A017330 -id:A017330 - cp's OEIS frontend

A017222 a(n) = (9*n + 5)^2.

25, 196, 529, 1024, 1681, 2500, 3481, 4624, 5929, 7396, 9025, 10816, 12769, 14884, 17161, 19600, 22201, 24964, 27889, 30976, 34225, 37636, 41209, 44944, 48841, 52900, 57121, 61504, 66049, 70756

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).

Sequences of the form (m*n+5)^2: A010864 (m=0), A000290 (m=1), A016754 (m=2), A016790 (m=3), A016814 (m=4), A016850 (m=5), A016970 (m=6), A017042 (m=7), A017126 (m=8), this sequence (m=9), A017330 (m=10), A017450 (m=11), A017582 (m=12).

Cf. A017221, A017246.

[(9*n+5)^2: n in [0..35]]; // Vincenzo Librandi, Jul 24 2011

(9Range[0,30]+5)^2 (* or *) LinearRecurrence[{3,-3,1},{25,196,529},30] (* Harvey P. Dale, May 22 2012 *)

a(n)=(9*n+5)^2 \\ Charles R Greathouse IV, Jun 17 2017

[(9*n+5)^2 for n in range(41)] # G. C. Greubel, Dec 29 2022

a(n) = A017221(n)^2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, May 22 2012
G.f.: (25 + 121*x + 16*x^2)/(1-x)^3. - R. J. Mathar, Mar 20 2018
From G. C. Greubel, Dec 29 2022: (Start)
a(2*n+1) = 4*A017246(n).
a(n) = a(n-1) + 9*(18*n + 1).
E.g.f.: (25 + 171*x + 81*x^2)*exp(x). (End)

A030485 Squares composed of digits {2, 5, 7}.

Original entry on oeis.org

25, 225, 7225, 27225, 55225, 2772225, 227557225, 277722225, 27777222225, 72272257225, 2777772222225, 25772527522225, 277777722222225, 2775552752755225, 27522257555772225, 27777777222222225, 77525222275255225, 257727727257277225, 722555225555275225, 2275752775775227225

Offset: 1

Patrick De Geest, Dec 11 1999

We can easily prove that, except for the first term, all terms are of the form 100*m^2 + 100*m + 25 where mod(m, 10) is one of the numbers 1, 3, 6 or 8. Also we can show that all numbers of the form ((5 * 10^n - 5)/3)^2 where n is a natural number, are in the sequence. - Farideh Firoozbakht, Dec 09 2008

Zhao Hui Du, Table of n, a(n) for n = 1..45
Patrick De Geest, Squares containing at most three distinct digits, Index entries for related sequences

Subsequence of A191486. Also subsequence of A017330. Cf. A030487.

Flatten[Table[Select[FromDigits/@Tuples[{2, 5, 7}, n], IntegerQ[Sqrt[#]] &], {n, 17}]] (* The program takes a long time to run *) (* Harvey P. Dale, Jan 18 2015 *)
Select[(5Range[1, 9999, 2])^2, Complement[IntegerDigits[#], {2, 5, 7}] == {} &] (* Alonso del Arte, Feb 19 2020 *)

fromTernary(n, d)=sum(i=0,d-1,[2,5,7][(n\3^i)%3+1]*10^i)
v=List([25]);for(d=0,16,for(n=0,3^d-1,if(issquare(t=225+1000*fromTernary(n,d)), listput(v,t); print1(t", ")))); Vec(v) \\ Charles R Greathouse IV, Dec 22 2012

a(n) = A030487(n)^2. - M. F. Hasler, Dec 23 2012

Extended and corrected by author, May 08 2000

a(17)-a(19) from Farideh Firoozbakht, Dec 09 2008

A273373 Squares ending in digit 6.

Original entry on oeis.org

16, 36, 196, 256, 576, 676, 1156, 1296, 1936, 2116, 2916, 3136, 4096, 4356, 5476, 5776, 7056, 7396, 8836, 9216, 10816, 11236, 12996, 13456, 15376, 15876, 17956, 18496, 20736, 21316, 23716, 24336, 26896, 27556, 30276, 30976, 33856, 34596, 37636, 38416, 41616

Offset: 1

Vincenzo Librandi, May 21 2016

These are the only squares whose second last digit is odd. This implies that the only squares whose last two digits are the same are those ending with 0 or 4; those ending with 1, 5, and 9 are paired with even second last digits. - Waldemar Puszkarz, May 24 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, A047221, A090773.
Cf. A017341 (numbers ending in 6), A017343 (cubes ending in 6).
Cf. squares with last digit k: A017270 (k=0), A273372 (k=1), A273375 (k=4), A017330 (k=5), this sequence (k=6), A273374 (k=9).

