A109344 -id:A109344 - 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

A102807 a(n) is the square of one plus the number consisting of n 3's.

Original entry on oeis.org

1, 16, 1156, 111556, 11115556, 1111155556, 111111555556, 11111115555556, 1111111155555556, 111111111555555556, 11111111115555555556, 1111111111155555555556, 111111111111555555555556, 11111111111115555555555556, 1111111111111155555555555556, 111111111111111555555555555556

Offset: 0

Ron Knott, Feb 27 2005

Old name was: The number (333...334)^2.
An infinite sequence of squares with no zeros in base 10.
a(n) = A104265(2n) for n > 0. - Chai Wah Wu, Mar 24 2020

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 31 at p. 61.
Italo Ghersi, Matematica dilettevole e curiosa, pp. 111-112, Hoepli, Milano, 1967. [Vincenzo Librandi, Dec 31 2008]

Seiichi Manyama, Table of n, a(n) for n = 0..500
Emile Fourrey, Récréations arithmétiques, Vuibert, 1899 and after, Paris, pages 72-73.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A052041, A093137, A075415, A109344, A104265.

a:= n-> (1+parse(cat(0, 3$n)))^2:
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 03 2018

Table[(10^n + 2)^2/9, {n, 0, 20}] (* Paolo Xausa, Jun 26 2024 *)

From R. J. Mathar, Jan 06 2009: (Start)
a(n) = (100^n + 4*10^n + 4)/9.
G.f.: (1 - 95*x + 490*x^2)/((1-x)*(100*x-1)*(10*x-1)). (End)
E.g.f.: exp(x)*(4 + 4*exp(9*x) + exp(99*x))/9. - Stefano Spezia, Jul 31 2024

New name from Alois P. Heinz, Sep 03 2018

A177769 a(n) = 111*n.

Original entry on oeis.org

111, 222, 333, 444, 555, 666, 777, 888, 999, 1110, 1221, 1332, 1443, 1554, 1665, 1776, 1887, 1998, 2109, 2220, 2331, 2442, 2553, 2664, 2775, 2886, 2997, 3108, 3219, 3330, 3441, 3552, 3663, 3774, 3885, 3996, 4107, 4218, 4329, 4440, 4551, 4662, 4773, 4884, 4995, 5106

Offset: 1

Paul Curtz, May 13 2010

The reference contains also sequences A102807, A109344, A075415, and A109492.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Émile Fourrey, Le nombre 37, Récreations arithmétiques, 1899 and later, Vuibert, Paris, page 11.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A075415, A102807, A109344, A109492.

[111*n: n in [1..50]]; // Vincenzo Librandi, Aug 04 2011

G.f.: 111*x/(x-1)^2.
a(n) = 2*a(n-1) - a(n-2).
a(n) = a(n-1) + 111.
E.g.f.: 111*x*exp(x). - Stefano Spezia, Sep 15 2023

A181719 a(n) = A133473(n+1)^2.

Original entry on oeis.org

25, 1225, 112225, 11122225, 1111222225, 111112222225, 11111122222225, 1111111222222225, 111111112222222225, 11111111122222222225, 1111111111222222222225, 111111111112222222222225, 11111111111122222222222225, 1111111111111222222222222225, 111111111111112222222222222225

Offset: 1

Paul Curtz, Nov 17 2010

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A109344, A133473, A181718.

[(100^n+10*10^n+25)/9: n in [1..20]]; // Vincenzo Librandi, Jun 02 2011

(5+10^Range[30])^2/9 (* G. C. Greubel, Mar 25 2024 *)
LinearRecurrence[{111,-1110,1000},{25,1225,112225},20] (* Harvey P. Dale, Feb 22 2025 *)

a(n)=(100^n+10*10^n+25)/9 \\ Charles R Greathouse IV, Jun 01 2011

Vec(5*x*(1 - 62*x + 160*x^2) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^17)) \\ Colin Barker, Aug 21 2019

[(5+10^n)^2//9 for n in range(1,31)] # G. C. Greubel, Mar 25 2024

a(n) = 100 * A181718(n-1) + 25.
a(n) = 25 * A109344(n-1), for n > 1.
From Colin Barker, Aug 21 2019: (Start)
G.f.: x*(1 - 62*x + 160*x^2) / ((1 - x)*(1 - 10*x)*(1 - 100*x)).
a(n) = (5 + 10^n)^2 / 9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>3. (End)
E.g.f.: (1/9)*exp(x)*(25 + 10*exp(9*x) + exp(99*x)). - Stefano Spezia, Aug 21 2019 after Colin Barker

Formulas edited by Eric M. Schmidt, Oct 29 2012

A309827 a(n) is the square of the number consisting of one 1 and n 6's: (166...6)^2.

