A199682 -id:A199682 - cp's OEIS frontend

A067275 Number of Fibonacci numbers A000045(k), k <= 10^n, which end in 4.

0, 1, 7, 67, 667, 6667, 66667, 666667, 6666667, 66666667, 666666667, 6666666667, 66666666667, 666666666667, 6666666666667, 66666666666667, 666666666666667, 6666666666666667, 66666666666666667, 666666666666666667, 6666666666666666667, 66666666666666666667, 666666666666666666667

Offset: 0

Joseph L. Pe, Feb 21 2002

Numbers n such that the digits of A000326(n), the n-th pentagonal number, begin with n. [original definition of this sequence]
The sequence 1,7,67.... has a(n) = 6*10^n/9+3/9. It is the second binomial transform of 6*A001045(3n)/3 + (-1)^n. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is k*10^n/9 + (1-k/9) = 1, 1+k, 1+11k, 1+111k, ... - Paul Barry, Mar 24 2004
Except for the first two terms, these are the 3-automorphic numbers ending in 7. - Eric M. Schmidt, Aug 28 2012
From Wolfdieter Lang, Feb 08 2017: (Start)
This sequence appears in curious identities based on the Armstrong numbers 370 = A005188(11), 371 = A005188(12) and 407 = A005188(13).
For such identities based on 153 = A005188(10) see a comment in A246057 with the van der Poorten et al. reference.
For 370 these identities are A002277(n)^3 + a(n+1)^3  + 0(n)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n) = A281858(n), where 0(n) means n 0's.
For 371 these identities are A002277(n)^3 + a(n+1)^3 + (0(n-1)1)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n-1)1 = A281860(n), where 0(n-1)1 means n-1 0's followed by a 1.
For 407 these identities are A093137(n)^3 + 0(n)^3 + a(n+1)^3 = concatenation(A093137(n), 0(n), a(n+1)) = A281859(n). (End)

			a(2) = 7 because 7 of the first 10^2 Fibonacci numbers end in 4.
From _Wolfdieter Lang_, Feb 08 2017: (Start)
Curious cubic identities:
3^3 + 7^3 + 0^3 = 370, 33^3 + 67^3 + (00)^3 = 336700, 333^3 + 667^3 + (000)^3 = 333667000, ...
3^3 + 7^3 + 1^3 = 371, 33^3 + 67^3 + (01)^3 = 336701, 333^3 + 667^3 + (001)^3 = 333667001, ...
4^3 + 0^3 + 7^3 = 407, 34^3 + (00)^3 + 67^3 = 340067 , 334^3 + (000)^3 + 677^3 = 334000677, ... (End)

Seiichi Manyama, Table of n, a(n) for n = 0..1001
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A072702, A093137, A109344, A246057, A281858, A281859, A281860.

Cf. A000045, A000326, A001045, A002277, A005188, A073553, A199682.

s = Fibonacci@ Range[10^5]; Table[Count[Take[s, 10^n], m_ /; Mod[m, 10] == 4], {n, 0, Floor@ Log10@ Length@ s}] (* or *) Table[Boole[n > 0] Ceiling[10^n/15], {n, 0, 20}] (* or *) CoefficientList[Series[x (1 - 4 x)/((1 - x) (1 - 10 x)), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)

a(n)=(10^n+13)\15 \\ Charles R Greathouse IV, Jun 05 2011

a(n) = ceiling((2/30)*10^n) - Benoit Cloitre, Aug 27 2002
From Paul Barry, Mar 24 2004: (Start)
G.f.: x*(1 - 4*x)/((1 - x)*(1 - 10*x)).
a(n) = 10^n/15 + 1/3 for n>0. (End)
a(n) = 10*a(n-1) - 3 for n>1. - Vincenzo Librandi, Dec 07 2010 [Immediate consequence of the previous formula, R. J. Mathar]
From Eric M. Schmidt, Oct 28 2012: (Start)
For n>0, a(n) = A199682(n-1)/3 = (2*10^(n-1) + 1)/3.
For n>=2, a(n+1) = a(n) + 6*10^n. (End)
From Elmo R. Oliveira, Jul 22 2025: (Start)
E.g.f.: (-6 + 5*exp(x) + exp(10*x))/15.
a(n) = 11*a(n-1) - 10*a(n-2) for n >= 3.
a(n) = A073553(n)/2 for n >= 1. (End)

A073552 merged into this sequence by Eric M. Schmidt, Oct 28 2012

A254338 Initial digits of A254143 in decimal representation.

