A017377 -id:A017377 - cp's OEIS frontend

A348055 Number of positive integers with n digits that are the product of two integers ending with 7.

0, 1, 20, 255, 3064, 34743, 380939, 4089499, 43282317, 453472867, 4715695283, 48760330737, 501941505404, 5148657883067, 52659616820819

Offset: 1

Stefano Spezia, Sep 26 2021

a(n) is the number of n-digit numbers in A348054.

Cf. A017377, A052268, A348054.

Cf. A346509 (ending with 1), A346629 (ending with 2), A346952 (ending with 3), A347255 (ending with 4), A337855 (ending with 5), A337856 (ending with 6), A348549 (ending with 8).

def a(n):
  lo, hi = 10**(n-1), 10**n
  return len(set(a*b for a in range(7, hi//7+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Sep 26 2021

a(n) < A052268(n).

Conjecture: lim_{n->infinity} a(n)/a(n-1) = 10.

a(9)-a(11) from Michael S. Branicky, Sep 26 2021

a(12)-a(15) from Martin Ehrenstein, Oct 25 2021

A348487 Positive numbers whose square starts and ends with exactly one 1.

Original entry on oeis.org

1, 11, 39, 41, 101, 111, 119, 121, 129, 131, 139, 141, 319, 321, 329, 331, 349, 351, 359, 361, 369, 371, 379, 381, 389, 391, 399, 401, 409, 411, 419, 421, 429, 431, 439, 441, 1001, 1009, 1011, 1019, 1021, 1029, 1031, 1039, 1041, 1099, 1101, 1109, 1111, 1119, 1121, 1129, 1131, 1139

Offset: 1

Bernard Schott, Oct 21 2021

When a square ends with 1, this square ends with exactly one 1.

Sequences A000533 and A253213 show that there are an infinity of terms. The square of their terms, for n >= 3, starts and ends with exactly one 1. Also, the numbers 119, 1119, 11119, ..., ((10^k + 71) / 9)^2, (k >= 3) are terms. The squares ((10^k + 71) / 9)^2, have the last digit 1 and because 12*10^(2*k - 3) < ((10^k + 71) / 9)^2 <13*10^(2*k - 3), for k >= 3, the squares ((10^k + 71) / 9)^2, k >= 4, start with 12. - Marius A. Burtea, Oct 21 2021

			39 is a term since 39^2 = 1521.
109 is not a term since 109^2 = 11881.
119 is a term since 119^2 = 14161.

Cf. A045855, A090771, A253213, A273372 (squares ending with 1), A017281, A017377.
Cf. A000533, A253213 for n >= 2 (subsequences).
Subsequence of A305719.

[1] cat [n:n in [2..1200]|Intseq(n*n)[1] eq 1 and Intseq(n*n)[#Intseq(n*n)] eq 1 and Intseq(n*n)[-1+#Intseq(n*n)] ne 1]; // Marius A. Burtea, Oct 21 2021

Join[{1}, Select[Range[11, 1200], (d = IntegerDigits[#^2])[[1]] == d[[-1]] == 1 && d[[2]] != 1 &]] (* Amiram Eldar, Oct 21 2021 *)

isok(k) = my(d=digits(sqr(k))); (d[1]==1) && (d[#d]==1) && if (#d>2, (d[2]!=1) && (d[#d-1]!=1), 1); \\ Michel Marcus, Oct 21 2021

from itertools import count, takewhile
def ok(n):
  s = str(n*n); return len(s.rstrip("1")) == len(s.lstrip("1")) == len(s)-1
def aupto(N):
  r = takewhile(lambda x: x<=N, (10*i+d for i in count(0) for d in [1, 9]))
  return [k for k in r if ok(k)]
print(aupto(1140)) # Michael S. Branicky, Oct 21 2021

A017378 a(n) = (10*n + 9)^2.

Original entry on oeis.org

81, 361, 841, 1521, 2401, 3481, 4761, 6241, 7921, 9801, 11881, 14161, 16641, 19321, 22201, 25281, 28561, 32041, 35721, 39601, 43681, 47961, 52441, 57121, 62001, 67081, 72361, 77841, 83521, 89401, 95481, 101761, 108241, 114921, 121801, 128881, 136161, 143641

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A017377.

