A166311 -id:A166311 - cp's OEIS frontend

A279776 Numbers n such that the sum of digits of 6n equals 12.

8, 11, 14, 23, 26, 29, 32, 38, 41, 44, 47, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 86, 89, 92, 95, 101, 104, 107, 110, 119, 122, 125, 134, 137, 140, 152, 155, 173, 176, 179, 182, 188, 191, 194, 197, 203, 206, 209, 212, 215, 218, 221, 224, 227, 230, 236

Offset: 1

M. F. Hasler, Dec 23 2016

Inspired by A088408 = A062768/6 and A279769 (the analog for 9).

Cf. A007953 (digital sum), A279772 (sumdigits(2n) = 4), A279773 (sumdigits(3n) = 6), A279774 (sumdigits(4n) = 8), A279775 (sumdigits(5n) = 10), A279770 (sumdigits(7n) = 14), A279768 (sumdigits(8n) = 16), A279769 (sumdigits(9n) = 18), A279777 (sumdigits(9n) = 27).
Digital sum of m*n equals m: A088404 = A069537/2, A088405 = A052217/3, A088406 = A063997/4, A088407 = A069540/5, A088408 = A062768/6, A088409 = A063416/7, A088410 = A069543/8.
Cf. A005349 (Niven or Harshad numbers), A245062 (arranged in rows by digit sums).
Numbers with given digital sum: A011557 (1), A052216 (2), A052217 (3), A052218 (4), A052219 (5), A052220 (6), A052221 (7), A052222 (8), A052223 (9), A052224 (10), A166311 (11), A235151 (12), A143164 (13), A235225 (14), A235226 (15), A235227 (16), A166370 (17), A235228 (18), A166459 (19), A235229 (20).

Select[Range@ 240, Total@ IntegerDigits[6 #] == 12 &] (* Michael De Vlieger, Dec 23 2016 *)

PARI
```
is(n)=sumdigits(6*n)==12
```

A283742 Numbers with digit sum 11 that are multiples of 11.

Original entry on oeis.org

209, 308, 407, 506, 605, 704, 803, 902, 2090, 3080, 4070, 5060, 6050, 7040, 8030, 9020, 10109, 10208, 10307, 10406, 10505, 10604, 10703, 10802, 10901, 20009, 20108, 20207, 20306, 20405, 20504, 20603, 20702, 20801, 20900, 30008, 30107, 30206, 30305, 30404

Offset: 1

Zak Seidov, Mar 15 2017

Subsequence of A166311.
Numbers with digit sum 11 and even-numbered digits all 0. - Robert Israel, Mar 16 2017
Intersection of A008593 and A166311. - Michel Marcus, Mar 17 2017
If m is a term, so is 10*m. - Zak Seidov, Mar 17 2017

			a(1) = A166311(18) = 209 = 19*11,
a(40) = A166311(9881) = 30404 = 2764*11.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A008593 (multiples of 11), A166311 (numbers with digit sum 11).

F:= proc(d,t) option remember;
if d <= 1 then
  if t <= 9 then return [t*10^d] else return [] fi;
fi;
[seq(seq(j*10^d+s, s = procname(d-2, t-j)), j=0..min(9, t))]
end proc:
sort([op(F(4,11)),op(F(5,11))]); # Robert Israel, Mar 16 2017

Reap[Do[If[11==Total[IntegerDigits[m=11*k]],Sow[m]],{k,3000}]][[2,1]]
Select[Range[209,30404,11],11==Total[IntegerDigits[#]]&]

A375460 Lexicographically earliest sequence of distinct nonnegative terms arranged in successive chunks whose digitsum = 10.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 10, 11, 20, 6, 12, 100, 7, 21, 8, 101, 9, 1000, 13, 14, 10000, 15, 22, 16, 30, 17, 110, 18, 100000, 19, 23, 31, 1000000, 24, 40, 25, 102, 26, 200, 27, 10000000, 28, 32, 41, 33, 103, 34, 111, 35, 1001, 36, 100000000, 37, 42, 112, 43, 120, 44, 1010, 45, 1000000000