Original entry on oeis.org

1, 4, 7, 1, 2, 3, 3, 4, 6, 1, 1, 2, 2, 2, 3, 3, 3, 4, 6, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2

Offset: 1

Reinhard Zumkeller, Feb 27 2015

a(n) = A000030(A254143(n));
also initial digits of A254323: a(n) = A000030(A254323(n)).
all terms are of the form u*v mod 10, where u <= v and belonging to {1,3,4,6,7}, the distinct elements of A254397:
length of k-th run of consecutive 1s = A005993(k-2), k > 1;
length of k-th run of consecutive 2s = k*(k+1)/2 = A000217(k), k >= 1;
length of k-th run of consecutive 3s = k+1, k >= 1;
length of k-th run of consecutive 4s = A065033(k-1);
n with a(n) = 4: A237424(n) = (10^a+10^b+1)/3 with b = 0, see also A093137, A133384;
n with a(n) = 6: A237424(n) = (10^a+10^b+1)/3 with a = b; A005994(a(n)) = 6 for n > 1; see also A199682;

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A254143, A254323, A000030, A254339, A005993, A000217, A065033.

Haskell
```
a254338 = a000030 . a254143
```

listA237424(lim)=my(v=List(),a,t); while(1, for(b=0,a, t=(10^a+10^b+1)/3; if(t>lim, return(Set(v))); listput(v, t)); a++)
do(lim)=my(v=List(),u=listA237424(lim),t); for(i=1,#u, for(j=1,i, t=u[i]*u[j]; if(t>lim,break); listput(v,t))); apply(n->digits(n)[1], Set(v)) \\ Charles R Greathouse IV, May 13 2015

A349278 a(n) is the product of the sum of the last i digits of n, with i going from 1 to the total number of digits of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 0, 4, 10, 18, 28, 40, 54, 70, 88, 108, 0, 5, 12, 21, 32, 45, 60, 77, 96, 117, 0, 6, 14, 24, 36, 50, 66, 84, 104, 126, 0, 7, 16, 27, 40, 55, 72, 91, 112, 135, 0

Offset: 1

Michel Marcus, Nov 13 2021

This is similar to A349194 but with digits taken in reversed order.
The only primes in the sequence are 2, 3, 5 and 7. - Bernard Schott, Dec 04 2021
The positive terms form a subsequence of A349194. - Bernard Schott, Dec 19 2021

			For n=256, a(256) = 6*(6+5)*(6+5+2) = 858.

Michel Marcus, Table of n, a(n) for n = 1..10000

Cf. A349194, A349279 (fixed points).

Cf. A349864, A349865.

a[n_] := Times @@ Accumulate @ Reverse @ IntegerDigits[n]; Array[a, 70] (* Amiram Eldar, Nov 13 2021 *)

a(n) = my(d=Vecrev(digits(n))); prod(i=1, #d, sum(j=1, i, d[j]));

from math import prod
from itertools import accumulate
def a(n): return 0 if n%10==0 else prod(accumulate(map(int, str(n)[::-1])))
print([a(n) for n in range(1, 71)]) # Michael S. Branicky, Nov 13 2021

From Bernard Schott, Dec 04 2021: (Start)
a(n) = 0 iff n is a multiple of 10 (A008592).
a(n) = 1 iff n = 1.
a(n) = 2 (resp. 3, 4, 5, 7, 9) iff n = 10^k+1 (A000533) (resp. 2*10^k+1 (A199682), 3*10^k+1 (A199683), 4*10^k+1 (A199684), 6*10^k+1 (A199686), 8*10^k+1 (A199689)).
a(R_n) = n! where R_n = A002275(n) is repunit > 0, and n! = A000142(n).
a(n) = A349194(n) if n is palindrome (A002113). (End)

A319170 Triangular numbers of the form 2..21..1; n_times 2 followed with n_times 1; n >= 1.

Original entry on oeis.org

21, 2211, 222111, 22221111, 2222211111, 222222111111, 22222221111111, 2222222211111111, 222222222111111111, 22222222221111111111, 2222222222211111111111, 222222222222111111111111, 22222222222221111111111111, 2222222222222211111111111111, 222222222222222111111111111111, 22222222222222221111111111111111

Offset: 1

Ctibor O. Zizka, Sep 12 2018

Triangular numbers of the form (5^(2x)*2^(2x+1)-10^x-1)/9. - Harvey P. Dale, Sep 16 2019

			a(1) = A000217(6) = 21; a(2) = A000217(66) = 2211; a(3) = A000217(666) = 222111.

