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.

A256152 Numbers k such that k is the product of two distinct primes and sigma(k) is a square number.

Original entry on oeis.org

22, 94, 115, 119, 214, 217, 265, 382, 497, 517, 527, 679, 745, 862, 889, 1174, 1177, 1207, 1219, 1393, 1465, 1501, 1649, 1687, 1915, 1942, 2101, 2159, 2201, 2359, 2899, 2902, 2995, 3007, 3143, 3383, 3401, 3427, 3937, 4039, 4054, 4097, 4315, 4529, 4537, 4702, 4741, 5029, 5065, 5398, 5587
Offset: 1

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Author

Antonio Roldán, Mar 16 2015

Keywords

Comments

This sequence is the intersection of A006881 and A006532.

Examples

			199 is in the sequence because 119=7*17 (the product of two distinct primes) and sigma(119)=8*18=144=12^2 (a square number).
		

Crossrefs

Programs

  • Haskell
    a256152 n = a256152_list !! (n-1)
    256152_list = filter f a006881_list where
       f x = a010052' ((spf + 1) * (x `div` spf + 1)) == 1
             where spf = a020639 x
    -- Reinhard Zumkeller, Apr 06 2015
  • Mathematica
    f[n_] := Block[{pf = FactorInteger@ n}, Max @@ Last /@ pf == 1 && Length@ pf == 2]; Select[Range@ 6000, IntegerQ@ Sqrt@ DivisorSigma[1, #] && f@ # &] (* Michael De Vlieger, Mar 17 2015 *)
  • PARI
    {for(i=1,10^4,if(omega(i)==2&&issquarefree(i)&&issquare(sigma(i)),print1(i,", ")))}
    

A256149 Square numbers n such that sigma(n) is a triangular number.

Original entry on oeis.org

1, 36, 441, 5625, 6084, 407044, 8444836, 17388900, 35070084, 40729924, 57790404, 80138304, 537822481, 588159504, 659821969, 918999225, 1820387556, 2179862721, 2599062361, 5110963081, 28816420516, 36144473689, 46082779561, 55145598561, 147225690000, 163405126756
Offset: 1

Views

Author

Antonio Roldán, Mar 16 2015

Keywords

Comments

This sequence is the intersection of A000290 and A045746.

Examples

			441 is in the sequence because 441 = 21^2 is square number, and sigma(441) = 441 + 147 + 63 + 49 + 21 + 9 + 7 + 3 + 1 = 741 = 38*39/2 is triangular number.
		

Crossrefs

Programs

  • Mathematica
    t = Accumulate[Range@ 10000]; Select[Range[10000]^2, MemberQ[t, DivisorSigma[1, #]] &] (* Michael De Vlieger, Mar 17 2015 *)
    Select[Range[500000]^2,OddQ[Sqrt[8DivisorSigma[1,#]+1]]&] (* Harvey P. Dale, Feb 25 2017 *)
  • PARI
    {for(i=1,10^6,n=i*i;if(ispolygonal(sigma(n), 3),print1(n,", ")))}

A256150 Oblong numbers n such that sigma(n) is a triangular number.

Original entry on oeis.org

2, 12, 56, 342, 992, 16256, 17822, 169332, 628056, 1189190, 2720850, 11085570, 35599122, 67100672, 1147210770, 1317435912, 1707135806, 7800334080, 11208986256, 13366943840, 17109032402, 17179738112, 46343540900, 58413331032, 83717924940, 204574837700, 274877382656, 445968192672, 589130699852
Offset: 1

Views

Author

Antonio Roldán, Mar 16 2015

Keywords

Comments

The numbers 12, 56, 992, 16256, 67100672, ... (A139256(n), twice even perfect numbers) are in the sequence because they are oblong (A139256(n) = 2^k*(2^k-1)) and sigma(A139256(n)) = sigma(2^k*(2^k-1)) = sigma(2^k)*sigma(2^k-1) = (2^(k+1)-1)*2^(k+1)/2 is a triangular number.
This sequence is the intersection of A002378 and A045746.

