A003059 -id:A003059 - cp's OEIS frontend

A257213 Least d>0 such that floor(n/d) = floor(n/(d+1)).

1, 2, 3, 2, 3, 3, 4, 4, 3, 5, 4, 4, 5, 5, 5, 4, 6, 6, 5, 5, 7, 6, 6, 6, 5, 7, 7, 7, 6, 6, 8, 8, 7, 7, 7, 6, 8, 8, 8, 8, 7, 7, 9, 9, 9, 8, 8, 8, 7, 10, 9, 9, 9, 9, 8, 8, 10, 10, 10, 10, 9, 9, 9, 8, 11, 11, 10, 10, 10, 10, 9, 9, 11, 11, 11, 11, 11, 10, 10, 10

Offset: 0

M. F. Hasler, Apr 18 2015

For n > 1: a(A043548(n)) = n. - Reinhard Zumkeller, Apr 19 2015

			a(0)=1 because 0/1 = 0/2.
a(1)=2 because [1/1] = 1 > [1/2] = 0 = [1/3], where [x] := floor(x).
a(2)=3 because [2/1] = 2 > [2/2] = 1 > [2/3] = 0 = [2/4].

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A003059, A257212.

Cf. A043548.

a257213 n = head [d | d <- [1..], div n d == div n (d + 1)]
-- Reinhard Zumkeller, Apr 19 2015

f[n_] := Block[{d, k}, Reap@ For[k = 0, k <= n, k++, d = 1; While[Floor[k/d] != Floor[k/(d + 1)], d++]; Sow@ d] // Flatten // Rest]; f@ 79 (* Michael De Vlieger, Apr 18 2015 *)
Table[Module[{d=1},While[Floor[n/d]!=Floor[n/(d+1)],d++];d],{n,0,80}] (* Harvey P. Dale, Aug 16 2025 *)

A257213(n)=for(d=sqrtint(n)+1,n+1,n\d==n\(d+1)&&return(d))

a(n) >= A003059(n+1) = floor(sqrt(n))+1 >= A003059(n) = ceiling(sqrt(n)) >= A257212(n), with strict inequality (in the middle relation) when n is a square.
a(k^2-1) = k for k > 1. Proof: For n=k^2-1=(k-1)*(k+1), floor(n/k) = k-1 = n/(k+1), but n/(k-1)=k+1 and when denominators decrease further, this keeps increasing.
a(k^2) >= k+d when k > d*(d-1). Proof: This follows from k^2/(k+d) = k-d+d^2/(k+d), which shows that a(k) >= d when k > d*(d-1).
a(n) = A259361(n) + 1 + floor(sqrt((A232091(n+1) - 1 - n) + A079813(n+1)) + A079813(n+1)/2) = floor((sqrt(4*n+1)+1)/2) + floor(sqrt(ceiling((n+1) / ceiling(sqrt(n+1)) + 1) * ceiling(sqrt(n+1)) - round(sqrt(n+1)) - n - 1) + (ceiling(sqrt(n+1)) - round(sqrt(n+1)))/2). - Haofen Liang, Aug 25 2021
a(n) = floor(sqrt(p*q - n) + (p + q)/2), where p = floor(sqrt(n)) and q = floor(sqrt(n+1) + 3/2). - Ridouane Oudra, Jan 24 2023

A038759 a(n) = ceiling(sqrt(n))*floor(sqrt(n)).

Original entry on oeis.org

0, 1, 2, 2, 4, 6, 6, 6, 6, 9, 12, 12, 12, 12, 12, 12, 16, 20, 20, 20, 20, 20, 20, 20, 20, 25, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 36, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 49, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 64, 72, 72, 72, 72, 72, 72, 72

Offset: 0

Henry Bottomley, May 03 2000

a(n) = n iff n is a square or a pronic (or heteromecic) number of form k(k+1). The sequence interleaves individual squares with 2k copies of each pronic.

			a(31) = 30 since 6 and 5 are on either side of the square root of 31 and 6*5 = 30.

Cf. A000196, A002378, A002620, A003059, A038760, A053187, A072691.

a[n_] := Ceiling[Sqrt[n]]*Floor[Sqrt[n]]; Array[a, 70, 0] (* Amiram Eldar, Dec 04 2022 *)

a(n) = my(r,s=sqrtint(n,&r)); if(r, n-r+s, n); \\ Kevin Ryde, Jul 30 2022

from math import isqrt
def A038759(n): return m+n+k if (m:=(k:=isqrt(n))**2-n) else n # Chai Wah Wu, Jul 28 2022

a(n) = A003059(n)*A000196(n) = n - A038760(n).
a(A002620(n)) = A002620(n). - Bernard Schott, Nov 06 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/12 (A072691). - Amiram Eldar, Dec 04 2022

