A046090 -id:A046090 - cp's OEIS frontend

A081065 Numbers n such that n^2 = (1/3)(n+floor(sqrt(3)nfloor(sqrt(3)n))).

2, 24, 330, 4592, 63954, 890760, 12406682, 172802784, 2406832290, 33522849272, 466913057514, 6503259955920, 90578726325362, 1261598908599144, 17571805994062650, 244743685008277952, 3408839784121828674, 47479013292697323480, 661297346313640700042

Offset: 1

Benoit Cloitre, Apr 15 2003

a(n)/2 gives indices of pentagonal numbers which are also triangular.

a(n) itself gives x-values solving the Diophantine equation 2*x^2 + (x-1)^2 = y^2.

Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A046090, A046174, A046175.

LinearRecurrence[{15,-15,1},{2,24,330},20] (* Harvey P. Dale, Mar 14 2016 *)

Vec(2*(1-3*x)/((1-x)*(1-14*x+x^2)) + O(x^40)) \\ Michel Marcus, Nov 17 2014

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = 14*a(n-1) - a(n-2) - 4. [Sture Sjöstedt, May 02 2011]
G.f.: 2*(1-3*x)/((1-x)*(1-14*x+x^2)). - Bruno Berselli, Nov 11 2011

A257274 Numbers whose square can be written as sum of at least 3 consecutive triangular numbers (A000217).

Original entry on oeis.org

8, 10, 19, 26, 44, 58, 79, 84, 91, 105, 111, 121, 135, 140, 154, 172, 184, 188, 195, 203, 208, 212, 217, 222, 230, 240, 246, 265, 276, 286, 292, 316, 322, 329, 338, 364, 390, 426, 429, 462, 490, 498, 506, 548, 550, 605, 704, 714, 715, 763, 770, 780, 782, 807

Offset: 1

M. F. Hasler, May 02 2015

nonn

Any square can trivially be written as sum of two consecutive triangular numbers T = A000217, since T(n-1) + T(n) = n(n-1)/2 + n(n+1)/2 = n*2n/2 = n^2. But it seems nontrivial to determine the squares that can be written as sum of more than 2 consecutive triangular numbers.
Some of these have two different decompositions of this form, e.g., 286^2 = T(13)+...+T(78) = T(75)+...+T(96), 826^2 = T(13)+...+T(159) = T(43)+...+T(160). What is the sequence of these numbers?
The terms > 2 of A129445, numbers k > 0 such that k^2 is a centered triangular number, form a subsequence: they correspond to k^2 = T(n-2) + T(n-1) + T(n), with n in sequence A129444: numbers n such that centered triangular number A005448(n) = 3n(n-1)/2 + 1 is a perfect square.
Terms > 2 of sequence A075870 also form a subsequence, namely the numbers whose square is the sum of four triangular numbers T(n-3)+...+T(n), with n given by twice the terms > 1 of A046090 or A182435.

			8^2 = T(5)+T(6)+T(7), 10^2 = T(5)+T(6)+T(7)+T(8), 19^2 = T(14)+T(15)+T(16), 26^2 = T(3)+...+T(15), 44^2 = T(13)+...+T(23), ...

Chai Wah Wu, Table of n, a(n) for n = 1..10000

{a=[];(S(n)=binomial(n+2,3)); for(n=1,999,for(k=1,n-3,issquare(S(n)-S(k))&&a=concat(a,sqrtint(S(n)-S(k))))); Set(a)[1..50]}

a(14), a(43)-a(54) from Chai Wah Wu, Jan 20 2016

A114336 Pythagorean triples of nearly isosceles triangle.

Original entry on oeis.org

3, 4, 5, 20, 21, 29, 119, 120, 169, 696, 697, 985, 4059, 4060, 5741, 23660, 23661, 33461, 137903, 137904, 195025, 803760, 803761, 1136689, 4684659, 4684660, 6625109, 27304196, 27304197, 38613965, 159140519, 159140520, 225058681, 927538920, 927538921, 1311738121, 5406093003, 5406093004, 7645370045

Offset: 1

Heinrich Baldauf (heinbald25(AT)web.de), Feb 07 2006

Pythagorean triples of exact isosceles triangles do not exist because 2a^2 = c^2 has no integer solution. a^2 + (a+1)^2 = c^2 are nearly isosceles triangles and give a recursive series.

			119^2 + 120^2 = 169^2.
Triples begin:
  n=1:   3,   4,   5;
  n=2:  20,  21,  29;
  n=3: 119, 120, 169;
  n=4: 696, 697, 985;
  ...

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

C.C. Chen and T.A. Peng, Classroom note: Almost-isosceles right-angled triangles, Australasian Journal of Combinatorics, Volume 11(1995), pp. 263-267. See p. 266.

Cf. A001652, A001653, A046090.

a(1):= 3
c(1):= 5
for n:=2 until 10 step 1
a(n):= 3*a(n-1) + 2*c(n-1) + 1
c(n):= 4*a(n-1) + 3*c(n-1) + 2
print a(n),a(n)+1,c(n)
next n
end

a=Table[(LucasL[2*n+1,2]-2)/4,{n,1,13}];Apply[Join,Map[{#,#+1,Sqrt[2#^2+2#+1]}&,a]] (* Miguel-Ángel Pérez García-Ortega, Nov 06 2024 *)

a^2 + (a+1)^2 = c^2, a(n) = 3a(n-1) + 2c(n-1) + 1, c(n) = 4a(n-1) + 3c(n-1) + 2.

a(n) = (A002315(n) - 1)/2. - Miguel-Ángel Pérez García-Ortega, Nov 06 2024

More terms from Robert Hutchins, Jun 10 2009

More terms from Miguel-Ángel Pérez García-Ortega, Nov 06 2024

A355182 a(n) = t(n) - s(n) where s(n) = n*(n-1)/2 is the sum of the first n nonnegative integers and t(n) is the smallest sum of consecutive integers starting from n such that t(n) > s(n).

