A074992 -id:A074992 - cp's OEIS frontend

A036377 Floor[concatenation of seven consecutive numbers from n to n+6 divided by 7].

17636, 176366, 335096, 493827, 6525558, 81127287, 969871587, 11272873030, 127287303044, 1300158875916, 1444459020216, 1588759164516, 1733059308816, 1877359453117, 2021659597417, 2165959741717, 2310259886017, 2454560030317

Offset: 0

Amarnath Murthy, Aug 31 2002

Cf. A074991, A074992, A074993, A074994, A074995, A074996, A073086, A074999.

Floor[#/7]&/@(FromDigits[Flatten[IntegerDigits/@#]]&/@Partition[Range[0,25],7,1])  (* Harvey P. Dale, Feb 03 2011 *)

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 23 2003

A075001 Smallest k such that the concatenation of n consecutive numbers starting with k (from k to n+k-1) is a multiple of n; or 0 if no such number exists.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 9, 1, 4, 7, 1, 5, 23, 1, 14, 1, 9, 9, 13, 5, 1, 21, 1, 13, 12, 1, 36, 21, 9, 3, 41, 1, 34, 33, 9, 21, 12, 9, 33, 9, 1, 13, 28, 5, 48, 1, 23, 21, 3, 1, 11, 13, 14, 41, 28, 1, 114, 115, 9, 41, 21, 9, 23, 69, 1, 61, 73, 5, 14, 43, 1, 145, 13, 9, 127, 41, 9, 95

Offset: 1

Amarnath Murthy, Aug 31 2002

Conjecture: For every n there exists a k.
First occurrence of k where a(n)=k: 1, 103, 4, 13, 8, 105, 14, 87, 11, 699, 55, 29, 23, 19, 114, 261, 102, 97, 178, 219, 26, 121, 17, 151, 92, ..., . - Robert G. Wilson v
a(n)=1 iff n is in A029455. - Robert G. Wilson v
Increasing a(n)'s: 1, 3, 5, 9, 23, 36, 41, 48, 114, 115, 145, 166, 175, 221, 251, ..., at n = 1, 4, 8, 11, 17, 31, 35, 49, 61, 62, 76, 85, 122, 133, 170, 179, 217, 229, ..., . - Robert G. Wilson v

			a(11) = 9 as 910111213141516171819 the concatenation of 11 numbers from 9 to 19 is divisible by 11 (11*82737383012865106529).

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 2500 terms from Robert G. Wilson v)

Cf. A058183, A074991, A074992, A074993, A074994, A074995, A074996, A036377, A073086, A074999, A075000, A077306, A075000.

Cf. A029455 (indices of 1's).

f[n_] := Block[{c = 1, id = Range@n}, While[k = FromDigits@Flatten@IntegerDigits@id/n; ! IntegerQ@k, id++; c++ ]; c]; Array[f, 82] (* Robert G. Wilson v, Oct 20 2007 *)

/* The following program assumes the conjecture is true. */ /* It has found nonzero a(n) for n up to 500. */ {for(n=1,500, k=0; until(s%n==0,k++; s=0; for(m=k,k+n-1, s=s*(10^length(Str(m)))+m)); print1(k,","))}

a(n) = {my(ld = 1, hd = n, qd, m = Mod(1, n), pow10, qdn = #digits(n), t=log(10*n+.5)\log(10)); qd = n*t+t-10^t\9; pow10 = Mod(10, n)^(qd-1); for(i = 2, n, m = m * Mod(10, n)^#digits(i) + i; ); while(1, if(lift(m) == 0, return(ld)); m -= ld * pow10; hd++; m = m * Mod(10, n)^#digits(hd) + hd; ld++; pow10*=10^(#digits(hd) - #digits(ld)); ) } \\ David A. Corneth, Aug 23 2020

More terms from Rick L. Shepherd, Sep 03 2002

A181718 a(n) = (1/9)(10^(2n) + 10^n - 2).

Original entry on oeis.org

0, 12, 1122, 111222, 11112222, 1111122222, 111111222222, 11111112222222, 1111111122222222, 111111111222222222, 11111111112222222222, 1111111111122222222222, 111111111111222222222222, 11111111111112222222222222, 1111111111111122222222222222

Offset: 0

Paul Curtz, Nov 17 2010

In decimal, n times 1 followed by n times 2.

a(n) = 3 + 3*3, 33 + 33*33, 333 + 333*333, written with 3,6,9,12,... = A008585(n+1) 3's.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A002277, A008585, A074992, A093137.

[(1/9)*(10^(2*n) + 10^n - 2): n in [0..20]]; // Vincenzo Librandi, Aug 04 2011

Table[FromDigits[Join[PadRight[{},n,1],PadRight[{},n,2]]],{n,0,20}] (* or *) LinearRecurrence[{111,-1110,1000},{0,12,1122},20] (* Harvey P. Dale, Jul 31 2013 *)

vector(30, n, n--; (10^(2*n) + 10^n - 2)/9) \\ G. C. Greubel, Nov 02 2018

for n in range(30):
    print((10**(2*n)+10**n-2)//9, end=', ')
# Stefano Spezia, Nov 02 2018

[(100^n +10^n -2)//9 for n in range(31)] # G. C. Greubel, Mar 25 2024

a(n) = A002277(n)*A093137(n).
G.f.: 6*x*(2-35*x) / ( (1-x)*(1-10*x)*(1-100*x) ). - R. J. Mathar, Feb 28 2011
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3), a(0)=0, a(1)=12, a(2)=1122. - Harvey P. Dale, Jul 31 2013
a(n) = (A074992(n) - 1)/3. - Michel Marcus, Sep 14 2013
E.g.f.: (1/9)*(-2*exp(x) + exp(10*x) + exp(100*x)). - G. C. Greubel, Mar 25 2024

A243162 Numbers n such that n^2 divides n.n.n where dot "." means concatenation.

