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 1-6 of 6 results.

A344332 Side s of squares of type 2 that can be tiled with squares of two different sizes so that the number of large or small squares is the same.

Original entry on oeis.org

15, 30, 45, 60, 65, 75, 90, 105, 120, 130, 135, 150, 165, 175, 180, 195, 210, 225, 240, 255, 260, 270, 285, 300, 315, 325, 330, 345, 350, 360, 369, 375, 390, 405, 420, 435, 450, 455, 465, 480, 495, 510, 520, 525, 540, 555, 570, 585, 600, 615, 630, 645, 650, 660, 671, 675, 690, 700, 705, 715, 720, 735, 738, 750, 765, 780, 795, 810, 825, 840, 845, 855, 870, 875, 885, 900
Offset: 1

Views

Author

Bernard Schott, May 20 2021

Keywords

Comments

This sequence is relative to the generalization of the 4th problem proposed for the pupils in grade 6 during the 19th Mathematical Festival at Moscow in 2008 (see A344330).
There are two types of solutions, the second one is proposed here, while type 1 is described in A344331.
If m is a term and k > 1, k * m is another term.
Every term (primitive or not primitive) is the side of an elementary square of type 2 (see A346263).
Some notations: s = side of the tiled square, a = side of small squares, b = side of large squares, and z = number of small squares = number of large squares.
-> Primitive squares
Side s of primitive squares of type 2 must satisfy the Diophantine equation s^2 = z * (a^2+b^2) with the conditions a^2+b^2 = c^2 and gcd(a, b, c) = 1.
In this case, q = a/(c-b) must be odd, and side s = q*c = a*c/(c-b) = (a+b)*c/a with a number of squares z = q^2 = (a/(c-b))^2 = ((b+c)/a)^2.
Indeed, these conditions give exactly the following solutions for n >= 2: s = n^4-(n-1)^4 (A005917), a = 2*n-1 (A005408), b = 2*n*(n-1) (A046092), c = 2*n*(n-1)+1 (A001844), z = (2*n-1)^2 (A016754); this results come from the identity:
(n^4 - (n-1)^4)^2 = (2*n-1)^2 * ((2*n-1)^2 + (2*n*(n-1))^2).
For n >= 2, every primitive square is composed by a square S1 of z = (2*n-1)^2 large squares with side b = 2*n*(n-1), then an edge on two sides of this square S1 of z = (2*n-1)^2 small squares with side a = 2*n-1.
See example with design of square of side s = 15 with a = 3, b = 4, c = 5, q = 3, z = 9, obtained with n= 2.
-> Non-primitive squares
If s is the side of a primitive square of type 2, then every k * s, k > 1 is a non-primitive term that gives two distinct tilings of type 2.
The square ks X ks can be tiled with z = q^2 = (2n-1)^2 = (a/(c-b))^2 = ((b+c)/a)^2 squares of side ka and of side kb, but also,
The square ks X ks can be tiled with z = k^2*q^2 = ((2n-1)*k)^2 = (k*a/(c-b))^2 = (k*(b+c)/a)^2 squares of side a and of side b (see example).

Examples

			Primitive square with s = 15:
  a = 3, b = 4, c = 5, s = 15, z = 9; s = 15 is the side of primitive square, with  z = 9 squares of size 3 x 3 and 9 squares of size 4 x 4
Non-primitive square k*s = 2*15 = 30:
  a = 3, b = 4, c = 5, s = 30, z = 36, this square is obtained with 4 copies of the primitive square as below.
  a = 6, b = 8, c = 10, s = 30, z = 9, this square and its tiling are exactly as the primitive square with scale 2.
               b = 4 (or = 8)     a = 3 (or = 6)
            ________ ________ ________ ______ ______________________________
           |        |        |        |      |                              |
           |        |        |        |      |                              |
           |        |        |        |______|                              |
           |_______ |________|________|      |                              |
           |        |        |        |      |                              |
           |        |        |        |______|                              |
           |        |        |        |      |                              |
           |________|________|________|      |                              |
           |        |        |        |______|                              |
           |        |        |        |      |                              |
           |        |        |        |      |                              |
           |_____ __|___ ____|_ ______|______|                              |
           |     |      |      |      |      |                              |
           |     |      |      |      |      |                              |
           |_____|______|______|______|______|______________________________|
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |_________________________________|______________________________|
                      s = 15               s = 30

References

  • Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013.