Colin Barker, Table of n, a(n) for n = 1..500
Jiri Sedlacek, Trojuhelnikova cisla, In: Jiří Sedláček (author): Faktoriály a kombinační čísla. (Czech). Praha: Mladá fronta, 1964. pp. 60-71.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A000217, A002280, A002283, A199682.

Select[Table[FromDigits[Join[PadRight[{},n,2],PadRight[{},n,1]]],{n,20}], OddQ[ Sqrt[8#+1]]&] (& or *) Select[Table[(5^(2x) 2^(2x+1)-10^x-1)/9,{x,20}],OddQ[Sqrt[8#+1]]&] (* Harvey P. Dale, Sep 16 2019 *)

Vec(3*x*(7 - 40*x) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^20)) \\ Colin Barker, Sep 13 2018

For n >= 1, a(n) = 2..21..1; n_times 2 followed with n_times 1.
a(n) = A000217(n_times 6), that is a(n) = A000217(A002280(n)).
a(n) = 1/9 * (2*10^n + 1) * (10^n - 1), that is a(n) = 1/9 * A199682(n) * A002283(n).
From Colin Barker, Sep 13 2018: (Start)
G.f.: 3*x*(7 - 40*x) / ((1 - x)*(1 - 10*x)*(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>3
(End)

A367593 Nonnegative integers k such that (R(k) - 1)/(k + 1) is an integer, where R(k) is the digit reversal of k.

Original entry on oeis.org

0, 1, 10, 100, 147, 1000, 1099, 1407, 10000, 14007, 100000, 140007, 1000000, 1400007, 2124736, 10000000, 14000007, 100000000, 123456789, 140000007, 1000000000, 1234506789, 1400000007, 10000000000, 12345006789, 14000000007, 21247524736, 100000000000, 123450006789

Offset: 1

Stefano Spezia, Nov 24 2023

Among the terms of this sequence, there are:
  the powers of 10 (cf. A011557);
  the numbers of the form 14*10^h + 7 = 7*A199682(h) for h > 0;
  the numbers of the form 12345*10^m + 6789 = A367650(m) for m > 3.
From Chai Wah Wu, Dec 01 2023: (Start)
First digit of positive terms must be either 1 or 2. If first digit is 1, then last digit must be 0,1,4,5,7 or 9. If first digit is 2, then last digit is 6. In particular, if a, b, c are the first digit, last digit and (R(k)-1)/(k+1) of a term k>0, then (a, b, c) must take on values from one of the following triples:
  (a, b, c): (1, 0, 0), (1, 1, 0), (1, 4, 2), (1, 4, 4), (1, 5, 5), (1, 7, 5),
  (1, 9, 4), (1, 9, 5), (1, 9, 6), (1, 9, 7), (1, 9, 8), (1, 9, 9), (2, 6, 3).
Numbers of the form 21 [2475]* 24736 are terms of this sequence, where [2475]* denote a (possibly zero) repetition of the digits 2475. The first few terms of this form are: 2124736, 21247524736, 212475247524736, ...
Similarly numbers of the form 21261 [2475]* 24738736 are terms of this sequence:
  2126124738736, 21261247524738736, 212612475247524738736, ... (End)
More generally, numbers of the form 21 [261]^k [2475]* 2473 [873]^k 6 are terms of this sequence, where [261]^k denote the digits '261' repeated k times with k>=0: e.g. 21261261247524752475247524738738736, ... It appears that all terms with first digit 2 and last digit 6 are of this form. - Chai Wah Wu, Dec 02 2023

			123456789 is a term since (987654321 - 1)/(123456789 + 1) = 8, which is an integer.

Chai Wah Wu, Table of n, a(n) for n = 1..44

Cf. A004086, A011557, A199682, A367650.

