cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A072862 The smallest k>1 such that k divides sigma(k*n) is equal to 6.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 64, 81, 100, 121, 144, 225, 256, 289, 324, 400, 484, 529, 576, 625, 729, 841, 900, 1024, 1089, 1156, 1296, 1600, 1681, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2809, 2916, 3025, 3364, 3481, 3600, 4096, 4356, 4624, 4761, 5041
Offset: 1

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Author

Benoit Cloitre, Jul 26 2002

Keywords

Comments

The smallest k>1 such that k divides sigma(k*n) is always 2, 3 or 6. Sequence generates squares but not all the squares (49, 169, 196, 361, 441, 676, 784, ... are the first squares not in the sequence).
This is not the sequence of solutions to sigma(n) == 1 (mod 6), because 2401 is in this sequence but sigma(2401) == 5 (mod 6). See A074384 for more examples. - R. J. Mathar, May 19 2020

Formula

{n: A049605(n) = 6}. - R. J. Mathar, May 19 2020

A062295 A B_2 sequence: a(n) is the smallest square such that pairwise sums of not necessarily distinct elements are all distinct.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 64, 81, 100, 169, 256, 289, 441, 484, 576, 625, 841, 1089, 1296, 1444, 1936, 2025, 2401, 2601, 3136, 4225, 4356, 4624, 5329, 5476, 5776, 6084, 7569, 9025, 10201, 11449, 11664, 12321, 12996, 13456, 14400, 16129, 17956, 20164, 22201
Offset: 1

Views

Author

Labos Elemer, Jul 02 2001

Keywords

Examples

			36 is in the sequence since the pairwise sums of {1, 4, 9, 16, 25, 36} are all distinct: 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, 34, 37, 40, 41, 45, 50, 52, 61, 72.
49 is not in the sequence since 1 + 49 = 25 + 25.
		

Crossrefs

Programs

  • Python
    from itertools import count, islice
    def A062295_gen(): # generator of terms
        aset1, aset2, alist = set(), set(), []
        for k in (n**2 for n in count(1)):
            bset2 = {k<<1}
            if (k<<1) not in aset2:
                for d in aset1:
                    if (m:=d+k) in aset2:
                        break
                    bset2.add(m)
                else:
                    yield k
                    alist.append(k)
                    aset1.add(k)
                    aset2 |= bset2
    A062295_list = list(islice(A062295_gen(),30)) # Chai Wah Wu, Sep 05 2023

Extensions

Edited, corrected and extended by Klaus Brockhaus, Sep 24 2007

A068802 Smaller of two consecutive squares which have no common digits.

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 36, 64, 196, 361, 484, 841, 1936, 5929, 8836, 69696, 1999396, 29997529
Offset: 0

Views

Author

Amarnath Murthy, Mar 06 2002

Keywords

Comments

There are no more terms. Sketch of proof: Suppose n^2 and (n+1)^2 have no common digits. Then their first digits differ, so n+1 = ceiling(sqrt(d*10^r)) with 1<=d<=9 and r>=0. In other words,
n+1 = ceiling(sqrt(e*100^s)) with e in {1,2,...,9,10,20,...,90} and s>=0. The cases e=1, 4 and 9 are easy. Otherwise note that about half the digits of (n+1)^2 equal 0, so n^2 has no 0's. We have
(n+1)^2 - n^2 = 2n+1 ~ 10^s*2*sqrt(e). For e=20, this is about 10^s * 8.94427190999916. So for s>=10, the decimal expansion of (n+1)^2 - n^2 has 2 consecutive 9's. (In fact 4 for large s, but 2 is enough.) Since n^2 has no 0's this implies that n^2 and (n+1)^2 have the same digit in the position of the first of the 2 9's. The same idea works for other values of e, but the consecutive 9's occur later.

Examples

			29997529 is a term since 29997529 and 30008484 are two consecutive squares with no common digits.
		