[(10*n+9)^2: n in [0..40]]; // Vincenzo Librandi, Aug 31 2011

A017378:=n->(10*n+9)^2: seq(A017378(n), n=0..50); # Wesley Ivan Hurt, Mar 30 2017

Table[(10 n + 9)^2, {n, 0, 37}] (* or *)
LinearRecurrence[{3, -3, 1}, {81, 361, 841}, 38] (* or *)
CoefficientList[Series[(81 + 118 x + x^2)/(1 - x)^3, {x, 0, 37}], x] (* Michael De Vlieger, Mar 30 2017 *)

a(n) = (10*n+9)^2; \\ Michel Marcus, Aug 26 2015

Vec((81 + 118*x + x^2) / (1 - x)^3 + O(x^40)) \\ Colin Barker, Mar 30 2017

From Colin Barker, Mar 30 2017: (Start)
G.f.: (81 + 118*x + x^2) / (1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2.
(End)

A262389 Numbers whose last digit is composite.

Original entry on oeis.org

4, 6, 8, 9, 14, 16, 18, 19, 24, 26, 28, 29, 34, 36, 38, 39, 44, 46, 48, 49, 54, 56, 58, 59, 64, 66, 68, 69, 74, 76, 78, 79, 84, 86, 88, 89, 94, 96, 98, 99, 104, 106, 108, 109, 114, 116, 118, 119, 124, 126, 128, 129, 134, 136, 138, 139, 144, 146, 148, 149

Offset: 1

Wesley Ivan Hurt, Sep 21 2015

Numbers ending in 4, 6, 8 or 9.
Union of A017317, A017341, A017365 and A017377.
Subsequence of A118951 (numbers containing at least one composite digit).
Complement of (A197652 Union A260181).

Gerald Hillier and Didier Lachieze, Last Digit Composite.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A017317, A017341, A017365, A017377.

Cf. A118951, A197652, A260181 (last digit is prime).

[(5*n+1-(-1)^n+(3+(-1)^n)*(-1)^((2*n-3-(-1)^n) div 4) div 2) div 2: n in [1..70]]; // Vincenzo Librandi, Sep 21 2015

A262389:=n->(5*n+1-(-1)^n+(3+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4)/2)/2: seq(A262389(n), n=1..100);

Table[(5n+1-(-1)^n+(3+(-1)^n)*(-1)^((2n-3-(-1)^n)/4)/2)/2, {n, 100}]
LinearRecurrence[{1, 0, 0, 1, -1}, {4, 6, 8, 9, 14}, 80] (* Vincenzo Librandi, Sep 21 2015 *)
CoefficientList[Series[(4 + 2*x + 2*x^2 + x^3 + x^4)/((x - 1)^2*(1 + x + x^2 + x^3)), {x, 0, 80}], x] (* Wesley Ivan Hurt, Sep 21 2015 *)
Select[Range[200],CompositeQ[Mod[#,10]]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 21 2019 *)

G.f.: x*(4+2*x+2*x^2+x^3+x^4)/((x-1)^2*(1+x+x^2+x^3)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (5*n+1-(-1)^n+(3+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4)/2)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(10-2*sqrt(5))*Pi - sqrt(5)*arccoth(3/sqrt(5)) - 4*log(2))/20. - Amiram Eldar, Jul 30 2024

Name edited by Jon E. Schoenfield, Feb 15 2018

A273372 Squares ending in digit 1.

Original entry on oeis.org

1, 81, 121, 361, 441, 841, 961, 1521, 1681, 2401, 2601, 3481, 3721, 4761, 5041, 6241, 6561, 7921, 8281, 9801, 10201, 11881, 12321, 14161, 14641, 16641, 17161, 19321, 19881, 22201, 22801, 25281, 25921, 28561, 29241, 32041, 32761, 35721, 36481, 39601, 40401

Offset: 1

Vincenzo Librandi, May 21 2016

Intersection of A000290 and A017281; also, union of A017282 and A017378. The square roots are in A017281 or in A017377 (numbers ending in 1 or 9, respectively). - David A. Corneth, May 22 2016

Seiichi Manyama, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000290, A090771, A132356.
Cf. A017281 (numbers ending in 1), A017283 (cubes ending in 1).
Cf. similar sequences listed in A273373.

/* By definition: */ [n^2: n in [0..200] | Modexp(n,2,10) eq 1];

[5*(10*n+(-1)^n-5)*(2*n+(-1)^n-1)/4+1: n in [1..50]];

Table[5 (10 n + (-1)^n - 5) (2 n + (-1)^n - 1)/4 + 1, {n, 1, 50}]

A273372_list = [(10*n+m)**2 for n in range(10**3) for m in (1,9)] # Chai Wah Wu, May 24 2016

p (1..(n + 1) / 2).inject([]){|s, i| s + [(10 * i - 9) ** 2, (10 * i - 1) ** 2]}[0..n - 1] # Seiichi Manyama, May 24 2016