Cf. A367727, A367728, A367740.

a={}; For[k=0, k<=10^10, k++,If[IntegerQ[(FromDigits[Reverse[IntegerDigits[k]]]-1)/(k+1)],AppendTo[a,k]]]; a
Select[Range[0,10^6],IntegerQ[(IntegerReverse[#]-1)/(#+1)]&] (* The program generates the first 13 terms of the sequence. *) (* Harvey P. Dale, Aug 11 2024 *)

isok(k) = denominator((fromdigits(Vecrev(digits(k))) - 1)/(k + 1)) == 1; \\ Michel Marcus, Nov 30 2023

def digit_reversal(n):
    return int(str(n)[::-1])
def find_integers():
    result = []
    for k in range(0, 10**10):
        reversed_k = digit_reversal(k)
        if (reversed_k - 1) % (k + 1) == 0:
            result.append(k)
    return result
integers_list = find_integers()
print(integers_list)

from itertools import product, count, islice
def A367593_gen(): # generator of terms
    yield from (0,1,10)
    for l in count(1):
        m = 10**(l+1)
        for d in product('0123456789',repeat=l):
            for a, b, c in ((1, 0, 0), (1, 1, 0), (1, 4, 2), (1, 5, 5), (1, 7, 5)):
                k = a*m+int(s:=''.join(d))*10+b
                r = b*m+int(s[::-1])*10+a
                if c*(k+1)==r-1:
                    yield k
            a,b = 1,9
            k = a*m+int(s:=''.join(d))*10+b
            r = b*m+int(s[::-1])*10+a
            if not (r-1)%(k+1):
                yield k
        a,b,c=2,6,3
        for d in product('0123456789',repeat=l):
            k = a*m+int(s:=''.join(d))*10+b
            r = b*m+int(s[::-1])*10+a
            if c*(k+1)==r-1:
                yield k
A367593_list = list(islice(A367593_gen(),20)) # Chai Wah Wu, Dec 01 2023

A367728(a(n)) = 1.

A356349 Primitive Niven numbers: terms of A005349 that are not ten times another term of A005349.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 18, 21, 24, 27, 36, 42, 45, 48, 54, 63, 72, 81, 84, 102, 108, 110, 111, 112, 114, 117, 126, 132, 133, 135, 140, 144, 150, 152, 153, 156, 162, 171, 190, 192, 195, 198, 201, 204, 207, 209, 216, 220, 222, 224, 225, 228, 230, 234

Offset: 1

Bernard Schott and Rémy Sigrist, Oct 15 2022

A005349(k) belongs to this sequence iff A113315(k) is not a multiple of 10.
This sequence is infinite as it contains A133384 and A199682.
Each Niven number can be uniquely written as a(m)*10^z for some m > 0 and z >= 0.
This sequence contains numbers with k trailing zeros for any k >= 0; for example R(2^k) * 10^k (where R = A002275).

			190 is a term as 190 is a Niven number and 19 is not a Niven number.
192 is a term as 192 is a Niven number and 192 is not divisible by 10.

Index entries for sequences related to decimal expansion of n

Cf. A002275, A005349, A113315, A133384, A199682.

is(n, base=10) = my (s=sumdigits(n, base)); n%s==0 && (n%base || (n/base)%s)

def ok(n):
    sd = sum(map(int, str(n)))
    return sd and not n%sd and (n%10 or (n//10)%sd)
print([k for k in range(235) if ok(k)]) # Michael S. Branicky, Oct 16 2022

A254397 Initial digits of A237424 in decimal representation.

Original entry on oeis.org

1, 4, 7, 3, 3, 6, 3, 3, 3, 6, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 6, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 1

Reinhard Zumkeller, Feb 23 2015

a(n) = A000030(A237424(n));
n with a(n) = 3: A237424(n) = (10^a+10^b+1)/3 with 0 <= b < a, see also A014132;
n with a(n) = 6: A237424(n) = (10^a+10^b+1)/3 with a = b, see also A199682, A254338;
length of k-th run of consecutive 3s = k+1, k > 0;
digits 0, 2, 5, 8 and 9 do not occur.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A237424, A000030, A254339.

Haskell
```
a254397 = a000030 . a237424
```

A260702 Numbers n such that 3*n and n^2 have the same digit sum.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 21, 30, 33, 39, 45, 48, 51, 60, 66, 90, 96, 99, 102, 105, 111, 120, 123, 129, 132, 150, 153, 156, 159, 162, 165, 180, 189, 195, 198, 201, 210, 225, 231, 246, 252, 255, 261, 285, 300, 330, 333, 348, 351, 390, 399, 429, 450, 453, 459, 462

Offset: 1

Vincenzo Librandi, Nov 17 2015

All terms are multiple of 3.