A157466 Numbers of numbers <= n, that are squares when inserting at most one digit in decimal representation (see A157465).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 12, 13, 13, 14, 14, 14, 15, 15, 16, 17, 17, 18, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 25, 26, 26, 26, 26, 27, 28, 28, 28, 29, 29, 29, 29, 30, 31, 31, 31, 32, 32, 32, 32, 33, 34, 34, 35, 35, 36, 37, 37, 38, 39, 40, 41, 41, 42, 42, 42, 43

Offset: 0

Reinhard Zumkeller, Mar 01 2009

a(n) > n/2 for n <= 131;
a(n) <= n/2 for n > 131;
a(n) <= a(n+1); a(n)/n --> 0 for n --> oo;
a(A157465(n+1)) = a(A157465(n)) + 1.

R. Zumkeller, Table of n, a(n) for n = 0..10000

A003059, A000196, A157464.

A220693 Irregular triangle where the k-th item in the n-th row (both starting from 1) tells in how many ways we can add 2 distinct integers from 1 to n in such a way that the sum is divisible by k. Row n has 2n-1 terms.

Original entry on oeis.org

0, 1, 0, 1, 3, 1, 1, 1, 1, 6, 2, 2, 1, 2, 1, 1, 10, 4, 4, 2, 2, 2, 2, 1, 1, 15, 6, 5, 3, 3, 2, 3, 2, 2, 1, 1, 21, 9, 7, 5, 4, 3, 3, 3, 3, 2, 2, 1, 1, 28, 12, 10, 6, 6, 4, 4, 3, 4, 3, 3, 2, 2, 1, 1, 36, 16, 12, 8, 8, 5, 5, 4, 4, 4, 4, 3, 3, 2, 2, 1, 1

Offset: 1

Antti Karttunen, Feb 18 2013. Proposed by Robert Israel, May 07 2012

After the first two rows, this irregular table gives all the nonzero terms from the beginning of each row of A220691. See the comments there and at A061857.

			Row n (starting from row 1) has 2n-1 terms in this irregular table:
   0;
   1, 0, 1;
   3, 1, 1, 1, 1;
   6, 2, 2, 1, 2, 1, 1;
  10, 4, 4, 2, 2, 2, 2, 1, 1;
  15, 6, 5, 3, 3, 2, 3, 2, 2, 1, 1;
  etc.
See A220691 and A061857 for the meaning of each term.

Antti Karttunen, Rows n = 1..100 of table, flattened
Index entries for sequences related to subset sums modulo m

(define (A220693 n) (A220691bi (A003059 n) (A071797 n)))
;; For A220691bi see A220691.

See Robert Israel's formula at A061857.

A269501 Subsequence immediately following the instances of n in the sequence is n, n-1, ..., 1, n+1, n+2, ....

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 3, 2, 3, 1, 4, 4, 3, 4, 2, 4, 1, 5, 5, 4, 5, 3, 5, 2, 5, 1, 6, 6, 5, 6, 4, 6, 3, 6, 2, 6, 1, 7, 7, 6, 7, 5, 7, 4, 7, 3, 7, 2, 7, 1, 8, 8, 7, 8, 6, 8, 5, 8, 4, 8, 3, 8, 2, 8, 1, 9, 9, 8, 9, 7, 9, 6, 9, 5, 9, 4, 9, 3, 9, 2, 9, 1, 10, 10, 9, 10, 8, 10, 7, 10, 6, 10, 5, 10, 4, 10, 3, 10, 2, 10, 1

Offset: 0

Franklin T. Adams-Watters, Mar 04 2016

The sequence includes every ordered pair of positive integers exactly once as consecutive terms of the sequence. Through n = k^2, it has every pair i,j with 0 < i,j <= k.
Can be regarded as an irregular triangle where row k contains 1, k, k, k-1, k, k-2, ..., 2, k, with 2n-1 terms.
See A305615 for an essentially identical sequence: a(n) = A305615(n)+1. - N. J. A. Sloane, Jul 03 2018

			The first 3 occurs as a(5), so a(6) = 3, the first term of 3, 2, 1, 4, 5, 6, .... The second 3 is thus a(6), so a(7) = 2. The third 3 is a(8), so a(9) = 1. The fourth 3 is a(12), now we start incrementing, and a(13) = 4.
The triangle starts:
  1
  1, 2, 2
  1, 3, 3, 2, 3
  1, 4, 4, 3, 4, 2, 4
  1, 5, 5, 4, 5, 3, 5, 2, 5

Cf. A003059, A060747 (row lengths), A000326 (row sums), A097291, A269780.

A295006 Numbers n such that the largest digit of n^2 is 6.