Examples

			2 is in the sequence because 2=1*2 is oblong, and sigma(2) = 1+2 = 3 = 2*3/2 is triangular.
		

Crossrefs

Programs

  • Mathematica
    Select[2 Accumulate[Range@10000], MemberQ[Accumulate[Range@10000], DivisorSigma[1, #]] &] (* Michael De Vlieger, Mar 17 2015 *)
  • PARI
    {for (i=1,i=10^6,n=i*(i+1);if(ispolygonal(sigma(n), 3),print(n)))}

A287473 Triangular numbers k such that phi(k) is a square number, where phi(k) is the Euler totient function (A000010).

Original entry on oeis.org

1, 10, 136, 630, 2016, 7875, 9180, 18915, 32896, 37128, 46056, 58311, 66430, 103740, 131841, 198135, 225456, 301476, 323610, 332520, 408156, 499500, 738720, 786885, 839160, 862641, 922761, 924120, 1065070, 1079715, 1183491, 1385280, 1851850, 1906128, 1925703
Offset: 1

Views

Author

Amiram Eldar, May 25 2017

Keywords

Comments

The indices of these triangular numbers are: 1, 4, 16, 35, 63, 125, 135, 194, 256, 272, 303, 341, 364, 455, 513, 629, 671, 776, 804, 815, 903, 999, 1215, 1254, 1295, 1313, 1358, 1359, 1459, 1469, 1538, 1664, 1924, 1952, 1962, ... and their phi values are the squares of: 1, 2, 8, 12, 24, 60, 48, 96, 128, 96, 120, 180, 144, 144, 288, 288, 240, 288, 264, 288, 336, 360, 432, 600, 432, 720, 720, 480, 648, 672, 864, 576, 720, 720, 1080, ...
Similar to A115910, since A115910(n)^2 are squares whose phi is a triangular number.

Examples

			136=16*17/2 is triangular, phi(136)=64=8^2 is a square, thus 136 is in the sequence.
		

Crossrefs

Intersection of A000217 and A039770.

Programs

  • Mathematica
    Select[Accumulate[Range[1000]],IntegerQ[Sqrt[EulerPhi[#]]]&]
  • PARI
    isok(n) = ispolygonal(n, 3) && issquare(eulerphi(n)); \\ Michel Marcus, May 25 2017

A339599 Quadratically perfect numbers: numbers k such that (sigma(k) - 2k)^2 = sigma(k).

Original entry on oeis.org

1, 3, 66, 491536
Offset: 1

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Author

Tomohiro Yamada, Dec 09 2020

Keywords

Comments

Paul Lescot observed that if m is a perfect number and 2m - 1 is prime, then m * (2m - 1) belongs to this sequence.
Moreover, Lescot proved that an even integer k belongs to this sequences if and only if k = m * (2m - 1) for an even perfect number m = 2^(p-1) * (2^p - 1) and 2m - 1 is prime. Among all primes p in A000043 (Mersenne exponents) up to 23209, only p = 2 and 5, which give 66 and 491536 respectively, satisfy this condition.
An even term k in this sequence belongs to A256151 since k = m * (2m - 1) and sigma(k) is square.
Lescot also confirms that there exists no odd integer between 5 and 10^6 in this sequence and conjectures that there exists no further odd term in this sequence.
There exists no further term in this sequence in n <= 2^32.
Question: are there only four terms in this sequence?

Examples

			sigma(1) = 1 = (sigma(1) - 2)^2;
sigma(3) = 4 = (sigma(3) - 6)^2;
sigma(66) = 144 = 12^2 = (sigma(66) - 132)^2.
		

Crossrefs

Programs

  • PARI
    is(k) = {sigma(k) == (sigma(k)-2*k)^2}
Showing 1-5 of 5 results.