Original entry on oeis.org

1, 1, 4, 3, 1, 6, 3, 10, 6, 1, 10, 4, 15, 8, 21, 13, 4, 19, 9, 26, 15, 3, 22, 9, 30, 16, 1, 24, 8, 33, 16, 43, 25, 6, 35, 15, 46, 25, 3, 36, 13, 48, 24, 61, 36, 10, 49, 22, 63, 35, 6, 49, 19, 64, 33, 1, 48, 15, 64, 30, 81, 46, 10, 63, 26, 81, 43, 4, 61, 21, 80, 39, 100, 58, 15, 78, 34, 99

Offset: 1

Andrea La Rosa, Jun 23 2022

nonn

Record high values of a(n)/n approach sqrt(2) and occur at values of n that are terms of A011900; a(A011900(k)) = A046090(k). - Jon E. Schoenfield, Jun 23 2022

It appears that the sequence 1,2,4,5,6,8,... (the largest integer in the t(n) sum) is A288998. - Michel Marcus, Jun 24 2022

			a(6) = -s(6) + t(6):
s(6) is the sum of the first 6 nonnegative integers = 6*5 / 2 = 15.
t(6) is the smallest sum k of consecutive integers starting from n = 6 such that k > s(6) = 15.
The first few sets of consecutive integers starting from 6 are
  {6}, whose elements add up to 6,
  {6, 7}, whose elements add up to 13,
  {6, 7, 8}, whose elements add up to 21,
  {6, 7, 8, 9}, whose elements add up to 30,
  ...
the smallest sum that is > 15 is 21, so t(6) = 21, so a(6) = -15 + 21 = 6.

Cf. A000217, A001477, A093001.

Cf. A288998.

function a(n) {
    var sum = 0;
    for (var i = 0; i < n; i++)
        sum -= i;
    while (sum <= 0)
        sum += i++;
    return sum;
}

a(n) = my(t=0, s=n*(n-1)/2, k=n); until (t > s, t += k; k ++); t-s; \\ Michel Marcus, Jun 24 2022

from math import isqrt
def A355182(n): return ((m:=(isqrt(((k:=n*(n-1))<<3)+1)+1)>>1)*(m+1)>>1)-k # Chai Wah Wu, Jul 14 2022

From Jon E. Schoenfield, Jun 23 2022: (Start)
a(n) = t(n) - s(n) where
   s(n) = n*(n-1)/2,
      j = floor(sqrt(8*n^2 - 8*n + 1)),
      m = ceiling(j/2) - n + 1, and
   t(n) = (m*(m + 2*n - 1))/2. (End)

A076715 Duplicate of A029549.

Original entry on oeis.org

0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680

Offset: 1

dead

a(n) = Binomial(A046090(n), 2). - Mitch Harris, Apr 19 2007

A081097 Numbers n such that n^2= (1/5)(n+floor(sqrt(5)nfloor(sqrt(5)n))).

Original entry on oeis.org

1, 6, 23, 40, 273, 1870, 7343, 12816, 87841, 602070, 2364359, 4126648, 28284465, 193864606, 761316191, 1328767776, 9107509825, 62423800998, 245141449079, 427859097160, 2932589879121, 20100270056686, 78934785287183

Offset: 1

Benoit Cloitre, Apr 15 2003

nonn

Conjecture: if m is an integer and sqrt(m) is irrational, the sequence of n such that n^2 = (1/m)*(n + floor(sqrt(m)*n*floor(sqrt(m)*n))) always satisfies a recurrence of order m. For example: if m=6, the sequence n=b(k) satisfies: b(6k)=4*b(6k-1)+4*b(6k-2)-b(6k-3)-1; b(6k+1)=.... etc.

Cf. A046090.

x=1; y=6; z=23; u=40; for(n=5,50,v=if((n%4-1)*(n%4-2),if(n%4,4*u-z/2-1/2,2*u-z),if(n%4-1,7*u-z-1,7*u-y-1)); x=y; y=z; z=u; u=v; print1(v,","))

a(1)=1; a(2)=6; a(3)=23; a(4)=40; a(4n)=2*a(4n-1)-a(4n-2); a(4n+1)=7*a(4n)-a(4n-2)-1; a(4n+2)=7*a(4n+1)-a(4n-1)-1; a(4n+3)=4*a(4n+2)-a(4n+1)/2-1/2.

Empirical g.f.: x*(x^7+x^6+13*x^5+89*x^4-17*x^3-17*x^2-5*x-1) / ((x-1)*(x^2-4*x-1)*(x^2+4*x-1)*(x^4+18*x^2+1)). - Colin Barker, Jun 24 2013

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A081065 Numbers n such that n^2 = (1/3)*(n+floor(sqrt(3)*n*floor(sqrt(3)*n))).

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A257274 Numbers whose square can be written as sum of at least 3 consecutive triangular numbers (A000217).

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A114336 Pythagorean triples of nearly isosceles triangle.

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A355182 a(n) = t(n) - s(n) where s(n) = n*(n-1)/2 is the sum of the first n nonnegative integers and t(n) is the smallest sum of consecutive integers starting from n such that t(n) > s(n).

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A076715 Duplicate of A029549.

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A081097 Numbers n such that n^2= (1/5)*(n+floor(sqrt(5)*n*floor(sqrt(5)*n))).

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A081065 Numbers n such that n^2 = (1/3)(n+floor(sqrt(3)nfloor(sqrt(3)n))).

A081097 Numbers n such that n^2= (1/5)(n+floor(sqrt(5)nfloor(sqrt(5)n))).