Original entry on oeis.org

4, 6, 8, 16, 19, 25, 34, 40, 46, 51, 56, 58, 60, 66, 68, 75, 79, 80, 81, 106, 108, 116, 119, 121, 125, 129, 142, 146, 156, 160, 162, 175, 190, 204, 206, 208, 215, 216, 225, 231, 238, 245, 246, 248, 249, 250, 251, 252, 254, 255, 256, 258, 325, 334, 340, 354, 355, 369, 375, 379

Offset: 1

M. F. Hasler, Nov 12 2017

			19 is in this sequence because 19^2 = 361 has 6 as largest digit.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A295016 (the corresponding squares), A277959, A277960, A277961, A295005 .. A295009 (analog for digits 2 through 9), A294996 (analog for cubes).

Cf. A000290 (the squares).

Select[Range[400],Max[IntegerDigits[#^2]]==6&] (* Harvey P. Dale, Mar 30 2024 *)

select( is_A295006(n)=n&&vecmax(digits(n^2))==6 , [0..999]) \\ The "n&&" avoids an error message for n=0.

a(n) = sqrt(A295016(n)), where sqrt = A000196 or A000194 or A003059.

A295008 Numbers whose square has largest digit 8.

Original entry on oeis.org

9, 22, 28, 29, 41, 59, 62, 72, 78, 90, 91, 92, 94, 104, 109, 122, 126, 128, 135, 151, 159, 168, 169, 178, 184, 191, 192, 195, 196, 202, 209, 220, 221, 232, 241, 242, 259, 261, 262, 268, 278, 279, 280, 284, 285, 289, 290, 291, 292, 294, 295, 296, 298, 322, 328, 329, 341, 344, 349

Offset: 1

M. F. Hasler, Nov 12 2017

Includes a*10^n+b for n >= 2 and [a,b] in {[4, 1], [9, 1], [2, 2], [9, 2], [1, 4], [6, 4], [9, 4], [8, 5], [4, 6], [9, 6], [5, 8], [8, 8], [9, 8], [1, 9], [2, 9], [4, 9], [6, 9], [8, 9], [9, 9]}. - Robert Israel, Nov 13 2017

			28 is in this sequence because 28^2 = 784 has 8 as largest digit.

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

Cf. A295018 (the corresponding squares), A277959 .. A277961 (same for digit 2 .. 4), A295005 .. A295009 (same for digit 5 .. 9).

Cf. A000290 (the squares).

select(t -> max(convert(t^2,base,10))=8, [$1..1000]); # Robert Israel, Nov 13 2017

Select[Range[400],Max[IntegerDigits[#^2]]==8&] (* Harvey P. Dale, Jun 02 2019 *)

select( is_A295008(n)=n&&vecmax(digits(n^2))==8 , [0..999]) \\ The "n&&" avoids an error message for n=0.

def ok(n): return max(int(d) for d in str(n*n)) == 8
print(list(filter(ok, range(350)))) # Michael S. Branicky, Sep 22 2021

a(n) = sqrt(A295018(n)), where sqrt = A000196 or A000194 or A003059.

A321489 Numbers m such that both m and m+1 have at least 7 distinct prime factors.

Original entry on oeis.org

965009045, 1068044054, 1168008204, 1177173074, 1209907985, 1218115535, 1240268490, 1338753129, 1344185205, 1408520805, 1477640450, 1487720234, 1509981395, 1663654629, 1693460405, 1731986894, 1758259425, 1819458354, 1821278459, 1826445984, 1857332840

Offset: 1

Amiram Eldar and M. F. Hasler, Nov 12 2018

nonn

The first 300 terms of this sequence are such that m and m+1 both have exactly 7 prime divisors. See A321497 for the terms m such that m or m+1 has more than 7 prime factors: the smallest such term is 5163068910.
Numbers m and m+1 can never have a common prime factor (consider them mod p), therefore the terms are > sqrt(p(7+7)#) = A003059(A002110(7+7)). (Here we see that sqrt(p(7+8)#) is a more realistic estimate of a(1), but for smaller values of k we may have sqrt(p(2k+1)#) > m(k) > sqrt(p(2k)#), where m(k) is the smallest of two consecutive integers each having at least k prime divisors. For example, A321503(1) < sqrt(p(3+4)#) ~ A321493(1).)
From M. F. Hasler, Nov 28 2018: (Start)
The first 100 terms and beyond are all congruent to one of {14, 20, 35, 49, 50, 69, 84, 90, 104, 105, 110, 119, 125, 129, 134, 140, 144, 170, 174, 189, 195} mod 210. Here, 35, 195, 189, 14 140, 20 and 174 (in order of decreasing frequency) occur between 6 and 13 times, and {49, 50, 110, 129, 134, 144, 170} occur only once.
However, as observed by Charles R Greathouse IV, one can construct a term of this sequence congruent to any given m > 0, modulo any given n > 0.
The first terms of this sequence which are multiples of 210 are in A321497. An example of a term that is a multiple of 210 but not in A321497 is 29759526510, due to Charles R Greathouse IV. Such examples can be constructed by solving A*210 + 1 = B for A having 3 distinct prime factors not among {2, 3, 5, 7}, B having 7 distinct prime factors and gcd(B, 210*A) = 1. (End)