G.f.: x*(1 + 80*x + 38*x^2 + 80*x^3 + x^4) / ((1 + x)^2*(1 - x)^3).
a(n) = 10*A132356(n-1) + 1 = 5*(10*n+(-1)^n-5)*(2*n+(-1)^n-1)/4+1.
a(n) = (5*n - 5/2 + (3/2)*(-1)^n)^2 = 25*n^2 - 25*n + 17/2 + 15*n*(-1)^n - (15/2)*(-1)^n. - David A. Corneth, May 21 2016
a(n) = A090771(n)^2. - Michel Marcus, May 25 2016
Sum_{n>=1} 1/a(n) = Pi^2*(3+sqrt(5))/50. - Amiram Eldar, Feb 16 2023

Edited by Bruno Berselli, May 24 2016

A322489 Numbers k such that k^k ends with 4.

Original entry on oeis.org

2, 18, 22, 38, 42, 58, 62, 78, 82, 98, 102, 118, 122, 138, 142, 158, 162, 178, 182, 198, 202, 218, 222, 238, 242, 258, 262, 278, 282, 298, 302, 318, 322, 338, 342, 358, 362, 378, 382, 398, 402, 418, 422, 438, 442, 458, 462, 478, 482, 498, 502, 518, 522, 538, 542, 558

Offset: 1

Bruno Berselli, Dec 12 2018

Also numbers k == 2 (mod 4) such that 2^k and k^2 end with the same digit.

Numbers congruent to {2, 18} mod 20. - Amiram Eldar, Feb 27 2023

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004526, A056849.
Subsequence of A139544, A235700.
Numbers k such that k^k ends with d: A008592 (d=0), A017281 (d=1), A067870 (d=3), this sequence (d=4), A017329 (d=5), A271346 (d=6), A322490 (d=7), A017377 (d=9).

GAP
```
List([1..70], n -> 10*n+3*(-1)^n-5);
```

[10*n+3*(-1)^n-5 for n in 1:70] |> println

Magma
```
[10*n+3*(-1)^n-5: n in [1..70]];
```

select(n->n^n mod 10=4,[$1..558]); # Paolo P. Lava, Dec 18 2018

Mathematica
```
Table[10 n + 3 (-1)^n - 5, {n, 1, 60}]
```
Maxima
```
makelist(10*n+3*(-1)^n-5, n, 1, 70);
```

apply(A322489(n)=10*n+3*(-1)^n-5, [1..70]) \\ M. F. Hasler, Dec 14 2018

Vec(2*x*(1 + 8*x + x^2) / ((1 - x)^2*(1 + x)) + O(x^70)) \\ Colin Barker, Dec 13 2018

[10*n+3*(-1)**n-5 for n in range(1, 70)]

Sage
```
[10*n+3*(-1)^n-5 for n in (1..70)]
```

O.g.f.: 2*x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).
E.g.f.: 2 + 3*exp(-x) + 5*(2*x - 1)*exp(x).
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 10*n + 3*(-1)^n - 5. Therefore:
a(n) = 10*n - 8 for odd n;
a(n) = 10*n - 2 for even n.
a(n+2*k) = a(n) + 20*k.
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(2*Pi/5)*Pi/20 = sqrt(5+2*sqrt(5))*Pi/20. - Amiram Eldar, Feb 27 2023

A346951 Positive integers k such that 10*k+9 is equal to the product of two integers ending with 3 (A346950).

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 16, 18, 21, 24, 27, 29, 30, 33, 36, 39, 42, 45, 48, 51, 52, 54, 55, 57, 60, 63, 66, 68, 69, 72, 75, 78, 81, 84, 87, 90, 93, 94, 96, 98, 99, 102, 105, 107, 108, 111, 114, 117, 120, 121, 123, 126, 129, 132, 133, 135, 138, 141, 144, 146, 147, 150

Offset: 1

Stefano Spezia, Aug 08 2021

			15 is a term because 3*53 = 159 = 15*10 + 9.

Cf. A016873 (ending with 5), A017377, A324298 (ending with 6), A346950, A346952, A346953.

a={}; For[n=0, n<=150, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+3]==0 && Mod[(10*n+9)/(10*k+3), 10]==3&& 10*n+9>Max[10a+9], AppendTo[a, n]]]]; a

def aupto(lim): return sorted(set(a*b//10 for a in range(3, 10*lim//3+4, 10) for b in range(a, 10*lim//a+4, 10) if a*b//10 <= lim))
print(aupto(150)) # Michael S. Branicky, Aug 11 2021

a(n) = (A346950(n) - 9)/10.

Lim_{n->infinity} a(n)/a(n-1) = 1.

A346953 a(n) is the number of divisors of A346950(n) ending with 3.

