A034710 -id:A034710 - cp's OEIS frontend

A113761 Numbers k such that the number of divisors of k equals both the sum and the product of digits of k in base 10.

1, 2, 22, 2114, 11222, 21122, 22211, 112116, 121116, 1111143, 1413111, 3411111, 11111128, 11111821, 11112118, 11121231, 11811112, 13111212, 18111112, 21111118, 21111181, 21121113, 23111121, 111112119, 111119211, 192111111

Offset: 1

Giovanni Resta, Jan 18 2006

Intersection of A074312 and A057531.

			2114 is a term since 2+1+1+4 = 2*1*1*4 = 8 and 2114 has 8 divisors, {1, 2, 7, 14, 151, 302, 1057, 2114}.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A034710, A057531, A074312.

L={};Do[d=IntegerDigits@n; p=Times@@d; If[p==Plus@@d && p==DivisorSigma[0, n], AppendTo[L, n];Print[n]], {n, 1000000}];L
lst = {}; fQ[n_] := (id = IntegerDigits@n; Plus @@ id == Times @@ id == DivisorSigma[0, n]); Do[ If[ fQ@n, AppendTo[lst, n]], {n, 2*10^8}]; lst

a(13)-a(26) from Robert G. Wilson v, Jan 19 2006

A162840 Numbers k such that the cube of the sum of digits of k equals the product of digits of k.

Original entry on oeis.org

0, 1, 666666, 1377789, 1377798, 1377879, 1377897, 1377978, 1377987, 1378779, 1378797, 1378977, 1379778, 1379787, 1379877, 1387779, 1387797, 1387977, 1389777, 1397778, 1397787, 1397877, 1398777, 1555888, 1558588, 1558858

Offset: 0

Boris Hostnik (megpplus(AT)siol.net), Jul 14 2009

			666666 is in the sequence because (1) cubed sum of its digits is (6+6+6+6+6+6)^3 = 46656, (2) the product of its digits is 6*6*6*6*6*6=46656; 46656=46656.

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Cf. A007954, A118880, A117720, A034710. - R. J. Mathar, Jul 19 2009

A007953 := proc(n) add(d,d=convert(n,base,10)) ; end: A007954 := proc(n) mul(d,d=convert(n,base,10)) ; end: A118880 := proc(n) (A007953(n))^3; end: for n from 1 to 2000000 do if A118880(n) = A007954(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Jul 19 2009

Select[Range[0,156*10^4],Total[IntegerDigits[#]]^3==Times@@IntegerDigits[#]&] (* Harvey P. Dale, Jul 07 2022 *)

{n: A118880(n)=A007954(n)}. - R. J. Mathar, Jul 19 2009

A232709 Nonnegative integers such that the sum of digits mod 10 equals the product of digits mod 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 22, 48, 84, 109, 123, 132, 137, 145, 154, 159, 173, 178, 187, 190, 195, 208, 213, 228, 231, 233, 235, 237, 239, 248, 253, 268, 273, 280, 282, 284, 286, 288, 293, 307, 312, 317, 321, 323, 325, 327, 329, 332, 337, 347, 352, 357, 367, 370, 371, 372, 373, 374, 375, 376, 377

Offset: 1

Jon Perry, Nov 28 2013

			293 is in the sequence because 2+9+3 = 14 == 4 mod 10 and 2*9*3 = 54 == 4 mod 10.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A034710.

for (i=0;i<1000;i++) {
s=i.toString().split("");
sl=s.length;
c=0;d=1;
for (j=0;j

Select[Range[0,400],Mod[Total[IntegerDigits[#]],10]==Mod[Times@@ IntegerDigits[ #],10]&] (* Harvey P. Dale, Oct 15 2021 *)

is(n) = my(d=digits(n));vecsum(d)%10==vecprod(d)%10 \\ David A. Corneth, Oct 15 2021

from math import prod
def ok(n): d = list(map(int, str(n))); return sum(d)%10 == prod(d)%10
print([k for k in range(378) if ok(k)]) # Michael S. Branicky, Oct 15 2021

Offset changed from 0 to 1 by N. J. A. Sloane, Oct 15 2021

A261778 Positive numbers n such that (digitsum(n))^2 equals (product of digits(n))^3.

Original entry on oeis.org

1, 11114, 11141, 11411, 14111, 41111, 111122, 111212, 111221, 112112, 112121, 112211, 121112, 121121, 121211, 122111, 211112, 211121, 211211, 212111, 221111, 1111111111111111119, 1111111111111111191, 1111111111111111911, 1111111111111119111, 1111111111111191111, 1111111111111911111

Offset: 1

K. D. Bajpai, Aug 31 2015

Sequence is infinite because it contains all the numbers made of k fours and 8^k-4k ones. - Giovanni Resta, Sep 01 2015

			11114 appears in the sequence because (1 + 1 + 1 + 1 + 4)^2 = (1*1*1*1*4)^3 = 64.
111122 appears in the sequence because (1 + 1 + 1 + 1 + 2 + 2)^2  = (1*1*1*1*2*2)^3 = 64.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Charles R Greathouse IV, GP script

Cf. A034710, A117720, A117723, A241846.