			a(1) = 5 * 7 * 11 * 13 * 23 * 83 * 101, a(1)+1 = 2 * 3 * 17 * 29 * 41 * 73 * 109.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 102 terms from M. F. Hasler)

Cf. A255346, A321503 .. A321506 (analog for k = 2, ..., 6 prime divisors).
Cf. A321502, A321493 .. A321497 (m and m+1 have at least but not both exactly k = 2, ..., 7 prime divisors).
Cf. A074851, A140077, A140078, A140079 (m and m+1 both have exactly k = 2, 3, 4, 5 prime divisors).
Cf. A006049, A006549, A093548.
Cf. A002110.

Select[Range[36000000], PrimeNu[#] > 6 && PrimeNu[# + 1] > 6 &]

is(n)=omega(n)>6&&omega(n+1)>6
A321489=List();for(n=965*10^6,1.8e9,is(n)&&listput(A321489,n))

a(n) ~ n. - Charles R Greathouse IV, Nov 29 2018

A134917 a(n) = ceiling(n^(4/3)).

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 14, 16, 19, 22, 25, 28, 31, 34, 37, 41, 44, 48, 51, 55, 58, 62, 66, 70, 74, 78, 81, 86, 90, 94, 98, 102, 106, 111, 115, 119, 124, 128, 133, 137, 142, 146, 151, 156, 161, 165, 170, 175, 180, 185, 190, 195, 200

Offset: 1

Mohammad K. Azarian, Nov 17 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A048766, A038605, A003059, A134914, A129011, A100196, A121536.

[Ceiling(n^(4/3)): n in [1..60]]; // Vincenzo Librandi, Feb 18 2013

Table[Ceiling[n^(4/3)], {n, 60}] (* Vincenzo Librandi, Feb 18 2013 *)

A341403 Number of nonnegative solutions to (x_1)^2 + (x_2)^2 + ... + (x_8)^2 <= n.

Original entry on oeis.org

1, 9, 37, 93, 171, 283, 479, 767, 1076, 1420, 1952, 2688, 3444, 4228, 5320, 6776, 8262, 9662, 11454, 13918, 16480, 18832, 21772, 25644, 29508, 33044, 37300, 42732, 48340, 53556, 59632, 67472, 75405, 82237, 90189, 100661, 111155, 120403, 131099, 144651, 158469, 170621

Offset: 0

Ilya Gutkovskiy, Feb 10 2021

nonn

Partial sums of A045850.

Index entries for sequences related to sums of squares

Cf. A000122, A000606, A003059, A045850, A055407, A055414, A224212, A224213, A302862, A341400, A341401, A341402, A341404, A341405.

b:= proc(n, k) option remember; `if`(n=0, 1, `if`(n<0 or k<1, 0,
      b(n, k-1)+add(b(n-j^2, k-1), j=1..isqrt(n))))
    end:
a:= proc(n) option remember; b(n, 8)+`if`(n>0, a(n-1), 0) end:
seq(a(n), n=0..41);  # Alois P. Heinz, Feb 10 2021

nmax = 41; CoefficientList[Series[(1 + EllipticTheta[3, 0, x])^8/(256 (1 - x)), {x, 0, nmax}], x]

G.f.: (1 + theta_3(x))^8 / (256 * (1 - x)).

a(n^2) = A055407(n).

cp's OEIS Frontend

A257213 Least d>0 such that floor(n/d) = floor(n/(d+1)).

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A038759 a(n) = ceiling(sqrt(n))*floor(sqrt(n)).

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A157466 Numbers of numbers <= n, that are squares when inserting at most one digit in decimal representation (see A157465).

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A220693 Irregular triangle where the k-th item in the n-th row (both starting from 1) tells in how many ways we can add 2 distinct integers from 1 to n in such a way that the sum is divisible by k. Row n has 2n-1 terms.

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A269501 Subsequence immediately following the instances of n in the sequence is n, n-1, ..., 1, n+1, n+2, ....

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A295006 Numbers n such that the largest digit of n^2 is 6.

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A295008 Numbers whose square has largest digit 8.

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A321489 Numbers m such that both m and m+1 have at least 7 distinct prime factors.

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