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.

A271625 a(n) = = 2*(n+1)^2 - 5.

Original entry on oeis.org

3, 13, 27, 45, 67, 93, 123, 157, 195, 237, 283, 333, 387, 445, 507, 573, 643, 717, 795, 877, 963, 1053, 1147, 1245, 1347, 1453, 1563, 1677, 1795, 1917, 2043, 2173, 2307, 2445, 2587, 2733, 2883, 3037, 3195, 3357, 3523, 3693, 3867, 4045, 4227, 4413, 4603, 4797, 4995, 5197, 5403, 5613, 5827
Offset: 1

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Author

Juri-Stepan Gerasimov, Apr 11 2016

Keywords

Comments

Numbers n such that 2*n + 10 is a perfect square.

Crossrefs

Numbers h such that 2*h + k is a perfect square: A294774 (k=-9), A255843 (k=-8), A271649 (k=-7), A093328 (k=-6), A097080 (k=-5), A271624 (k=-4), A051890 (k=-3), A058331 (k=-2), A001844 (k=-1), A001105 (k=0), A046092 (k=1), A056222 (k=2), A142463 (k=3), A054000 (k=4), A090288 (k=5), A268581 (k=6), A059993 (k=7), (-1)*A147973 (k=8), A139570 (k=9), this sequence (k=10), A222182 (k=11), A152811 (k=12), A181510 (k=13), A161532 (k=14), no sequence (k=15).

Programs

  • Magma
    [ 2*n^2 + 4*n - 3: n in [1..60]];
    
  • Magma
    [ n: n in [1..6000] | IsSquare(2*n+10)];
    
  • Mathematica
    Table[2 n^2 + 4 n - 3, {n, 53}] (* Michael De Vlieger, Apr 11 2016 *)
    LinearRecurrence[{3,-3,1},{3,13,27},60] (* Harvey P. Dale, Jun 08 2023 *)
    2*Range[2,60]^2 -5 (* G. C. Greubel, Jan 21 2025 *)
  • PARI
    x='x+O('x^99); Vec(x*(3+4*x-3*x^2)/(1-x)^3) \\ Altug Alkan, Apr 11 2016
    
  • Python
    def A271625(n): return 2*pow(n+1,2) - 5
    print([A271625(n) for n in range(1,61)]) # G. C. Greubel, Jan 21 2025

Formula

G.f.: x*(3 + 4*x - 3*x^2)/(1 - x)^3. - Ilya Gutkovskiy, Apr 11 2016
Sum_{n>=1} 1/a(n) = 13/30 - Pi*cot(sqrt(5/2)*Pi)/(2*sqrt(10)) = 0.5627678459924... . - Vaclav Kotesovec, Apr 11 2016
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(2*x^2 + 6*x - 3) + 3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)
a(n) = 2*A000290(n+1) - 5. - G. C. Greubel, Jan 21 2025

Extensions

Name simplified by G. C. Greubel, Jan 21 2025