Original entry on oeis.org

1, 3, 13, 21, 37, 39, 91, 1443, 3367, 9901, 157737, 333667, 999001, 3075403, 9226209, 14287143, 33336667, 99990001, 1171182883, 1224848037, 1286294191, 1397863441, 1428557143, 1469179621, 1535254357, 1568996211, 1753536967, 1792076241, 1839599913, 1891910811

Offset: 1

Hans Havermann, May 31 2014

Number of d-digit solutions for d = 1..100: 2, 5, 0, 3, 0, 3, 2, 3, 0, 39, 0, 2, 0, 106, 0, 3, 3, 2, 0, 441, 4, 14, 0, 5, 0, 15, 2, 283, 0, 23, 0, 61, 0, 24, 21, 4, 0, 22, 0, 240, 0, 34, 0, 96, 3, 30, 0, 6, 16, 281, 0, 216, 0, 22, 5, 3894, 2, 10, 0, 149, 2, 11, 0, 407, 0, 25, 0, 2136, 0, 53983, 0, 12, 1, 29, 11, 1872, 99, 20, 0, 6984, 0, 45, 0, 279, 32, 10, 5, 15928, 0, 213, 24, 791, 0, 20, 14, 44, 0, 713, 12, 89804.

Numbers n such that n divides 100^d+10^d+1, where 10^(d-1)<=n<10^d. - Robert Israel, Jan 11 2017

			21^2 divides 212121; 91^2 divides 919191; so both 21 and 91 are in the sequence.

Hans Havermann, Table of n, a(n) for n = 1..1366

Cf. A147553 (n^2 divides n.n), A147554 (primes in this sequence).

Contains A074992 and A168624.

Res:= {}:
for d from 1 to 15 do
  Res:= Res union select(t -> t >= 10^(d-1) and t < 10^d,
   numtheory:-divisors(100^d+10^d+1))
od:
sort(convert(Res,list)); # Robert Israel, Jan 11 2017

Do[d=Divisors[100^i+10^i+1];s=Select[d,Length[IntegerDigits[#]]==i&];If[Length[s]>0,Do[Print[s[[j]]],{j,Length[s]}]],{i,42}]

A304290 Numbers k such that k-1 is a substring of k^2.

Original entry on oeis.org

9, 37, 99, 370, 999, 3367, 9999, 22186, 99999, 221860, 333667, 625001, 625009, 859415, 926968, 999999, 1507152, 3125001, 3701562, 7012141, 9375009, 9999999, 20506249, 28658098, 33336667, 46875009, 78125001, 79632152, 86609391, 98089448, 99999999, 306481073

Offset: 1

Paolo P. Lava, May 24 2018

The repdigit sequence A002283, apart from the first term 0, is a subset.
In fact (999...9)^2 = (10^n-1)^2 = 10^n((10^n-1)-1)+1 = 10^n(999...9-1)+1 = 10^n(999...8)+1 = 999...8000...1.
The sequence A074992, apart from the first term 1, is a subsequence. - Michel Marcus, May 27 2018

			9^2 = 81 and 9-1 = 8 is a substring.
37^2 = 1369 and 37-1 = 36 is a substring.

Chai Wah Wu, Table of n, a(n) for n = 1..65 (n = 1..41 from Jon E. Schoenfield)

Cf. A002283, A074992, A282384.

P:=proc(q) local a,b,k,n; a:=2; b:=1;
for n from 1 to q do for k from 1 to ilog10(a^2)-ilog10(b)+1 do
if b=trunc(a^2/10^(k-1)) mod 10^(ilog10(b)+1) then print(a); fi; od;
b:=a; a:=a+1; od; print(); end: P(10^8);

Select[Range[10^6], SequenceCount[IntegerDigits[#^2], IntegerDigits[# - 1]] > 0 &] (* Michael De Vlieger, May 27 2018 *)

A304290_list = [k for k in range(10**6) if str(k-1) in str(k**2)] # Chai Wah Wu, Oct 22 2018

a(32) from Jon E. Schoenfield, Jun 01 2018

cp's OEIS Frontend

A036377 Floor[concatenation of seven consecutive numbers from n to n+6 divided by 7].

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A075001 Smallest k such that the concatenation of n consecutive numbers starting with k (from k to n+k-1) is a multiple of n; or 0 if no such number exists.

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A181718 a(n) = (1/9)*(10^(2*n) + 10^n - 2).

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A243162 Numbers n such that n^2 divides n.n.n where dot "." means concatenation.

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A304290 Numbers k such that k-1 is a substring of k^2.

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Maple

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Python

Extensions

A181718 a(n) = (1/9)(10^(2n) + 10^n - 2).