[n : n in [1..1000000] | (&+Intseq(n))^2 eq (&*Intseq(n))^3 ];

Select[Range[20000000], Plus @@ IntegerDigits[#]^2 == Times @@ IntegerDigits[#]^3 &]

for(n = 1,1000000, d = digits(n); if((sumdigits(n))^2 == prod(i = 1, #d, d[i])^3, print1(n, ", ")));

proddigits(n)=my(d=digits(n)); prod(i=1,#d,d[i])
is(n)=my(s=sumdigits(n)); if(!ispower(s,3), return(0)); s^2==proddigits(n)^3 \\ Charles R Greathouse IV, Aug 31 2015

a(22)-a(27) from Charles R Greathouse IV, Aug 31 2015

A272814 Palindromes such that sum of digits equals product of digits.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 22, 12221, 13131, 21212, 31113, 1111441111, 1114114111, 1141111411, 1411111141, 4111111114, 11112421111, 11121412111, 11211411211, 12111411121, 21111411112, 111122221111, 111212212111, 111221122111, 112112211211, 112121121211

Offset: 1

Altug Alkan, May 06 2016

Inspired by A272436.
Intersection of A002113 and A034710.
This sequence is obviously infinite.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A002113, A034710, A272436.

m[w_] := Flatten@Table[i, {i, 9}, {w[[i]]}]; palQ[n_] := n == FromDigits@ Reverse@ IntegerDigits@n; all[upd_] := Union@ Flatten@ Table[ FromDigits /@ Flatten[ Permutations /@ m /@ Select[ Flatten[Permutations /@ (IntegerPartitions[d + 9, {9}, Range[d+1]] -1), 1], Times @@ (Range[9]^#) == Total[# Range[9]] &], 1], {d, upd}]; Select[all@13, palQ] (* Giovanni Resta, May 06 2016 *)

isok(n) = { my(d = digits(n)); (vecsum(d) == prod(k=1, #d, d[k])) && (subst(Polrev(d), x, 10) == n);} \\ Michel Marcus, May 07 2016

a(15)-a(29) from Giovanni Resta, May 06 2016

A280355 Numbers that are divisible by the sum of their digits and for which the sum of digits equals the product of digits.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 132, 312, 4112, 11133, 11313, 11331, 13113, 13131, 13311, 22112, 31113, 31131, 31311, 33111, 111216, 111612, 112116, 116112, 121116, 161112, 211116, 611112, 1111712, 11111232, 11112132, 11112312, 11113212, 11118112, 11121132, 11121312, 11123112, 11131212, 11132112

Offset: 1

Ilya Gutkovskiy, Jan 01 2017

			132 is in the sequence because 1 + 3 + 2 = 1*3*2 = 6 and 6 divides 132.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Harshad Numbers

Intersection of A005349 and A034710.

Cf. A038186.

Select[Range[11300000], Divisible[#1, (Plus @@ IntegerDigits[#1])] && (Plus @@ IntegerDigits[#1]) == (Times @@ IntegerDigits[#1]) &]
nQ[n_]:=With[{idn=IntegerDigits[n]},Mod[n,Total[idn]]==0&&Total[idn]==Times@@idn]; Select[Range[112*10^5],nQ] (* Harvey P. Dale, Nov 02 2024 *)

isok(n) = (d=digits(n)) && ((n % vecsum(d)) == 0) && (vecsum(d) == prod(k=1, #d, d[k])); \\ Michel Marcus, Jan 02 2017

A281745 Numbers k with the property that the square root of the product of the digits of k is equal to the sum of the square roots of its digits.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 44, 149, 194, 228, 282, 333, 419, 491, 822, 914, 941, 11199, 11444, 11919, 11991, 14144, 14414, 14441, 19119, 19191, 19911, 41144, 41414, 41441, 44114, 44141, 44411, 91119, 91191, 91911, 99111, 11111449, 11111494, 11111944, 11114149

Offset: 1

José de Jesús Camacho Medina, Jan 28 2017

			    1 is a term because     sqrt(1) = sqrt(1);
   44 is a term because   sqrt(4*4) = sqrt(4) + sqrt(4);
  941 is a term because sqrt(9*4*1) = sqrt(9) + sqrt(4) + sqrt(1).

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A007953, A007954, A034710.

Select[Range[10^6], Sqrt[Times @@ #] == Total[Sqrt@ #] &@ IntegerDigits@ # &] (* Michael De Vlieger, Feb 02 2017 *)

isok(n) = my(d = vecsort(digits(n))); sqrt(prod(k=1, #d, d[k])) == sum(k=1, #d, sqrt(d[k])); \\ Michel Marcus, Jan 29 2017

More terms from Jon E. Schoenfield, Jan 30 2017

A297815 Number of positive integers with n digits whose digit sum is equal to its digit product.

