A060159 -id:A060159 - cp's OEIS frontend

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

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

A141769 Beginning of a run of 4 consecutive Niven (or Harshad) numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 510, 1014, 2022, 3030, 10307, 12102, 12255, 13110, 60398, 61215, 93040, 100302, 101310, 110175, 122415, 127533, 131052, 131053, 196447, 201102, 202110, 220335, 223167, 245725, 255045, 280824, 306015, 311232, 318800, 325600, 372112, 455422

Offset: 1

Sergio Pimentel, Sep 15 2008

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

			510 is in the sequence because 510, 511, 512 and 513 are all 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.
Eric Weisstein's World of Mathematics, Harshad Number.
Wikipedia, Harshad number.
Brad Wilson Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.

Cf. A005349, A330927, A154701, A330928, A330929, A330930, A060159 (start of run of 1, 2, ..., 7, exactly n consecutive Harshad numbers).

Cf. A330933, A328211, A328215 (analog for base 2, Zeckendorf- resp. Fibonacci-Niven variants).

f:=func; a:=[]; for k in [1..500000] 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

nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; niv = nivenQ /@ Range[4]; seq = {}; Do[niv = Join[Rest[niv], {nivenQ[k]}]; If[And @@ niv, AppendTo[seq, k - 3]], {k, 4, 5*10^5}]; seq (* Amiram Eldar, Jan 03 2020 *)

{A141769_first( N=50, L=4, a=List())= for(n=1,oo, n+=L; for(m=1,L, n--%sumdigits(n) && next(2)); listput(a,n); N--|| break);a} \\ M. F. Hasler, Jan 03 2022

from itertools import count, islice
def agen(): # generator of terms
    h1, h2, h3, h4 = 1, 2, 3, 4
    while True:
        if h4 - h1 == 3: yield h1
        h1, h2, h3, h4, = h2, h3, h4, next(k for k in count(h4+1) if k%sum(map(int, str(k))) == 0)
print(list(islice(agen(), 40))) # Michael S. Branicky, Mar 17 2024

This A141769 = { A005349(k) | A005349(k+3) = A005349(k)+3 }. - M. F. Hasler, Jan 03 2022

More terms from Amiram Eldar, Jan 03 2020

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

A330929 Starts of runs of 6 consecutive Niven (or Harshad) numbers (A005349).

Original entry on oeis.org

1, 2, 3, 4, 5, 10000095, 10000096, 12751220, 14250624, 22314620, 22604423, 25502420, 28501224, 35521222, 41441420, 41441421, 51004820, 56511023, 57002424, 70131620, 71042422, 71253024, 97740760, 102009620, 111573020, 114004824, 121136420, 124324220, 124324221

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.

			10000095 is a term since 10000095 is divisible by 1 + 0 + 0 + 0 + 0 + 0 + 9 + 5 = 15, 10000096 is divisible by 16, ..., and 10000100 is divisible by 2.

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..4000
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, A330927, A330928, A330930.

f:=func; a:=[]; for k in [1..30000000] do  if forall{m:m in [0..5]|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[6]; seq = {}; Do[niv = Join[Rest[niv], {nivenQ[k]}]; If[And @@ niv, AppendTo[seq, k - 5]], {k, 6, 10^7}]; seq

A330930 Starts of runs of 7 consecutive Niven (or Harshad) numbers (A005349).

Original entry on oeis.org

1, 2, 3, 4, 10000095, 41441420, 124324220, 124324221, 124324222, 207207020, 233735070, 331531220, 350602590, 409036350, 414414020, 467470110, 621621020, 621621021, 621621022, 1030302012, 1036035020, 1051807710, 1201800620, 1243242020, 1243242021, 1243242022

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.

			10000095 is a term since 10000095 is divisible by 1 + 0 + 0 + 0 + 0 + 0 + 9 + 5 = 15, 10000096 is divisible by 16, ..., and 10000101 is divisible by 3.

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..400
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, A330927, A330928, A330929.

nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; niv = nivenQ /@ Range[7]; seq = {}; Do[niv = Join[Rest[niv], {nivenQ[k]}]; If[And @@ niv, AppendTo[seq, k - 6]], {k, 7, 10^7}]; seq

A060288 Distinct (non-overlapping) twin Harshad numbers whose sum is prime.

