A154145 -id:A154145 - cp's OEIS frontend

A154153 Numbers k such that 28 plus the k-th triangular number is a perfect square.

6, 8, 47, 57, 278, 336, 1623, 1961, 9462, 11432, 55151, 66633, 321446, 388368, 1873527, 2263577, 10919718, 13193096, 63644783, 76895001, 370948982, 448176912, 2162049111, 2612166473, 12601345686, 15224821928, 73446025007, 88736765097, 428074804358, 517195768656

Offset: 1

R. J. Mathar, Oct 18 2009

nonn

			6, 8, 47, and 57 are terms:
   6* (6+1)/2 + 28 =  7^2,
   8* (8+1)/2 + 28 =  8^2,
  47*(47+1)/2 + 28 = 34^2,
  57*(57+1)/2 + 28 = 41^2.

F. T. Adams-Watters, SeqFan Discussion, Oct 2009
David A. Corneth, Conjectured formula for a(n)

Cf. A000217, A000290.

Cf. A001108 (0), A006451 (1), A154138 (3), A154139 (4), A154140 (6), A154141 (8), A154142 (9), A154143 (10), A154144 (13), A154145 (15), A154146 (16), A154147 (19), A154148 (21), A154149 (22), A154150(24), A154151 (25), A154151 (26), this sequence (28), A154154 (30).

Join[{6, 8}, Select[Range[0, 10^5], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 28 &]] (* G. C. Greubel, Sep 03 2016 *)

{for (n=0, 10^9, if ( issquare(n*(n+1)\2 + 28), print1(n, ", ") ) );}

{k: 28+k*(k+1)/2 in A000290}.
Conjectures: (Start)
a(n) = +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
G.f.: x*(-6-2*x-3*x^2+2*x^3+7*x^4)/((x-1) * (x^2-2*x-1) * (x^2+2*x-1)).
G.f.: ( 14 + 1/(x-1) + (14+29*x)/(x^2-2*x-1) + (-1-12*x)/(x^2+2*x-1) )/2. (End)
See also the Corneth link - David A. Corneth, Mar 18 2019

a(21)-a(30) from Amiram Eldar, Mar 18 2019

A175032 a(n) is the difference between the n-th triangular number and the next perfect square.

Original entry on oeis.org

0, 0, 1, 3, 6, 1, 4, 8, 0, 4, 9, 15, 3, 9, 16, 1, 8, 16, 25, 6, 15, 25, 3, 13, 24, 36, 10, 22, 35, 6, 19, 33, 1, 15, 30, 46, 10, 26, 43, 4, 21, 39, 58, 15, 34, 54, 8, 28, 49, 0, 21, 43, 66, 13, 36, 60, 4, 28, 53, 79, 19, 45, 72, 9, 36, 64, 93, 26, 55, 85, 15, 45, 76, 3, 34, 66, 99, 22

Offset: 0

Ctibor O. Zizka, Nov 09 2009

All terms are from {0} U A175035. No terms are from A175034.

The sequence consists of ascending runs of length 3 or 4. The first run starts at n = 1 and thereafter the k-th run starts at n = A214858(k - 1). - John Tyler Rascoe, Nov 05 2022

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A001109, A214858.
Cf. A000217, A080819.
Cf. A175034, A175035.
Cf. sequences where a(m)=k: A001108 (0), A006451 (1), A154138 (3), A154139 (4), A154140 (6), A154141 (8), A154142 (9), A154143 (10), A154144 (13), A154145 (15), A154146 (16), A154147 (19), A154148 (21), A154149 (22), A154150(24), A154151 (25), A154151 (26), A154153(28), A154154 (30).

Ceiling[Sqrt[#]]^2-#&/@Accumulate[Range[0,80]] (* Harvey P. Dale, Aug 25 2013 *)

a(n) = my(t=n*(n+1)/2); if (issquare(t), 0, (sqrtint(t)+1)^2 - t); \\ Michel Marcus, Nov 06 2022

a(n) = (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2. - Ctibor O. Zizka, Nov 09 2009

a(n) = A080819(n) - A000217(n). - R. J. Mathar, Aug 24 2010

Erroneous formula variant deleted and offset set to zero by R. J. Mathar, Aug 24 2010

A154137 Greatest number m such that the fractional part of (4/3)^A154133(n) >= 1-(1/m).

Original entry on oeis.org

1, 4, 88, 1228, 187, 4562, 8183, 167378, 35419, 77421, 5593723, 3306511, 83205705, 22413581, 24296709, 35457806, 26593355, 19945016, 80184972, 389460601

Offset: 1

Hieronymus Fischer, Jan 11 2009

			a(3)=88, since 1-(1/89)=0.988764...>fract((4/3)^A154133(3))=fract((4/3)^8)=0.988721...>0.988636...=1-(1/88).

Cf. A153668, A153676, A153708, A153716, A154145, A154153.

Cf. A002379, A064628.

a(n):=floor(1/(1-fract((4/3)^A154133(n)))), where fract(x) = x-floor(x).

a(11)-a(20) from Robert Price, May 10 2012

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A154153 Numbers k such that 28 plus the k-th triangular number is a perfect square.

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A175032 a(n) is the difference between the n-th triangular number and the next perfect square.

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Mathematica

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A154137 Greatest number m such that the fractional part of (4/3)^A154133(n) >= 1-(1/m).

Views

Author

Keywords

Examples

Crossrefs

Formula

Extensions