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 1-2 of 2 results.

A014493 Odd triangular numbers.

Original entry on oeis.org

1, 3, 15, 21, 45, 55, 91, 105, 153, 171, 231, 253, 325, 351, 435, 465, 561, 595, 703, 741, 861, 903, 1035, 1081, 1225, 1275, 1431, 1485, 1653, 1711, 1891, 1953, 2145, 2211, 2415, 2485, 2701, 2775, 3003, 3081, 3321, 3403, 3655, 3741, 4005, 4095, 4371, 4465, 4753, 4851
Offset: 1

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Author

Keywords

Comments

Odd numbers of the form n*(n+1)/2.
For n such that n(n+1)/2 is odd see A042963 (congruent to 1 or 2 mod 4).
Even central polygonal numbers minus 1. - Omar E. Pol, Aug 17 2011
Odd generalized hexagonal numbers. - Omar E. Pol, Sep 24 2015

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 68.

Crossrefs

Programs

  • GAP
    List([1..50], n -> (2*n-1)*(2*n-1-(-1)^n)/2); # G. C. Greubel, Feb 09 2019
    
  • Magma
    [(2*n-1)*(2*n-1-(-1)^n)/2: n in [1..50]]; // Vincenzo Librandi, Aug 18 2011
    
  • Maple
    [(2*n-1)*(2*n-1-(-1)^n)/2$n=1..50]; # Muniru A Asiru, Mar 10 2019
  • Mathematica
    Select[ Table[n(n + 1)/2, {n, 93}], OddQ[ # ] &] (* Robert G. Wilson v, Nov 05 2004 *)
    LinearRecurrence[{1,2,-2,-1,1},{1,3,15,21,45},50] (* Harvey P. Dale, Jun 19 2011 *)
  • PARI
    a(n)=(2*n-1)*(2*n-1-(-1)^n)/2 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Python
    def A014493(n): return ((n<<1)-1)*(n-(n&1^1)) # Chai Wah Wu, Feb 12 2023
  • Sage
    [(2*n-1)*(2*n-1-(-1)^n)/2 for n in (1..50)] # G. C. Greubel, Feb 09 2019
    

Formula

From Ant King, Nov 17 2010: (Start)
a(n) = (2*n-1)*(2*n - 1 - (-1)^n)/2.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). (End)
G.f.: x*(1 + 2*x + 10*x^2 + 2*x^3 + x^4)/((1+x)^2*(1-x)^3). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009
a(n) = A000217(A042963(n)). - Reinhard Zumkeller, Feb 14 2012, Oct 04 2004
a(n) = A193868(n) - 1. - Omar E. Pol, Aug 17 2011
Let S = Sum_{n>=0} x^n/a(n), then S = Q(0) where Q(k) = 1 + x*(4*k+1)/(4*k + 3 - x*(2*k+1)*(4*k+3)^2/(x*(2*k+1)*(4*k+3) + (4*k+5)*(2*k+3)/Q(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 27 2013
E.g.f.: (2*x^2+x+1)*cosh(x)+x*(2*x-1)*sinh(x)-1. - Ilya Gutkovskiy, Apr 24 2016
Sum_{n>=1} 1/a(n) = Pi/2 (A019669). - Robert Bilinski, Jan 20 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2). - Amiram Eldar, Mar 06 2022

Extensions

More terms from Erich Friedman

A067589 Numbers k such that A067588(k) is an odd number.

Original entry on oeis.org

1, 5, 7, 15, 35, 51, 57, 77, 117, 145, 155, 187, 247, 287, 301, 345, 425, 477, 495, 551, 651, 715, 737, 805, 925, 1001, 1027, 1107, 1247, 1335, 1365, 1457, 1617, 1717, 1751, 1855, 2035, 2147, 2185, 2301, 2501, 2625, 2667, 2795, 3015, 3151, 3197, 3337
Offset: 1

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Author

Naohiro Nomoto, Jan 31 2002

Keywords

Comments

The terms are exactly the odd pentagonal numbers; that is, they are all the odd numbers of the form k*(3*k-1)/2 where k is an integer. - James Sellers, Jun 09 2007
Apparently groups of two odd pentagonal numbers (A000326, A014632) followed by two odd 2nd pentagonal numbers (A005449), which leads to the conjectured generating function x*(x^2+4*x+1)*(x^4-2*x^3+4*x^2-2*x+1)/((x^2+1)^2*(1-x)^3). - R. J. Mathar, Jul 26 2009
Odd generalized pentagonal numbers. - Omar E. Pol, Aug 19 2011
From Peter Bala, Jan 10 2025: (Start)
The sequence terms are the exponents in the expansion of Sum_{n >= 0} x^(2*n+1)/(Product_{k = 1..2*n+1} 1 + x^(2*k+1)) = x + x^5 - x^7 - x^15 + x^35 + x^51 - x^57 - x^77 + + - - ... (follows from Berndt et al., Theorem 3.3). Cf. A193828.
For positive integer m, define b_m(n) = Sum_{k = 1..n} k^(2*m+1)*A000009(k)*A000009(n-k). We conjecture that
i) for odd n, b(n)/ n is an integer
ii) b(2*n)/n is an integer, which is odd iff n is a member of this sequence.
Cf. A067567. (End)

Crossrefs

Programs

  • Mathematica
    With[{nn=50},Sort[Select[Table[(n(3n-1))/2,{n,-nn,nn}],OddQ]]] (* Harvey P. Dale, Feb 16 2014 *)

Formula

Sum_{n>=1} 1/a(n) = Pi/2. - Amiram Eldar, Aug 18 2022

Extensions

Corrected by T. D. Noe, Oct 25 2006
Showing 1-2 of 2 results.