Original entry on oeis.org

9, 1, 6, 12, 40, 30, 84, 224, 144, 45, 605, 495, 1170, 1092, 210, 240, 2448, 4896, 15846, 3420, 1750, 462, 15939, 0, 8100, 67925, 80730, 19656, 11774, 164430, 930, 29760, 197472, 0, 0, 1260, 23976, 50616, 54834, 395200, 1248860, 4253340, 75852, 0, 42570

Offset: 1

Reiner Moewald, Jan 06 2018

			The only term with two digits is 22: 2 * 2 = 2 + 2.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A034710, A061672.

cperm[w_] := Length[w]!/Times @@ ((Last /@ Tally[w])!); ric[s_, p_, w_, tg_] := Block[{d}, If[tg == 0, If[s == p, tot += cperm@ w], Do[ If[p*d > s + d + (tg-1)*9, Break[]]; ric[s+d, p*d, Append[w,d], tg-1], {d, Last@ w, 9}]]]; a[n_] := (tot=0; ric[#, #, {#}, n-1] & /@ Range[9]; tot); Array[a, 45] (* Giovanni Resta, Feb 05 2018 *)

import math
def digitProd(natNumber):
    digitProd = 1
    for letter in str(natNumber):
        digitProd *= int(letter)
    return digitProd
def digitSum(natNumber):
    digitSum = 0
    for letter in str(natNumber):
        digitSum += int(letter)
    return digitSum
for n in range(24):
    count = 0
    for a in range(int(math.pow(10,n)), int(math.pow(10, n+1))):
        if digitProd(a) == digitSum(a):
            count += 1
    print(n+1, count)

from sympy.utilities.iterables import combinations_with_replacement
from sympy import prod, factorial
def A297815(n):
    f = factorial(n)
    return sum(f//prod(factorial(d.count(a)) for a in set(d)) for d in combinations_with_replacement(range(1,10),n) if prod(d) == sum(d)) # Chai Wah Wu, Feb 06 2018

a(10) and a(23) corrected by and a(25)-a(45) from Giovanni Resta, Feb 05 2018

A337296 Number whose sum and product of ternary representation digits are equal.

Original entry on oeis.org

0, 1, 2, 8, 134, 152, 158, 160, 206, 212, 214, 230, 232, 238, 265760, 265814, 265832, 265838, 265840, 265976, 265994, 266000, 266002, 266048, 266054, 266056, 266072, 266074, 266080, 266462, 266480, 266486, 266488, 266534, 266540, 266542, 266558, 266560, 266566

Offset: 1

Amiram Eldar, Aug 21 2020

In ternary representation all the terms except 0 are zeroless (A032924).
If k is the number of digits 2 of a term, then the number of digits 1 is 2^k - 2*k, and the total number of digits is thus 2^k - k (A000325).
The total number of terms with k digits 2, for k = 1, 2, ..., is binomial(2^k-k,k) = 1, 1, 10, 495, 80730, 40475358, ...

			8 is a term since in base 3 the representation of 8 is 22 and 2 + 2 = 2 * 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

The ternary version of A034710.

Cf. A000325, A032924.

Select[Range[0, 266566], Times @@ (d = IntegerDigits[#, 3]) == Plus @@ d &]
(* or *)
f[k_] := FromDigits[#, 3] & /@ Permutations[Join[Table[1, {2^k - 2*k}], Table[2, k]]]; Flatten@ Join[{0}, Table[f[k], {k, 0, 4}]] (* Amiram Eldar, Oct 16 2023 *)

isok(m) = my(d=digits(m,3)); vecsum(d) == vecprod(d); \\ Michel Marcus, Aug 22 2020

A338214 Positive numbers for which the sum of some digits equals the product of the remaining ones.

Original entry on oeis.org

11, 22, 33, 44, 55, 66, 77, 88, 99, 100, 101, 110, 111, 112, 121, 122, 123, 132, 133, 134, 143, 144, 145, 154, 155, 156, 165, 166, 167, 176, 177, 178, 187, 188, 189, 198, 199, 200, 202, 211, 212, 213, 220, 221, 224, 231, 235, 236, 242, 246, 248, 253, 257, 263, 264, 268, 275, 279, 284, 286, 297, 300, 303, 312

Offset: 1

Eric Angelini and Carole Dubois, Oct 16 2020

123456789059 is in the sequence because (1+2+3+4+5+6+7+8+9+0) = (5*9) = 45.

Carole Dubois, Table of n, a(n) for n = 1..5000

Cf. A034710.

cp's OEIS Frontend

A113761 Numbers k such that the number of divisors of k equals both the sum and the product of digits of k in base 10.

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A162840 Numbers k such that the cube of the sum of digits of k equals the product of digits of k.

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A232709 Nonnegative integers such that the sum of digits mod 10 equals the product of digits mod 10.

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A261778 Positive numbers n such that (digitsum(n))^2 equals (product of digits(n))^3.

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A272814 Palindromes such that sum of digits equals product of digits.

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A280355 Numbers that are divisible by the sum of their digits and for which the sum of digits equals the product of digits.

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A281745 Numbers k with the property that the square root of the product of the digits of k is equal to the sum of the square roots of its digits.

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