/* By definition: */ [n^2: n in [0..200] | Modexp(n,2,10) eq 6];

[(10*n - 3*(-1)^n - 5)^2/4: n in [1..50]];

seq(seq((10*i+j)^2,j=[4,6]),i=0..20); # Robert Israel, May 24 2016

Table[(10 n - 3 (-1)^n - 5)^2/4, {n, 1, 50}]
CoefficientList[Series[4 (4 + 5 x + 32 x^2 + 5 x^3 + 4 x^4) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x]
Select[Range[250]^2,Mod[#,10]==6&] (* Harvey P. Dale, May 31 2020 *)

G.f.: 4*x*(4 + 5*x + 32*x^2 + 5*x^3 + 4*x^4)/((1 + x)^2*(1 - x)^3).
a(n) = 4*A047221(n)^2 = (10*n - 3*(-1)^n - 5)^2/4.
a(n) = A090773(n)^2. - Michel Marcus, May 25 2016
Sum_{n>=1} 1/a(n) = 2*Pi^2/(25*(5+sqrt(5))). - Amiram Eldar, Feb 16 2023

Corrected and extended by Bruno Berselli, May 23 2016

A348489 Positive numbers whose square starts and ends with exactly one 5.

Original entry on oeis.org

75, 225, 715, 725, 735, 755, 765, 2245, 2255, 2265, 2275, 2285, 2295, 2305, 2315, 2325, 2335, 2345, 2375, 2385, 2395, 2405, 2415, 2425, 2435, 2445, 7075, 7085, 7095, 7105, 7115, 7125, 7135, 7145, 7155, 7165, 7175, 7185, 7195, 7205, 7215, 7225, 7235, 7245

Offset: 1

Bernard Schott, Oct 25 2021

When a square ends with 5, it ends with 25.
From  Marius A. Burtea, Oct 25 2021: (Start)
Numbers 75, 765, 7665, 76665, ..., (23*10^k -5) / 3, k >= 1, are terms and have no digits 0, because their squares are 5625, 585225, 58752225, 5877522225, 587775222225, 58777752222225, ...
Also 75, 735, 7335, 73335, ..., (22*10^n+5) / 3, k >= 1, are terms and have no digits 0, because their squares are 5625, 540225, 53802225, 5378022225, 537780222225, 53777802222225, ... (End)

			75^2 = 5625, hence 75 is a term.
235^2 = 55225, hence 235 is not a term.

Cf. A045859, A017330 (squares ending with 5).
Similar to: A348487 (k=1), A348488 (k=4), this sequence (k=5), A348490 (k=6), A348491 (k=9).
Subsequence of A305719.

[n:n in [4..7500]|Intseq(n*n)[1] eq 5 and Intseq(n*n)[#Intseq(n*n)] eq 5 and Intseq(n*n)[-1+#Intseq(n*n)] ne 5 ]; // Marius A. Burtea, Oct 25 2021

Select[5 * Range[2, 1500], (d = IntegerDigits[#^2])[[1]] == d[[-1]] == 5 && d[[2]] != 5 &] (* Amiram Eldar, Oct 25 2021 *)

isok(k) = my(d=digits(sqr(k))); (d[1]==5) && (d[#d]==5) && if (#d>2, (d[2]!=5) && (d[#d-1]!=5), 1); \\ Michel Marcus, Oct 25 2021

from itertools import count, takewhile
def ok(n):
  s = str(n*n); return len(s.rstrip("5")) == len(s.lstrip("5")) == len(s)-1
def aupto(N):
  r = takewhile(lambda x: x<=N, (10*i+5 for i in count(0)))
  return [k for k in r if ok(k)]
print(aupto(7245)) # Michael S. Branicky, Oct 26 2021

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A017222 a(n) = (9*n + 5)^2.

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A030485 Squares composed of digits {2, 5, 7}.

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A273373 Squares ending in digit 6.

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A348489 Positive numbers whose square starts and ends with exactly one 5.

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