A141769 -id:A141769 - cp's OEIS frontend

A005349 Niven (or Harshad, or harshad) numbers: numbers that are divisible by the sum of their digits.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81, 84, 90, 100, 102, 108, 110, 111, 112, 114, 117, 120, 126, 132, 133, 135, 140, 144, 150, 152, 153, 156, 162, 171, 180, 190, 192, 195, 198, 200, 201, 204

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

Both spellings, "Harshad" or "harshad", are in use. It is a Sanskrit word, and in Sanskrit there is no distinction between upper- and lower-case letters. - N. J. A. Sloane, Jan 04 2022
z-Niven numbers are numbers n which are divisible by (A*s(n) + B) where A, B are integers and s(n) is sum of digits of n. Niven numbers have A = 1, B = 0. - Ctibor O. Zizka, Feb 23 2008
A070635(a(n)) = 0. A038186 is a subsequence. - Reinhard Zumkeller, Mar 10 2008
A049445 is a subsequence of this sequence. - Ctibor O. Zizka, Sep 06 2010
Complement of A065877; A188641(a(n)) = 1; A070635(a(n)) = 0. - Reinhard Zumkeller, Apr 07 2011
A001101, the Moran numbers, are a subsequence. - Reinhard Zumkeller, Jun 16 2011
A140866 gives the number of terms <= 10^k. - Robert G. Wilson v, Oct 16 2012
The asymptotic density of this sequence is 0 (Cooper and Kennedy, 1984). - Amiram Eldar, Jul 10 2020
From Amiram Eldar, Oct 02 2023: (Start)
Named "Harshad numbers" by the Indian recreational mathematician Dattatreya Ramchandra Kaprekar (1905-1986) in 1955. The meaning of the word is "giving joy" in Sanskrit.
Named "Niven numbers" by Kennedy et al. (1980) after the Canadian-American mathematician Ivan Morton Niven (1915-1999). During a lecture given at the 5th Annual Miami University Conference on Number Theory in 1977, Niven mentioned a question of finding a number that equals twice the sum of its digits, which appeared in the children's pages of a newspaper. (End)

			195 is a term of the sequence because it is divisible by 15 (= 1 + 9 + 5).

Paul Dahlenberg and T. Edgar, Consecutive factorial base Niven numbers, Fib. Q., 56:2 (2018), 163-166.
D. R. Kaprekar, Multidigital Numbers, Scripta Math., Vol. 21 (1955), p. 27.
Robert E. Kennedy and Curtis N. Cooper, On the natural density of the Niven numbers, Abstract 816-11-219, Abstracts Amer. Math. Soc., 6 (1985), 17.
Robert E. Kennedy, Terry A. Goodman, and Clarence H. Best, Mathematical Discovery and Niven Numbers, The MATYC Journal, Vol. 14, No. 1 (1980), pp. 21-25.
József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, p. 381.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 171.

N. J. A. Sloane, Table of n, a(n) for n = 1..11872 (all a(n) <= 100000)
Bob Albrecht, Don Albers, and Jim Conlan, Problem #22 Two-digit Niven numbers, Programming Problems, Recreational Computing Magazine, Vol. 9, No. 1, Issue 46 (1980), p. 59.
Curtis N. Cooper and Robert E. Kennedy, On an asymptotic formula for the Niven numbers, International Journal of Mathematics and Mathematical Sciences, Vol. 8, No. 3 (1985), pp. 537-543.
Curtis N. Cooper and Robert E. Kennedy, Chebyshev's inequality and natural density, Amer. Math. Monthly 96 (1989), no. 2, 118-124.
Paul Dalenberg and Tom Edgar, Consecutive factorial base Niven numbers, Fibonacci Quart. (2018) Vol. 56, No. 2, 163-166.
Jean-Marie De Koninck and Nicolas Doyon, Large and Small Gaps Between Consecutive Niven Numbers, J. Integer Seqs., Vol. 6, 2003, Article 03.2.5.
Nicolas Doyon, Les fascinants nombres de Niven, Thèse de la Faculté des Sciences et de Génie de l'Université Laval, Québec, Novembre 2006 (in French).
Ömer Eğecioğlu and Bünyamin Şahin, On twin EP numbers, Transact. Comb. (2025) Vol. 14, Iss. 4, Art. No. 4, 261-270. See p. 262.
Richard K. Guy, The Second Strong Law of Small Numbers, Math. Mag, 63 (1990), no. 1, 3-20.
Richard K. Guy, The Second Strong Law of Small Numbers, Math. Mag, 63 (1990), no. 1, 3-20. [Annotated scanned copy]
Robert E. Kennedy, Digital sums, Niven numbers, and natural density, Crux Mathematicorum, Vol. 8 (1982), pp. 131-135.
Robert E. Kennedy and Curtis N. Cooper, On the natural density of the Niven numbers, The College Mathematics Journal, Vol. 15, No. 4 (Sep., 1984), pp. 309-312.
Project Euler, Harshad Numbers: Problem 387.
Terry Trotter, Niven Numbers for Fun and Profit. [archived page]
Gérard Villemin, Nombres de Harshad (French).
Elaine E. Visitacion, Renalyn T. Boado, Mary Ann V. Doria, and Eduard M. Albay, On Harshad Number, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 134-138. [archived]
Eric Weisstein's World of Mathematics, Digit and Harshad Numbers.
Wikipedia, Harshad number.