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 4, 2, 2, 4, 2, 2

Offset: 1

Stefano Spezia, Aug 08 2021

a(n) = 1 if A346950(n) = k^2 where k is either a prime ending with 3 or the product of a prime ending with 7 and a prime ending with 9. - Robert Israel, Nov 03 2024

			a(17) = 4 since there are 4 divisors of A346950(17) = 429 ending with 3: 3, 13, 33 and 143.

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

Cf. A000005, A017377, A346388 (ending with 5), A346389 (ending with 6), A346950, A346951, A346952.

N:= 10000: # for a(1) .. a(M) where the last term of A346950 less than N is A346950(M)
S:= {}:
for n from 3 to floor(sqrt(N)) by 10 do
  S:= S union map(`*`, {seq(i,i= n .. floor(N/n), 10)},n)
od:
S:= sort(convert(S,list)):
map(t -> nops(select(t -> t mod 10 = 3, numtheory:-divisors(t))), S); # Robert Israel, Nov 03 2024

b={}; For[n=0, n<=450, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+3]==0 && Mod[(10*n+9)/(10*k+3), 10]==3 && 10*n+9>Max[b], AppendTo[b, 10*n+9]]]]; (* A346950 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#, 10]==3&)]]]]]; a

from sympy import divisors
def f(n): return sum(d%10 == 3 for d in divisors(n)[1:-1])
def A346950upto(lim): return sorted(set(a*b for a in range(3, lim//3+1, 10) for b in range(a, lim//a+1, 10)))
print(list(map(f, A346950upto(2129)))) # Michael S. Branicky, Aug 11 2021

A017379 a(n) = (10*n + 9)^3.

Original entry on oeis.org

729, 6859, 24389, 59319, 117649, 205379, 328509, 493039, 704969, 970299, 1295029, 1685159, 2146689, 2685619, 3307949, 4019679, 4826809, 5735339, 6751269, 7880599, 9129329, 10503459, 12008989

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Cf. A017377 (10n+9), A000578 (n^3).

[(10*n+9)^3: n in [0..40]]; // Vincenzo Librandi, Aug 31 2011

(10Range[0,30]+9)^3 (* or *) LinearRecurrence[{4,-6,4,-1},{729,6859,24389,59319},30] (* Harvey P. Dale, Nov 01 2011 *)

vector(40, n, n--; (10*n + 9)^3) \\ G. C. Greubel, Nov 10 2018

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=729, a(1)=6859, a(2)=24389, a(3)=59319. - Harvey P. Dale, Nov 01 2011

A053181 Composite numbers ending in 9.

Original entry on oeis.org

9, 39, 49, 69, 99, 119, 129, 159, 169, 189, 209, 219, 249, 259, 279, 289, 299, 309, 319, 329, 339, 369, 399, 429, 459, 469, 489, 519, 529, 539, 549, 559, 579, 589, 609, 629, 639, 649, 669, 679, 689, 699, 729, 749, 759, 779, 789, 799, 819, 849, 869, 879, 889, 899

Offset: 1

Enoch Haga, Feb 29 2000

Numbers of the form 10n + 9 (A017377) not in A030433. Obviously multiples of 2 or 5 can't be in A017377 nor this sequence. All other primes occur as factors of these terms infinitely often. For example: If n is a multiple of 3, then 10n + 9 is in this sequence; if n = 4 mod 7, then 10n + 9 is in this sequence; if n = 9 mod 11, then 10n + 9 is in this sequence; and so on and so forth. The first of these congruences to be satisfied determines the least prime factor of 10n + 9. - Alonso del Arte, Jun 23 2011

G. C. Greubel, Table of n, a(n) for n = 1..10000

Complement[Range[9, 1000, 10], Prime[Range[PrimePi[1000]]]] (* Alonso del Arte, Jun 23 2011 *)
Select[Range[1000],CompositeQ[#]&&Mod[#,10]==9&] (* Harvey P. Dale, May 01 2016 *)

forstep(n=9,1e3,10,if(!isprime(n),print1(n", "))) \\ Charles R Greathouse IV, Jun 24 2011

a(n) ~ 10n. - Charles R Greathouse IV, Jun 24 2011

9 added by Kausthub Gudipati, Jun 16 2011

cp's OEIS Frontend

A348055 Number of positive integers with n digits that are the product of two integers ending with 7.

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A348487 Positive numbers whose square starts and ends with exactly one 1.

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A017378 a(n) = (10*n + 9)^2.

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A262389 Numbers whose last digit is composite.

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A273372 Squares ending in digit 1.

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A322489 Numbers k such that k^k ends with 4.

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