=0); forvec(i=vector(n, j, [1, #X]), s+=prod(k=1, n, X[i[k]]), 2); s ;
f(n) = my(d=digits(n)); sum(k=1, #d, sympol(d, k));
lista(nn) = {for (n=1, nn, my(q = n/f(n)); if (denominator(q) == 1, print1(q, ", ")););} \\ Michel Marcus, Apr 08 2021

a(n) = A343131(n)/A061486(A343131(n)).

cp's OEIS Frontend

A061016 Duplicate of A038366.

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A061762 a(n) = (sum of digits of n) + (product of digits of n).

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A061763 Numbers k such that k is divisible by A061762(k) and the product of digits of k (A007954(k)) is not zero.

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A328864 For any three-digit number k = hdu, f(k) = (h+d+u) + (h*d+d*u+u*h) + (h*d*u). This sequence consists of the numbers k for which the ratio k/f(k) is an integer.

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Maple

Mathematica

A343131 For m >= 1, the m-digit number k = d_m||...||d_2||d_1 is a term if it is divisible by f_m(k) that is the sum of the m elementary symmetric polynomials in m variables e_i(k): f_m(k) = Sum_{i=1..m} e_i(d_1, ..., d_m).

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A343132 a(n) is the quotient obtained when integer A343131(n) = k is divided by A061486(k).

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A328864 For any three-digit number k = hdu, f(k) = (h+d+u) + (hd+du+uh) + (hd*u). This sequence consists of the numbers k for which the ratio k/f(k) is an integer.