Cf. A001101, A007602, A007953, A028834, A038186, A049445, A052018, A052019, A052020, A052021, A052022, A065877, A070635, A113315, A188641.
Cf. A001102 (a subsequence).
Cf. A118363 (for factorial-base analog).
Cf. A330927, A154701, A141769, A330928, A330929, A330930 (start of runs of 2, 3, ..., 7 consecutive Niven numbers).

Filtered([1..230],n-> n mod List(List([1..n],ListOfDigits),Sum)[n]=0); # Muniru A Asiru

a005349 n = a005349_list !! (n-1)
a005349_list = filter ((== 0) . a070635) [1..]
-- Reinhard Zumkeller, Aug 17 2011, Apr 07 2011

[n: n in [1..250] | n mod &+Intseq(n) eq 0];  // Bruno Berselli, May 28 2011

[n: n in [1..250] | IsIntegral(n/&+Intseq(n))];  // Bruno Berselli, Feb 09 2016

s:=proc(n) local N:N:=convert(n,base,10):sum(N[j],j=1..nops(N)) end:p:=proc(n) if floor(n/s(n))=n/s(n) then n else fi end: seq(p(n),n=1..210); # Emeric Deutsch

harshadQ[n_] := Mod[n, Plus @@ IntegerDigits@ n] == 0; Select[ Range[1000], harshadQ] (* Alonso del Arte, Aug 04 2004 and modified by Robert G. Wilson v, Oct 16 2012 *)
Select[Range[300],Divisible[#,Total[IntegerDigits[#]]]&] (* Harvey P. Dale, Sep 07 2015 *)

is(n)=n%sumdigits(n)==0 \\ Charles R Greathouse IV, Oct 16 2012

A005349 = [n for n in range(1,10**6) if not n % sum([int(d) for d in str(n)])] # Chai Wah Wu, Aug 22 2014

[n for n in (1..10^4) if sum(n.digits(base=10)).divides(n)] # Freddy Barrera, Jul 27 2018

A154701 Numbers k such that k, k + 1 and k + 2 are 3 consecutive Harshad numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 110, 510, 511, 1010, 1014, 1015, 2022, 2023, 2464, 3030, 3031, 4912, 5054, 5831, 7360, 8203, 9854, 10010, 10094, 10307, 10308, 11645, 12102, 12103, 12255, 12256, 13110, 13111, 13116, 13880, 14704, 15134, 17152, 17575, 18238, 19600, 19682

Offset: 1

Avik Roy (avik_3.1416(AT)yahoo.co.in), Jan 14 2009, Jan 15 2009

Harshad numbers are also known as Niven numbers.

Cooper and Kennedy proved that there are infinitely many runs of 20 consecutive Niven numbers. Therefore this sequence is infinite. - Amiram Eldar, Jan 03 2020

			110 is a term since 110 is divisible by 1 + 1 + 0 = 2, 111 is divisible by 1 + 1 + 1 = 3, and 112 is divisible by 1 + 1 + 2 = 4.

Jean-Marie De Koninck, Those Fascinating Numbers, American Mathematical Society, 2009, p. 36, entry 110.