Original entry on oeis.org

1, 256, 27556, 2775556, 277755556, 27777555556, 2777775555556, 277777755555556, 27777777555555556, 2777777775555555556, 277777777755555555556, 27777777777555555555556, 2777777777775555555555556, 277777777777755555555555556, 27777777777777555555555555556

Offset: 0

Seiichi Manyama, Aug 23 2019

All terms are zeroless (element of A052382).

Seiichi Manyama, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

((k*10^n+3-k)/3)^2: A102807 (k=1), A109344 (k=2), A098608 (k=3), A309907 (k=4), this sequence (k=5).

Cf. A052382, A246057.

LinearRecurrence[{111,-1110,1000},{1,256,27556},20] (* Harvey P. Dale, Dec 13 2021 *)

PARI
```
{a(n) = ((5*10^n-2)/3)^2}
```

N=20; x='x+O('x^N); Vec((1+145*x+250*x^2)/((1-x)*(1-10*x)*(1-100*x)))

a(n) = A246057(n)^2 = ((5*10^n-2)/3)^2.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
G.f.: (1+145*x+250*x^2)/((1-x)*(1-10*x)*(1-100*x)).

A309828 Squares formed by concatenating k and 2*k+1.

Original entry on oeis.org

25, 49, 1225, 4489, 112225, 444889, 11122225, 44448889, 816416329, 1111222225, 1451229025, 3832476649, 4444488889, 111112222225, 444444888889, 10185602037121, 11111122222225, 44444448888889, 46355849271169, 997230019944601, 1111111222222225, 1231148024622961

Offset: 1

Marius A. Burtea, Aug 18 2019

The sequence is infinite. The squares of the form 66...67^2 = 4..48..89 are terms.

Another infinite family is the squares 33...35^2 = 1...122...25. - Robert Israel, Aug 20 2019

			5^2 = 25 = 2_(2 * 2 + 1);
7^2 = 49 = 4_(2 * 4 + 1);
35^2 = 1225 = 12_(2 * 12 + 1);
61907^2 = 3832476649 = 38324_(2 * 38324 + 1).

Ion Cucurezeanu, Perfect squares and cubes of integers, Ed. Gil, Zalău, (2007), ch. 4, p. 25, pr. 211, 212 (in Romanian).

Chai Wah Wu, Table of n, a(n) for n = 1..286

Cf. A000290, A030466, A054215, A109344, A181719, A309808, A309809.

[a:n in [1..30000000]|IsSquare(a) where a is 10^(#Intseq(2*n+1))*n+2*n+1];

F:= proc(m) local x,X,A;
  X:= [numtheory:-rootsunity(2,10^m+2)];
  A:= map(x -> (x^2-1)/(10^m+2), X);
  A:= sort(select(x -> 2*x+1>=10^(m-1) and 2*x+1<10^m, A));
  op(map(x -> x*10^m+2*x+1, A))
end proc:
subsop(1=NULL, [seq(F(m),m=1..10)]); # Robert Israel, Aug 20 2019

Select[Array[FromDigits@ Flatten@ IntegerDigits[{#, 2 # + 1}] &, 10^5],
IntegerQ@ Sqrt@ # &] (* Michael De Vlieger, Aug 19 2019 *)

def Test(n):
    s = str(n)
    ps, ss = s[0:len(s)//2], s[len(s)//2:len(s)]
    return int(ss) == 2*int(ps)+1 and s[len(s)//2] != "0"
n, a = 1, 4
while n < 23:
    if Test(a*a):
        print(n,a*a)
        n = n+1
    a = a+1 # A.H.M. Smeets, Aug 19 2019

from itertools import count, islice
from sympy.ntheory.primetest import is_square
def A309828_gen(): # generator of terms
    return filter(is_square,(int(str(k)+str((k<<1)+1)) for k in count(1)))
A309828_list = list(islice(A309828_gen(),20)) # Chai Wah Wu, Feb 20 2023

cp's OEIS Frontend

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

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A102807 a(n) is the square of one plus the number consisting of n 3's.

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A177769 a(n) = 111*n.

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A181719 a(n) = A133473(n+1)^2.

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A309827 a(n) is the square of the number consisting of one 1 and n 6's: (166...6)^2.

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A309828 Squares formed by concatenating k and 2*k+1.

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