Links

Crossrefs

Cf. A001844, A005408, A005917, A046092, A046754, A344330, A344331.

Programs

  • PARI
    pts(lim) = my(v=List(), m2, s2, h2, h); for(middle=4, lim-1, m2=middle^2; for(small=1, middle, s2=small^2; if(issquare(h2=m2+s2, &h), if(h>lim, break); listput(v, [small, middle, h])))); vecsort(Vec(v)); \\ A009000
isdp4(s) = my(k=1, x); while(((x=k^4 - (k-1)^4) <= s), if (x == s, return (1)); k++); return(0);
isokp2(s) = {if (!isdp4(s), return(0)); if (s % 2, my(vp = pts(s)); for (i=1, #vp, my(vpi = vp[i], a = vpi[1], b = vpi[2], c = vpi[3]); if (a*c/(c-b) == s, return(1)); ); ); }
isok2(s) = {if (isokp2(s), return (1)); fordiv(s, d, if ((d>1) || (dMichel Marcus, Jun 04 2021

A046755 Numbers k such that d(k)^3 divides k.

Original entry on oeis.org

1, 625, 6561, 21952, 32768, 64000, 98304, 117649, 163840, 229376, 360448, 425984, 557056, 622592, 753664, 950272, 1015808, 1212416, 1343488, 1362944, 1409024, 1540096, 1736704, 1933312, 1998848, 2195456, 2326528, 2392064, 2588672
Offset: 1

Views

Author

Labos Elemer

Keywords

Comments

Proper subset of both A033950 and A046754. If two terms in the sequence are coprime then their product is also in the sequence.

Examples

			If k = 21952, d(k) = sigma(0,k) = 28. Its 3rd power is 21952, which divides k.
a(103) = 14385152 = (2^15)*439 with 32 divisors and 14385152/(32^3) = 439; 2^15*prime is a typical form of terms in the sequence that have 32 divisors.

Links

Crossrefs

Cf. A033950, A046754, A046756.

Programs

  • Mathematica
    Select[ Range[ 1, 14500000 ], IntegerQ[ #/(DivisorSigma[ 0, # ])^3 ]& ]

A046756 Numbers k such that d(k)^4 divides k.

Original entry on oeis.org

1, 625, 6561, 117649, 4100625, 31250000, 37515625, 73530625, 771895089, 1000000000, 2147483648, 6442450944, 10737418240, 15032385536, 23622320128, 25937424601, 27917287424, 32212254720, 33059881728, 36507222016, 40802189312, 45097156608, 49392123904
Offset: 1

Views

Author

Labos Elemer

Keywords

Comments

Proper subset of A033950, A046754 and A046755. Relatively prime terms are in the sequence together with their products like 73530625 or 771895089000000000.
2^31 is a term, as is every integer of the form 2^31*p, 2^31*p^3, and 2^31*p*q, where p and q are distinct odd primes; each of these has 32, 64, or 128 divisors. Of the first 10000 terms, 9609 are of one of those forms. Of the remaining 391 terms, 316 are of the form 2^8 * 3^17 * m, where m is 1, a prime > 3, or 5^4. - Jon E. Schoenfield, Aug 13 2022

Examples

			If k=625, d(k) = sigma(0,k) = 5. Its 4th power is 625, which divides k.

Links

Crossrefs

Cf. A033950, A046754, A046755.