Original entry on oeis.org

3, 7, 11, 19, 41, 401, 419, 449, 881, 1021, 1259, 1289, 1471, 1601, 1607, 1871, 1999, 2029, 2281, 2549, 2609, 2833, 3041, 3359, 3457, 4001, 4049, 4481, 4801, 5641, 6329, 7499, 7561, 8081, 8849, 8929, 9613, 9619, 10321, 11131, 12401, 12799, 13033

Offset: 1

Enoch Haga, Mar 23 2001

Suggested by Puzzle 129, The Prime Puzzles and Problems Connection.

			a(3)=19, a prime, because the first Harshad number is 9 and the second is 10 and 9+10=19. To find the Harshad numbers take H1=(p-1)/2 as the first Harshad and then the second Harshad, H2=H1+1. Harshad numbers are those which have integral quotients after division by the sum of their digits. Note that 2+3=5 is not included because 1+2=3 are the first twins whose sum is prime and the next twins, 3+4=7, must not overlap the preceding pair.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Carlos Rivera, Puzzle 129. Earliest sets of K consecutive Harshad Numbers, The Prime Puzzles and Problems Connection.

Cf. A005349, A060159, A060289, A060290, A060291.

harshadQ[n_] := Divisible[n, Plus @@ IntegerDigits[n]]; s = {}; q1 = True; Do[q2 = harshadQ[n]; If[q1 && q2, If[PrimeQ[2*n - 1], AppendTo[s, 2*n - 1]]; q1 = False, q1 = q2], {n, 2, 5000}]; s (* Amiram Eldar, Jan 19 2021 *)

20 A=0; 30 inc A; 40 if Ct=2 then Z=(A-1)+(A-2): if Z=prmdiv(Z) then print A-2; "+"; A-1; "="; Z; "/"; :inc Pt; 50 if Ct=2 then Ct=1:A=A-1; 60 X=1; 70 B=str(A); 80 L=len(B); 90 inc X; 100 S=mid(B,X,1); 110 V=val(S):W=W+V; 120 if XDt+1 then Ct=0:Dt=0; 150 Dt=Ct:W=0; 160 if A<10000001 then 30; 170 print Pt;

Offset corrected by Amiram Eldar, Jan 19 2021

A060290 Primes which are sums of twin Harshad numbers (includes overlaps).

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 41, 223, 401, 419, 449, 881, 1021, 1259, 1289, 1471, 1601, 1607, 1871, 1999, 2029, 2281, 2549, 2609, 2833, 3041, 3359, 3457, 4001, 4049, 4481, 4801, 4931, 5641, 6329, 7499, 7561, 8081, 8849, 8929, 9613, 9619, 10111, 10321

Offset: 1

Enoch Haga, Mar 24 2001

			a(5)=17, a prime because the first Harshad number is 8 and the second is 9 and 8+9=17. In this sequence overlapping Harshad's are permitted: 1+2=3 and 2+3=5.

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

Cf. A005349, A060159, A060288, A060289, A060291.

isA005349 := proc(n)
    if n mod digsum(n) = 0 then
        true;
    else
        false;
    end if;
end proc:
isA060290 := proc(n)
    local h1 ;
    if isprime(n) then
        h1 := (n-1)/2 ;
        if isA005349(h1) and isA005349(h1+1) then
            true;
        else
            false;
        end if;
    else
        false;
    end if;
end proc:
for n from 3 to 20000 do
    if isA060290(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Dec 20 2013

harshadQ[n_] := Divisible[n, Plus @@ IntegerDigits[n]]; s = {}; q1 = True; Do[q2 = harshadQ[n]; If[q1 && q2 && PrimeQ[2*n - 1], AppendTo[s, 2*n - 1]]; q1 = q2, {n, 2, 5000}]; s (* Amiram Eldar, Jan 19 2021 *)

20 A=0;
30 inc A;
40 if Ct=2 then Z=(A-1)+(A-2): if Z=prmdiv(Z) then print A-2; "+"; A-1; "="; Z; "/"; :inc Pt;
50 if Ct=2 then Ct=1:A=A-2;
60 X=1;
70 B=str(A);
80 L=len(B);
90 inc X;
100 S=mid(B,X,1);
110 V=val(S):W=W+V;
120 if XDt+1 then Ct=0:Dt=0;
150 Dt=Ct:W=0;
160 if A<=10 then 30;
170 print Pt;

{n in A000040: (n-1)/2 in A005349 and (n+1)/2 in A005349}. - R. J. Mathar, Dec 20 2013

Offset corrected by Amiram Eldar, Jan 19 2021

A235397 The first term of the least sequence of n consecutive Moran numbers.