Programs

  • Maple
    s := X->convert(convert(X,base,10),set); seq(`if`((s(n^2) intersect s((n+1)^2))={},n^2,printf("")),n=1..350000);
  • Mathematica
    For[lastn=-1; r=0, r<500, r++, For[d=1, d<10, d++, n=Ceiling[Sqrt[d*10^r]]-1; If[n>lastn, lastn=n; If[Intersection[IntegerDigits[n^2], IntegerDigits[(n+1)^2]]=={}, Print[n^2]]]]]
    First /@ Select[Partition[Range[0, 6000]^2, 2, 1], Intersection @@ IntegerDigits /@ # == {} &] (* Jayanta Basu, Aug 06 2013 *)

Extensions

Edited by Dean Hickerson and Francois Jooste (phukraut(AT)hotmail.com), Mar 19 2002

A238203 Squares s such that s^2+s+41 is prime.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 64, 100, 144, 169, 196, 225, 324, 400, 441, 484, 529, 576, 625, 841, 900, 961, 1089, 1444, 1521, 1849, 2209, 2601, 2704, 2809, 3025, 3136, 3249, 3364, 3721, 3844, 4096, 4225, 4356, 4489, 5476, 5625, 5776, 6241, 7056, 7921, 8464, 8836, 9025
Offset: 1

Views

Author

K. D. Bajpai, Feb 20 2014

Keywords

Comments

n^2+n+41: Euler’s prime generating polynomial.
First 6 terms in the sequence are first 6 consecutive squares.

Examples

			9 is in the sequence because 9 = 3^2 and 9^2+9+41 = 131 is prime.
36 is in the sequence because 36 = 6^2 and 36^2+36+41 = 1373 is prime.
		

Crossrefs

Programs

  • Maple
    with(numtheory):KD := proc() local a,b; a:=(n^2);b:=a^2+a+41; if isprime(b) then RETURN (a); fi; end: seq(KD(), n=1..500);
  • Mathematica
    Select[Table[k = n^2, {n, 100}], PrimeQ[#^2 + # + 41] &] (* or *) c = 0; Do[k = n^2; If[PrimeQ[k^2 + k + 41], c = c + 1; Print[c, " ", k]], {n, 1, 10000}];
    Select[Range[100]^2,PrimeQ[#^2+#+41]&] (* Harvey P. Dale, Dec 13 2021 *)

A300303 Squares that are not of the form x^2 + x*y + y^2, where x and y are positive integers.

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 36, 64, 81, 100, 121, 144, 225, 256, 289, 324, 400, 484, 529, 576, 625, 729, 841, 900, 1024, 1089, 1156, 1296, 1600, 1681, 1936, 2025, 2116, 2209, 2304, 2500, 2601, 2809, 2916, 3025, 3364, 3481, 3600, 4096, 4356, 4624, 4761, 5041, 5184, 5625, 6400, 6561, 6724, 6889, 7225, 7569
Offset: 1

Views

Author

Altug Alkan, Mar 02 2018

Keywords

Comments

Or Loeschian numbers (A003136) that are not in A024614.
Squares that are not in this sequence are 49, 169, 196, 361, 441, 676, ...
This is the list of squares not of the form A050931(k)^2. A number n is in this sequence iff n = m^2 with m having no prime factor == 1 (mod 6). - M. F. Hasler, Mar 04 2018

Examples

			Loeschian number 25 = 5^2 is a term because 25 = x^2 + x*y + y^2 has no solution for positive integers x, y.
		

Crossrefs

Programs

  • Maple
    isA024614:= proc(n) local x,y;
    for x from 1 to floor(sqrt(n-1)) do
       if issqr(4*n-3*x^2) then return true fi
    od:
    false
    end proc:
    isA024614(0):= false:
    remove(isA024614, [seq(i^2,i=0..200)]); # Robert Israel, Mar 02 2018
  • Mathematica
    sol[s_] := Solve[0 < x <= y && s == x^2 + x y + y^2, {x, y}, Integers];
    Select[Range[0, 100]^2, sol[#] == {}&] (* Jean-François Alcover, Oct 26 2020 *)
  • PARI
    is(n,m)=issquare(n,m)&&!setsearch(Set(factor(m)[,1]%6),1) \\ second part is equivalent to is_A230780(m), this is sufficient to test (e.g., to produce a list) if we know that n = m^2. - M. F. Hasler, Mar 04 2018

Formula

a(n) = A230780(n-1)^2 for n > 1.
Showing 1-5 of 5 results.