Offset: 1

Eric Angelini, Aug 15 2024

The first integer that will never appear in the sequence is 29, as its digitsum exceeds 10.
From Michael S. Branicky, Aug 16 2024: (Start)
Infinite since A052224 is infinite (as are all sequences with digital sum 1..10).
a(6492) has 1001 digits. (End)

			The first chunk of integers with digitsum 10 is (0,1,2,3,4);
the next one is (5,10,11,20),
the next one is (6,12,100),
the next one is (7,21),
the next one is (8,101),
the next one is (9,1000),
the next one is (13,14,10000), etc.
The concatenation of the above chunks produce the sequence.

Michael S. Branicky, Table of n, a(n) for n = 1..6491

Cf. A166311, A051885, A228915.

Numbers with digital sum 1..10: A011557 (1), A052216 (2), A052217 (3), A052218 (4), A052219 (5), A052220 (6), A052221 (7), A052222 (8), A052223 (9), A052224 (10).

from itertools import islice
def bgen(ds): # generator of terms with digital sum ds
    def A051885(n): return ((n%9)+1)*10**(n//9)-1 # due to Chai Wah Wu
    def A228915(n): # due to M. F. Hasler
        p = r = 0
        while True:
            d = n % 10
            if d < 9 and r: return (n+1)*10**p + A051885(r-1)
            n //= 10; r += d; p += 1
    k = A051885(ds)
    while True: yield k; k = A228915(k)
def agen(): # generator of terms
    an, ds_block = 0, 0
    dsg = [None] + [bgen(i) for i in range(1, 11)]
    dsi = [None] + [(next(dsg[i]), i) for i in range(1, 11)]
    while True:
        yield an
        an, ds_an = min(dsi[j] for j in range(1, 11-ds_block))
        ds_block = (ds_block + ds_an)%10
        dsi[ds_an] = (next(dsg[ds_an]), ds_an)
print(list(islice(agen(), 61))) # Michael S. Branicky, Aug 16 2024

a(46) and beyond from Michael S. Branicky, Aug 16 2024.

A279771 Numbers n such that the sum of digits of 11n equals 11.

Original entry on oeis.org

19, 28, 37, 46, 55, 64, 73, 82, 190, 280, 370, 460, 550, 640, 730, 820, 919, 928, 937, 946, 955, 964, 973, 982, 991, 1819, 1828, 1837, 1846, 1855, 1864, 1873, 1882, 1891, 1900, 2728, 2737, 2746, 2755, 2764, 2773, 2782, 2791, 2800, 3637, 3646, 3655, 3664

Offset: 1

M. F. Hasler, Dec 23 2016

Inspired by A088404 = A069537/2 through A088410 = A069543/8.

Cf. A007953 (digital sum), Digital sum of m*n equals m: A088404 = A069537/2, A088405 = A052217/3, A088406 = A063997/4, A088407 = A069540/5, A088408 = A062768/6, A088409 = A063416/7, A088410 = A069543/8.
Cf. A005349 (Niven or Harshad numbers), A245062 (arranged in rows by digit sums).
Numbers with given digital sum: A011557 (1), A052216 (2), A052217 (3), A052218 (4), A052219 (5), A052220 (6), A052221 (7), A052222 (8), A052223 (9), A052224 (10), A166311 (11), A235151 (12), A143164 (13), A235225 (14), A235226 (15), A235227 (16), A166370 (17), A235228 (18), A166459 (19), A235229 (20).
Cf. A279772 (sumdigits(2n) = 4), A279773 (sumdigits(3n) = 6), A279774 (sumdigits(4n) = 8), A279775 (sumdigits(5n) = 10), A279776 (sumdigits(6n) = 12), A279770 (sumdigits(7n) = 14), A279768 (sumdigits(8n) = 16), A279769 (sumdigits(9n) = 18), A279777 (sumdigits(9n) = 27).