Original entry on oeis.org

18, 152, 3031, 21481224, 25502420, 4007565001480, 2196125475223740, 905295493763807066010

Offset: 1

Carlos Rivera, Jan 09 2014

A number n is a Moran number if n divided by the sum of its decimal digits is prime.
From Amiram Eldar, Apr 25 2020: (Start)
Jens Kruse Andersen found that a(7) <= 2196125475223740 and a(8) <= 905295493763807066010 (see Rivera link).
Since Moran numbers (A001101) are also Niven numbers (A005349), this sequence is finite with no more than 20 terms (see A060159). (End)
a(9) <= 270140199032572375590810. - Giovanni Resta, Apr 30 2020

			a(6) = 4007565001480 because
4007565001480 = 40 * 100189125037,
4007565001481 = 41 * 97745487841,
4007565001482 = 42 * 95418214321,
4007565001483 = 43 * 93199186081,
4007565001484 = 44 * 91081022761,
4007565001485 = 45 * 89057000033.

Giovanni Resta, Moran numbers
Carlos Rivera, Puzzle 728. Consecutive Moran numbers, The Prime Puzzles & Problems Connection.

Cf. A001101, A060159, A085775.

isA001101(n)=(k->denominator(k)==1&&isprime(k))(n/sumdigits(n))
a(n)=my(k=n); while(1, forstep(i=k,k-n+1,-1, if(!isA001101(i), k=i+n; next(2))); return(k-n+1)) \\ Charles R Greathouse IV, Jan 10 2014

a(7)-a(8) from Giovanni Resta, Apr 27 2020

A060289 Number of distinct (non-overlapping) twin Harshad numbers whose sum is prime and where the 2nd Harshad is <= 10^n.

Original entry on oeis.org

4, 5, 17, 53, 250, 1404, 9013, 58608, 401614, 2908740, 21832530

Offset: 1

Enoch Haga, Mar 23 2001

			a(1)=4 because there are four pairs of Harshads whose sum is prime and the 2nd Harshad in the pair is <=10; these are 1+2=3, 3+4=7, 5+6=11, 9+10=19. 8+9=17 is not included because this pair overlaps 7+8=15, which also happens to be not prime. (Another sequence might include such overlapping pairs.)

Cf. A005349, A060159, A060288, A060290, A060291.

harshadQ[n_] := Divisible[n, Plus @@ IntegerDigits[n]]; c = 0; p = 10; s = {}; n = 0; k = 2; q1 = True; While[n <6, q2 = harshadQ[k]; If[q1 && q2, If[PrimeQ[2*k - 1],c++;If[k > p, n++; AppendTo[s, c-1]; p *= 10]]; q1 = False, q1 = q2];  k++]; s (* Amiram Eldar, Jan 19 2021 *)

Niven(n)=n%sumdigits(n)==0
a(n)=my(t,s); for(k=1,10^n,if(Niven(k), if(isprime(t+k), t=-10^n; s++); t=k)); s \\ Charles R Greathouse IV, Jan 23 2014

Generate the twin Harshads whose sum is prime. Count how many there are where the 2nd Harshad in the pair is <= a consecutive power of 10.

a(8)-a(11) from Amiram Eldar, Jan 19 2021

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

C

Magma

Maple

Mathematica

Python

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

Python

A141769 Beginning of a run of 4 consecutive Niven (or Harshad) numbers.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A330929 Starts of runs of 6 consecutive Niven (or Harshad) numbers (A005349).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

A330930 Starts of runs of 7 consecutive Niven (or Harshad) numbers (A005349).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs