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 1-4 of 4 results.

A256151 Triangular numbers n such that sigma(n) is a square number.

Original entry on oeis.org

1, 3, 66, 210, 820, 2346, 4278, 22578, 27966, 32131, 35511, 51681, 53956, 102378, 169653, 173755, 177906, 223446, 241860, 256686, 306153, 310866, 349866, 431056, 434778, 470935, 491536, 512578, 567645, 579426, 688551, 799480, 845650, 893116, 963966, 1031766, 1110795, 1200475, 1613706, 1719585
Offset: 1

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Author

Antonio Roldán, Mar 16 2015

Keywords

Comments

This sequence is the intersection of A000217 and A006532.
The corresponding triangular indices are in A116990. - Michel Marcus, Mar 17 2015

Examples

			3 is in the sequence because 3=2*3/2 is triangular, and sigma(3)=1+3=4=2^2 is square.
		

Crossrefs

Programs

  • Magma
    [n*(n+1) div 2: n in [1..2000] | IsSquare(SumOfDivisors(n*(n+1) div 2))]; // Vincenzo Librandi, Mar 17 2015
  • Mathematica
    Select[Accumulate[Range[0, 2000]], IntegerQ@Sqrt@DivisorSigma[1, #] &] (* Michael De Vlieger, Mar 17 2015 *)
  • PARI
    {for(i=1,2*10^3,n=i*(i+1)/2;if(issquare(sigma(n)),print1(n,", ")))}
    

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)))}

A327830 Numbers m such that the geometric mean of tau(m) and sigma(m) is an integer.

Original entry on oeis.org

1, 7, 17, 22, 30, 31, 71, 94, 97, 115, 119, 127, 138, 154, 164, 165, 199, 210, 214, 217, 232, 241, 260, 265, 318, 337, 343, 374, 382, 449, 497, 510, 513, 517, 527, 577, 647, 658, 668, 679, 682, 705, 745, 759, 805, 862, 881, 889, 894, 930, 966, 967, 996, 1102, 1125
Offset: 1

Views

Author

Bernard Schott, Sep 27 2019

Keywords

Comments

The first 20 terms of this sequence are also the first 20 terms of A144695: m such that sigma(m)/tau(m) is a square. Indeed, if sigma(m)/tau(m) is a square then sigma(m)*tau(m) is also a square, but the converse is false. These counterexamples are in A327831; the first one is a(21) = 232.
The primes p of the form 2*k^2 - 1: 7, 17, 31, 71, ... (A066436) form a subsequence because sigma(p) * tau(p) = (2*k)^2.
Another subsequence consists of the terms m such that sigma(m) and tau(m) are both squares; this occurs when m is the product of two distinct primes p*q, p < q where sigma(m) = (p+1)*(q+1) is a square and tau(m) = 4. The first few terms are 22, 94, 115, 119, 214, ... They are in A256152.

Examples

			sigma(30) = 72 and tau(30) = 8, sigma(30)*tau(30) = 576 = 24^2, hence 30 is a term.
		

Crossrefs

Cf. A000005 (tau), A000203 (sigma), A064840 (tau*sigma).
Cf. A011257 (similar, with phi(m) and sigma(m)), A144695 (sigma(m)/tau(m) is a square), A327831 (sigma(m) * tau(m) is a square but sigma(m)/tau(m) is not an integer).
Subsequences: A066436, A256152.

Programs

  • Magma
    [k:k in [1..1150]| IsSquare(#Divisors(k)*DivisorSigma(1,k))]; // Marius A. Burtea, Sep 27 2019
    
  • Maple
    filter:= s -> issqr(sigma(s)*tau(s)) : select(filter, [$1..2500]);
  • Mathematica
    Select[Range[1000], IntegerQ @ Sqrt[DivisorSigma[0, #] * DivisorSigma[1, #]] &] (* Amiram Eldar, Sep 27 2019 *)
  • PARI
    isok(m) = issquare(numdiv(m)*sigma(m)); \\ Michel Marcus, Sep 27 2019
Showing 1-4 of 4 results.