If n is in the sequence, then so is 10*n. - Robert Israel, Apr 05 2020

			159 is in the sequence because 159^2 = 25281 and 3*159 = 477 have the same digit sum: 18.

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

Cf. A000290, A007953, A008585, A049343, A058369.

Contains A133384 and A199682.

[n: n in [0..500] | &+Intseq(3*n) eq &+Intseq(n^2)];

select(n -> convert(convert(3*n,base,10),`+`)=convert(convert(n^2,base,10),`+`), [seq(i,i=0..1000,3)]); # Robert Israel, Apr 05 2020

Select[Range[0, 500], Total[IntegerDigits[3 #]] == Total[IntegerDigits[#^2]] &]

isok(n) = sumdigits(3*n) == sumdigits(n^2); \\ Michel Marcus, Nov 17 2015

[n for n in (0..500) if sum((3*n).digits())==sum((n^2).digits())] # Bruno Berselli, Nov 17 2015

A007953(A008585(a(n))) = A007953(A000290(a(n))).

A367740 a(n) = A367727(A367593(n)).

Original entry on oeis.org

-1, 0, 0, 0, 5, 0, 9, 5, 0, 5, 0, 5, 0, 5, 3, 0, 5, 0, 8, 5, 0, 8, 5, 0, 8, 5, 3, 0, 8, 5, 0, 8, 5, 3, 0, 8, 5, 0, 8, 5, 3, 0, 8, 5

Offset: 1

Stefano Spezia, Nov 29 2023

Restricted to integers k where k + 1 divides R(k) - 1 the terms are (R(k) - 1) / (k + 1), where R(k) denotes the digit reversal of k. - Peter Luschny, Dec 01 2023

The 0's correspond to the powers of 10 (cf. A011557), the 5's correspond to the numbers of the form 14*10^h + 7 = 7*A199682(h) for h > 0, and the 8's correspond to the numbers of the form 12345*10^m + 6789 = A367650(m) for m > 3. The only negative term -1 corresponds to A367593(1) = 0. - Stefano Spezia, Dec 01 2023

Cf. A367593, A367727, A367728.

Cf. A004086, A011557, A199682, A367650.

def A367740List(upto):
    L = []
    for n in range(upto):
        rev = int(str(n)[::-1])
        if (rev - 1) % (n + 1) == 0:
            L.append((rev - 1) // (n + 1))
    return L
print(A367740List(10**6))  # Peter Luschny, Dec 01 2023

from itertools import product, count, islice
def A367740_gen(): # generator of terms
    yield from (-1,0,0)
    for l in count(1):
        m = 10**(l+1)
        for d in product('0123456789',repeat=l):
            for a, b, c in ((1, 0, 0), (1, 1, 0), (1, 4, 2), (1, 5, 5), (1, 7, 5)):
                k = a*m+int(s:=''.join(d))*10+b
                r = b*m+int(s[::-1])*10+a
                if c*(k+1)==r-1:
                    yield c
            a,b = 1,9
            k = a*m+int(s:=''.join(d))*10+b
            r = b*m+int(s[::-1])*10+a
            p,q = divmod(r-1,k+1)
            if not q:
                yield p
        a,b,c=2,6,3
        for d in product('0123456789',repeat=l):
            k = a*m+int(s:=''.join(d))*10+b
            r = b*m+int(s[::-1])*10+a
            if c*(k+1)==r-1:
                yield c
A367740_list = list(islice(A367740_gen(),20)) # Chai Wah Wu, Dec 01 2023

-1 <= a(n) <= 9.

a(30)-a(44) from Chai Wah Wu, Dec 02 2023

cp's OEIS Frontend

A067275 Number of Fibonacci numbers A000045(k), k <= 10^n, which end in 4.

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A254338 Initial digits of A254143 in decimal representation.

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A349278 a(n) is the product of the sum of the last i digits of n, with i going from 1 to the total number of digits of n.

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A319170 Triangular numbers of the form 2..21..1; n_times 2 followed with n_times 1; n >= 1.

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A367593 Nonnegative integers k such that (R(k) - 1)/(k + 1) is an integer, where R(k) is the digit reversal of k.

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A356349 Primitive Niven numbers: terms of A005349 that are not ten times another term of A005349.

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A254397 Initial digits of A237424 in decimal representation.

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