Programs

  • Mathematica
    Select[ Range[ 1, 14500000 ], IntegerQ[ #/(DivisorSigma[ 0, # ])^2 ]& ]

Extensions

a(6)-a(20) from Donovan Johnson, Nov 30 2008
a(21)-a(23) from Donovan Johnson, Jun 08 2011

A048168 n is divisible by the square of the number of unitary divisors of n (A034444).

Original entry on oeis.org

1, 4, 8, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 256, 272, 288, 304, 320, 352, 368, 384, 400, 416, 432, 448, 464, 496, 512, 544, 576, 592, 608, 640, 656, 688, 704, 736, 752, 768, 784, 800, 832, 848, 864, 896, 928, 944, 960, 976, 992
Offset: 1

Views

Author

Labos Elemer

Keywords

Examples

			x=2032=16*127 has 4 unitary divisors: {1,16,127,2032} and ud[ x ]^2=16 divides 2032.

Links

Crossrefs

A034444, A046754, A048166.

Programs

  • Mathematica
    Select[Range[10^3], Divisible[#, 4^PrimeNu[#]] &] (* Amiram Eldar, Aug 05 2019 *)

A063921 Quotients arising when A046755(n) is divided by the cube of the number of its divisors.

Original entry on oeis.org

1, 5, 9, 1, 8, 1, 3, 343, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 16, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 16, 89, 97, 101, 103, 107, 109, 113, 6, 45, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 10, 211, 223, 227, 229, 64, 233, 239
Offset: 1

Views

Author

Labos Elemer, Sep 04 2001

Keywords

Examples

			Since (2^15)^p is in A046755 when p is a prime > 2, then p appears here at least once. Several terms breaking this regularity come from entries of A046755 of other categories. E.g. x=(2^10)*p*(11^3), d(x)=88, d(x)^3=(2^9)*(11^3) divides x and the quotient is 2p (p not equal to 11). Similar subsequences arise if 11 is replaced with different suitable primes.

Links

Crossrefs

Cf. A046755, A046754, A046756, A033950.

Programs

  • Magma
    [k/#Divisors(k)^3:k in [m:m in [1..9000000]|IsIntegral(m/#Divisors(m)^3)]]; // Marius A. Burtea, Aug 07 2019
  • Mathematica
    f[n_] := n/DivisorSigma[0, n]^3; Select[f /@ Range[10^5], IntegerQ] (* Amiram Eldar, Aug 07 2019 *)

Formula

a(n)= A046755(n)/(A000005(A046755(n))^3).

A226325 Integers of the form n/tau(n)^2 as n runs through the integers, where tau(n) is the number of divisors of n.

Original entry on oeis.org

1, 1, 2, 25, 3, 2, 5, 9, 25, 81, 6, 4, 16, 3, 5, 28, 36, 12, 128, 81, 81, 24, 7, 40, 8, 21, 28, 50, 12, 11, 16, 12, 96, 49, 13, 35, 44, 2401, 52, 45, 17, 36, 19, 160, 225, 68, 63, 23, 76, 30, 28, 36, 72, 21, 224, 92, 29, 77, 121, 31, 18, 27, 30, 91, 99, 116, 128, 37, 124
Offset: 1

Views

Author

Alex Ratushnyak, Jun 04 2013

Keywords

Examples

			A046754(3) = 128, A000005(128) = 8, so a(3) = 128 / 8^2 = 2.

Links

Crossrefs

Cf. A000005, A033950, A046754.
Cf. A036762 (if d(n) divides n, then n/d(n) is appended to the sequence).

Programs

  • Maple
    for n from 1 to 1000000 do
    r := n/(numtheory[tau](n))^2 ;
    if type(r,'integer') then
        printf("%d,",r);
    end if;
end do: # R. J. Mathar, Jun 07 2013
  • Mathematica
    Select[Table[n/DivisorSigma[0,n]^2,{n,10^6}],IntegerQ] (* Harvey P. Dale, Jan 01 2015 *)

Formula

a(n) = A046754(n) / A000005(A046754(n))^2.
Showing 1-6 of 6 results.

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