Select[Range@ 3664, Total@IntegerDigits[11 #] == 11 &] (* Michael De Vlieger, Dec 23 2016 *)

PARI
```
is(n)=sumdigits(11*n)==11
```

A346221 Palindromes that are multiples of 11 and whose digit sum is also a multiple of 11.

Original entry on oeis.org

2992, 3883, 4774, 5665, 6556, 7447, 8338, 9229, 10901, 20702, 30503, 40304, 50105, 70807, 80608, 90409, 119911, 128821, 137731, 146641, 155551, 164461, 173371, 182281, 191191, 209902, 218812, 227722, 236632, 245542, 254452, 263362, 272272, 281182, 290092, 308803

Offset: 1

Tanya Khovanova, Jul 11 2021

Palindromes in A216995.

			11 is a palindrome that is a multiple of 11, but its digit sum is not divisible by 11. Thus, 11 is not in this sequence.

Cf. A002113, A083513, A166311 (halves of even length terms), A216995.

Select[Range[400000], PalindromeQ[#] && IntegerQ[#/11] && IntegerQ[Total[IntegerDigits[#]]/11] &]

isok(m) = my(d=digits(m)); (Vecrev(d) == d) && !(m % 11) && !(vecsum(d) % 11); \\ Michel Marcus, Aug 06 2021

from itertools import product
def sd(n): return sum(map(int, str(n)))
def pals(d, base=10): # all positive d-digit palindromes
    digits = "".join(str(i) for i in range(base))
    for p in product(digits, repeat=d//2):
        if d > 1 and p[0] == "0": continue
        left = "".join(p); right = left[::-1]
        for mid in [[""], digits][d%2]:
            t = int(left + mid + right)
            if t > 0: yield t
def ok(pal): return pal%11 == 0 and sd(pal)%11 == 0
print([p for d in range(1, 7) for p in pals(d) if ok(p)]) # Michael S. Branicky, Jul 11 2021

A357935 Primes p such that the sum of digits of 11*p is 11.

Original entry on oeis.org

19, 37, 73, 919, 937, 991, 1873, 2791, 3637, 3673, 3691, 4591, 6373, 8191, 91837, 91873, 92737, 92791, 93637, 94573, 181837, 181873, 181891, 182773, 183637, 183691, 185491, 186391, 187273, 272737, 274591, 275491, 276373, 277273, 278191, 363691, 365473, 367273, 455473, 455491, 458191, 459091

Offset: 1

J. M. Bergot and Robert Israel, Oct 21 2022

Primes in A279771.
All terms == 1 (mod 18).
Includes (8*10^k+3)/11 for k = 2, 6, 60, 320, 600.  Any others?

			a(3) = 73 is a term because 73 is prime and 11*73 = 803 with 8+0+3 = 11.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A166311, A279771.

filter:= proc(p) convert(convert(11*p,base,10),`+`) = 11 and isprime(p):
select(filter, [seq(i,i=1..10^6,18)]);

Select[Prime[Range[40000]], Plus @@ IntegerDigits[11*#] == 11 &] (* Amiram Eldar, Oct 21 2022 *)

cp's OEIS Frontend

A279776 Numbers n such that the sum of digits of 6n equals 12.

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A283742 Numbers with digit sum 11 that are multiples of 11.

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A375460 Lexicographically earliest sequence of distinct nonnegative terms arranged in successive chunks whose digitsum = 10.

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A279771 Numbers n such that the sum of digits of 11n equals 11.

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PARI

A346221 Palindromes that are multiples of 11 and whose digit sum is also a multiple of 11.

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Mathematica

PARI

Python

A357935 Primes p such that the sum of digits of 11*p is 11.

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Author

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Maple

Mathematica