Robert Israel, Table of n, a(n) for n = 1..10000
Curtis Cooper and Robert E. Kennedy, On consecutive Niven numbers, Fibonacci Quarterly, Vol. 21, No. 2 (1993), pp. 146-151.
Helen G. Grundman, Sequences of consecutive Niven numbers, Fibonacci Quarterly, Vol. 32, No. 2 (1994), pp. 174-175.
Wikipedia, Harshad number
Brad Wilson, Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.

A subset of A005349.

Cf. A060159, A141769, A328210, A328214, A330927, A330928, A330929, A330930, A330932.

#include 
#include 
int is_harshad(int n){
  int i,j,count=0;
  i=n;
  while(i>0){
    count=count+i%10;
    i=i/10;
  }
  return n%count==0?1:0;
}
main(){
  int k;
  clrscr();
  for(k=1;k<=30000;k++)
    if(is_harshad(k)&&is_harshad(k+1)&&is_harshad(k+2))
      printf("%d,",k);
  getch();
  return 0;
}

f:=func; a:=[]; for k in [1..20000] do  if forall{m:m in [0..2]|f(k+m)} then Append(~a,k); end if; end for; a; // Marius A. Burtea, Jan 03 2020

Res:= NULL: count:= 0:
state:= 1:
L:= [1]:
for n from 2 while count < 100 do
  L[1]:=L[1]+1;
  for k from 1 while L[k]=10 do L[k]:= 0;
    if k = nops(L) then L:= [0$nops(L),1]; break
    else L[k+1]:= L[k+1]+1 fi
  od:
  s:= convert(L,`+`);
  if n mod s = 0 then
     state:= min(state+1,3);
     if state = 3 then count:= count+1; Res:= Res, n-2; fi
  else state:= 0
  fi
od:
Res; # Robert Israel, Feb 01 2019

nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; niv = nivenQ /@ Range[3]; seq = {}; Do[niv = Join[Rest[niv], {nivenQ[k]}]; If[And @@ niv, AppendTo[seq, k - 2]], {k, 3, 2*10^4}]; seq (* Amiram Eldar, Jan 03 2020 *)

from itertools import count, islice
def agen(): # generator of terms
    h1, h2, h3 = 1, 2, 3
    while True:
        if h3 - h1 == 2: yield h1
        h1, h2, h3 = h2, h3, next(k for k in count(h3+1) if k%sum(map(int, str(k))) == 0)
print(list(islice(agen(), 45))) # Michael S. Branicky, Mar 17 2024

A330927 Numbers k such that both k and k + 1 are Niven numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 20, 80, 110, 111, 132, 152, 200, 209, 224, 399, 407, 440, 480, 510, 511, 512, 629, 644, 735, 800, 803, 935, 999, 1010, 1011, 1014, 1015, 1016, 1100, 1140, 1160, 1232, 1274, 1304, 1386, 1416, 1455, 1520, 1547, 1651, 1679, 1728, 1853

Offset: 1

Amiram Eldar, Jan 03 2020

Cooper and Kennedy proved that there are infinitely many runs of 20 consecutive Niven numbers. Therefore this sequence is infinite.

			1 is a term since 1 and 1 + 1 = 2 are both Niven numbers.

Jean-Marie De Koninck, Those Fascinating Numbers, American Mathematical Society, 2009, p. 36, entry 110.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Curtis Cooper and Robert E. Kennedy, On consecutive Niven numbers, Fibonacci Quarterly, Vol. 21, No. 2 (1993), pp. 146-151.
Helen G. Grundman, Sequences of consecutive Niven numbers, Fibonacci Quarterly, Vol. 32, No. 2 (1994), pp. 174-175.
Wikipedia, Harshad number.
Brad Wilson, Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.

Cf. A005349, A060159, A141769, A154701, A328205, A328209, A328213, A330713, A330928, A330929, A330930, A330931.

f:=func; a:=[]; for k in [1..2000] do  if forall{m:m in [0..1]|f(k+m)} then Append(~a,k); end if; end for; a; // Marius A. Burtea, Jan 03 2020

nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; nq1 = nivenQ[1]; seq = {}; Do[nq2 = nivenQ[k]; If[nq1 && nq2, AppendTo[seq, k - 1]]; nq1 = nq2, {k, 2, 2000}]; seq
SequencePosition[Table[If[Divisible[n,Total[IntegerDigits[n]]],1,0],{n,2000}],{1,1}][[;;,1]] (* Harvey P. Dale, Dec 24 2023 *)

from itertools import count, islice
def agen(): # generator of terms
    h1, h2 = 1, 2
    while True:
        if h2 - h1 == 1: yield h1
        h1, h2 = h2, next(k for k in count(h2+1) if k%sum(map(int, str(k))) == 0)
print(list(islice(agen(), 52))) # Michael S. Branicky, Mar 17 2024

A330933 Starts of runs of 4 consecutive Niven numbers in base 2 (A049445).

Original entry on oeis.org

6222, 33102, 53262, 66702, 94830, 221550, 268302, 284910, 295182, 300750, 316590, 364110, 379950, 427470, 533950, 554190, 570030, 590862, 617550, 633390, 696750, 791790, 807630, 855150, 870990, 902670, 934350, 1081422, 1140270, 1282830, 1314510, 1330350, 1343502

Offset: 1

Amiram Eldar, Jan 03 2020

Cai proved that there are infinitely many runs of 4 consecutive Niven numbers in base 2.

Grundman proved that there are no runs of 5 or more consecutive Niven numbers in base 2.

			6222 is a term since 6222, 6223, 6224 and 6225 are all Niven numbers in base 2.

József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, p. 382.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Tianxin Cai, On 2-Niven numbers and 3-Niven numbers, Fibonacci Quarterly, Vol. 34, No. 2 (1996), pp. 118-120.
Helen G. Grundman, Sequences of consecutive Niven numbers, Fibonacci Quarterly, Vol. 32, No. 2 (1994), pp. 174-175.
Wikipedia, Harshad number.
Brad Wilson, Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.

Cf. A049445, A141769, A328211, A328215, A330931, A330932.

f:=func; a:=[]; for k in [1..1400000] do  if forall{m:m in [0..3]|f(k+m)} then Append(~a,k); end if; end for; a; // Marius A. Burtea, Jan 03 2020

binNivenQ[n_] := Divisible[n, Total @ IntegerDigits[n, 2]]; bin = binNivenQ /@ Range[4]; seq = {}; Do[bin = Join[Rest[bin], {binNivenQ[k]}]; If[And @@ bin, AppendTo[seq, k - 3]], {k, 4, 10^6}]; seq

A328211 Starts of runs of 4 consecutive Zeckendorf-Niven numbers (A328208).

Original entry on oeis.org

1, 2, 3, 123543, 124242, 545502, 1367583, 1856349, 2431230, 2465110, 2593590, 2783709, 3247389, 3479229, 3917823, 3942909, 4174749, 4303428, 4494390, 4920640, 5143830, 5710383, 6261309, 6493149, 6552903, 6956829, 7420509, 7470880, 8970948, 9107790, 9507069, 10952928

Offset: 1

Amiram Eldar, Oct 07 2019

nonn

Grundman proved that this sequence is infinite by showing the F(120k-6) + F(8) + F(6) + F(4) is a term for all k >= 1, where F(k) is the k-th Fibonacci number.

She also proved that the only starts of runs of 5 consecutive Zeckendorf-Niven numbers are 1 and 2.

			1 is in the sequence since 1, 2, 3 and 4 are in A328208: A007895(1) = 1 is a divisor of 1, A007895(2) = 1 is a divisor of 2, A007895(3) = 1 is a divisor of 3, and A007895(4) = 2 is a divisor of 4.

Amiram Eldar, Table of n, a(n) for n = 1..216
Helen G. Grundman, Consecutive Zeckendorf-Niven and lazy-Fibonacci-Niven numbers, Fibonacci Quarterly, Vol. 45, No. 3 (2007), pp. 272-276.

Cf. A005349, A007895, A141769, A328208.

z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; aQ[n_] := Divisible[n, z[n]]; c = 0; k = 1; s = {}; v = Table[-1, {4}]; While[c < 32, If[aQ[k], v = Join[Rest[v], {k}]; If[AllTrue[Differences[v], # == 1 &], c++; AppendTo[s, k - 3]]]; k++]; s (* after Alonso del Arte at A007895 *)

A328215 Starts of runs of 4 consecutive lazy-Fibonacci-Niven numbers (A328212).

Original entry on oeis.org

3674769, 17434975, 22711023, 26152125, 32784723, 41221725, 57846123, 93416568, 101681916, 122873490, 173504940, 225947148, 234209247, 259557450, 333681684, 377858544, 396241410, 413770056, 432640989, 443496447, 444571650, 484381323, 497625360, 556123167, 564869940

Offset: 1

Amiram Eldar, Oct 07 2019

nonn

Grundman found a(1) and proved that there are no runs of 5 consecutive lazy-Fibonacci-Niven numbers.

			3674769 is in the sequence since 3674769, 3674770, 3674771 and 3674772 are in A328212: A112310(3674769) = 21 is a divisor of 3674769, A112310(3674770) = 22 is a divisor of 3674770, A112310(3674771) = 17 is a divisor of 3674771, and A112310(3674772) = 18 is a divisor of 3674772.

Amiram Eldar, Table of n, a(n) for n = 1..72
Helen G. Grundman, Consecutive Zeckendorf-Niven and lazy-Fibonacci-Niven numbers, Fibonacci Quarterly, Vol. 45, No. 3 (2007), pp. 272-276.

Cf. A112310, A141769, A328212.

ooQ[n_] := Module[{k = n}, While[k > 3, If[Divisible[k, 4], Return[True], k = Quotient[k, 2]]]; False]; c = 0; cn = 0; k = 1; s = {}; v = Table[-1, {4}]; While[cn < 10, If[! ooQ[k], c++; d = Total@IntegerDigits[k, 2]; If[Divisible[c, d], v = Join[Rest[v], {c}]; If[AllTrue[Differences[v], # == 1 &], cn++; AppendTo[s, c - 3]]]]; k++]; s

More terms from Amiram Eldar, Oct 23 2019

A331824 Starts of runs of 4 consecutive positive negabinary-Niven numbers (A331728).

Original entry on oeis.org

1, 1264, 2104, 2944, 4624, 11888, 23768, 27312, 27728, 31688, 35648, 49144, 51488, 55448, 56704, 58384, 60072, 63424, 65104, 66784, 70144, 71288, 75248, 76452, 79208, 81904, 87128, 91088, 92832, 99008, 102968, 114848, 118808, 123904, 125592, 126728, 130624, 131044

Offset: 1

Amiram Eldar, Jan 27 2020

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

Cf. A027615, A039724, A141769, A328211, A328215, A330933, A331728.

negaBinWt[n_] := negaBinWt[n] = If[n == 0, 0, negaBinWt[Quotient[n - 1, -2]] + Mod[n, 2]]; negaBinNivenQ[n_] := Divisible[n, negaBinWt[n]]; nConsec = 4; neg = negaBinNivenQ /@ Range[nConsec]; seq = {}; c = 0; k = nConsec+1; While[c < 45, If[And @@ neg, c++; AppendTo[seq, k - nConsec]]; neg = Join[Rest[neg], {negaBinNivenQ[k]}]; k++]; seq

A342429 Starts of runs of 4 consecutive Niven numbers in base 3/2 (A342426).

Original entry on oeis.org

1649373, 4029519, 15281054, 31906263, 43387386, 58198173, 94468958, 100084949, 131393766, 131986502, 140282279, 156786124, 211004079, 246960048, 253000850, 278206663, 310135917, 330168203, 351204398, 363280904, 412296883, 504736647, 515831624, 537255647, 566300238

Offset: 1

Amiram Eldar, Mar 11 2021

Are there 5 consecutive Niven numbers in base 3/2? There are no such numbers below 3*10^9.

			1649373 is a term since 1649373, 1649374, 1649375 and 1649376 are all Niven numbers in base 3/2.

Amiram Eldar, Table of n, a(n) for n = 1..100

Subsequence of A342426, A342427 and A342428.

Similar sequences: A141769 (decimal), A328207 (factorial), A328211 (Zeckendorf), A328215 (lazy Fibonacci), A330933 (binary), A334311 (base phi), A331824 (negabinary).

s[0] = 0; s[n_] := s[n] = s[2*Floor[n/3]] + Mod[n, 3]; q[n_] := Divisible[n, s[n]]; v = q /@ Range[4]; seq = {}; Do[v = Join[Rest[v], {q[k]}]; If[And @@ v, AppendTo[seq, k - 3]], {k, 4, 10^7}]; seq

A330928 Starts of runs of 5 consecutive Niven (or harshad) numbers (A005349).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 131052, 491424, 1275140, 1310412, 1474224, 1614623, 1912700, 2031132, 2142014, 2457024, 2550260, 3229223, 3931224, 4422624, 4914024, 5405424, 5654912, 5920222, 7013180, 7125325, 7371024, 8073023, 8347710, 9424832, 10000095, 10000096, 10000097

Offset: 1

Amiram Eldar, Jan 03 2020

Cooper and Kennedy proved that there are infinitely many runs of 20 consecutive Niven numbers. Therefore this sequence is infinite.

			131052 is a term since 131052 is divisible by 1 + 3 + 1 + 0 + 5 + 2 = 12, 131053 is divisible by 13, 131054 is divisible by 14, 131055 is divisible by 15, and 131056 is divisible by 16.

Jean-Marie De Koninck, Those Fascinating Numbers, American Mathematical Society, 2009, p. 36, entry 110.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Curtis Cooper and Robert E. Kennedy, On consecutive Niven numbers, Fibonacci Quarterly, Vol. 21, No. 2 (1993), pp. 146-151.
Helen G. Grundman, Sequences of consecutive Niven numbers, Fibonacci Quarterly, Vol. 32, No. 2 (1994), pp. 174-175.
Wikipedia, Harshad number.
Brad Wilson, Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.

Cf. A005349, A060159; A330927, A154701, A141769, A330929, A330930 (same for 2, 3, 4, 6, 7 consecutive harshad numbers).

f:=func; a:=[]; for k in [1..11000000] do  if forall{m:m in [0..4]|f(k+m)} then Append(~a,k); end if; end for; a; // Marius A. Burtea, Jan 03 2020

nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; niv = nivenQ /@ Range[5]; seq = {}; Do[niv = Join[Rest[niv], {nivenQ[k]}]; If[And @@ niv, AppendTo[seq, k - 4]], {k, 5, 10^7}]; seq
SequencePosition[Table[If[Divisible[n,Total[IntegerDigits[n]]],1,0],{n,10^7+200}],{1,1,1,1,1}][[;;,1]] (* Harvey P. Dale, Dec 24 2023 *)

{first( N=50, LEN=5, L=List())= for(n=1,oo, n+=LEN; for(m=1,LEN, n--%sumdigits(n) && next(2)); listput(L,n); N--|| break);L} \\ M. F. Hasler, Jan 03 2022

This A330928 = { A005349(k) | A005349(k+4) = A005349(k)+4 }. - M. F. Hasler, Jan 03 2022

A344344 Starts of runs of 4 consecutive Gray-code Niven numbers (A344341).

Original entry on oeis.org

1, 6, 30, 126, 510, 543, 783, 903, 2046, 2093, 3773, 3903, 7133, 7743, 8190, 8223, 8703, 10087, 12303, 12543, 14343, 14463, 15423, 15903, 16143, 16263, 20167, 22687, 27727, 30247, 30653, 30783, 32766, 35629, 40327, 47509, 47887, 49133, 50407, 57533, 60071, 60487

Offset: 1

Amiram Eldar, May 15 2021

Are there 5 consecutive Gray-code Niven numbers? There are no such numbers below 10^10.

			1 is a term since 1, 2, 3 and 4 are all Gray-code Niven numbers.

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

Cf. A005811, A014550.
Subsequence of A344341, A344342 and A344343.
Similar sequences: A141769 (decimal), A328207 (factorial), A328211 (Zeckendorf), A328215 (lazy Fibonacci), A330933 (binary), A334311 (base phi), A331824 (negabinary), A342429 (base 3/2).

gcNivenQ[n_] := Divisible[n, DigitCount[BitXor[n, Floor[n/2]], 2, 1]]; Select[Range[60000], AllTrue[# + {0, 1, 2, 3}, gcNivenQ] &]

cp's OEIS Frontend

A005349 Niven (or Harshad, or harshad) numbers: numbers that are divisible by the sum of their digits.

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GAP

Haskell

Magma

Magma

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PARI

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Sage

A154701 Numbers k such that k, k + 1 and k + 2 are 3 consecutive Harshad numbers.

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C

Magma

Maple

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A330927 Numbers k such that both k and k + 1 are Niven numbers.

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Magma

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A330933 Starts of runs of 4 consecutive Niven numbers in base 2 (A049445).

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Magma

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A328211 Starts of runs of 4 consecutive Zeckendorf-Niven numbers (A328208).

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Mathematica

A328215 Starts of runs of 4 consecutive lazy-Fibonacci-Niven numbers (A328212).

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Mathematica