A008865 -id:A008865 - cp's OEIS frontend

A266956 Numbers m such that 9*m+7 is a square.

1, 2, 18, 21, 53, 58, 106, 113, 177, 186, 266, 277, 373, 386, 498, 513, 641, 658, 802, 821, 981, 1002, 1178, 1201, 1393, 1418, 1626, 1653, 1877, 1906, 2146, 2177, 2433, 2466, 2738, 2773, 3061, 3098, 3402, 3441, 3761, 3802, 4138, 4181, 4533, 4578, 4946, 4993, 5377, 5426

Offset: 1

Bruno Berselli, Jan 07 2016

Equivalently, numbers of the form h*(9*h+8)+1, where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ...
Also, integer values of k*(k+8)/9 plus 1.
It is easy to see that the Diophantine equation 9*x+3*j+1 = y^2 has infinitely many solutions in integers (x,y) for any j in Z. It follows a table with j = -5..5:
...
j = -5, x:  2, 7, 15, 30, 46, 71,  95, 130, 162, 207, 247, ...
j = -4, x:  3, 4, 20, 23, 55, 60, 108, 115, 179, 188, 268, ...
j = -3, x:  1, 8, 12, 33, 41, 76,  88, 137, 153, 216, 236, ...
j = -2, x:  1, 6, 14, 29, 45, 70,  94, 129, 161, 206, 246, ...
j = -1, x:  2, 3, 19, 22, 54, 59, 107, 114, 178, 187, 267, ...
j =  0, x:  0, 7, 11, 32, 40, 75,  87, 136, 152, 215, 235, ... (A132355)
j =  1, x:  0, 5, 13, 28, 44, 69,  93, 128, 160, 205, 245, ... (A185039)
j =  2, x:  1, 2, 18, 21, 53, 58, 106, 113, 177, 186, 266, ... (A266956)
j =  3, x: -1, 6, 10, 31, 39, 74,  86, 135, 151, 214, 234, ... (A266957)
j =  4, x: -1, 4, 12, 27, 43, 68,  92, 127, 159, 204, 244, ... (A266958)
j =  5, x:  0, 1, 17, 20, 52, 57, 105, 112, 176, 185, 265, ... (A218864)
...
The general closed form of these sequences is:
b(n,j) = (18*(n-1)*n + s(j)*(2*n-1)*(-1)^n + t(j))/8, where s(j) = 6*(-j) + 18*floor(j/3) - (-1)^floor(2*(j+1)/3) + 6 and t(j) = 4*(-j) + 4*floor((j+1)/3) + 5.
a(2m) - a(2m-1) gives the odd numbers (A005408); a(2m+1) - a(2m) gives the multiples of 16 (A008598).

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. numbers m such that 9*m+i: A132355 (i=1), A185039 (i=4), this sequence (i=7), A005563 (i=9), A266957 (i=10), A266958 (i=13), A218864 (i=16), A008865 (i=18, without -2).
Cf. A156638: square roots of 9*a(n)+7.
Cf. A005408, A008598.

Magma
```
[n: n in [0..6000] | IsSquare(9*n+7)];
```

[(18*(n-1)*n-7*(2*n-1)*(-1)^n+1)/8: n in [1..50]];

Select[Range[0, 6000], IntegerQ[Sqrt[9 # + 7]] &]
Table[(18 (n - 1) n - 7 (2 n - 1) (-1)^n + 1)/8, {n, 1, 50}]

for(n=0, 6000, if(issquare(9*n+7), print1(n, ", ")))

vector(50, n, n; (18*(n-1)*n-7*(2*n-1)*(-1)^n+1)/8)

from gmpy2 import is_square
[n for n in range(6000) if is_square(9*n+7)]

[(18*(n-1)*n-7*(2*n-1)*(-1)**n+1)/8 for n in range(1, 60)]

[n for n in (0..6000) if is_square(9*n+7)]

[(18*(n-1)*n-7*(2*n-1)*(-1)^n+1)/8 for n in (1..50)]

G.f.: x*(1 + x + 14*x^2 + x^3 + x^4)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (18*(n-1)*n - 7*(2*n-1)*(-1)^n + 1)/8.
a(n) = A218864(n) + 1.

A270109 a(n) = n^3 + (n+1)*(n+2).

Original entry on oeis.org

2, 7, 20, 47, 94, 167, 272, 415, 602, 839, 1132, 1487, 1910, 2407, 2984, 3647, 4402, 5255, 6212, 7279, 8462, 9767, 11200, 12767, 14474, 16327, 18332, 20495, 22822, 25319, 27992, 30847, 33890, 37127, 40564, 44207, 48062, 52135, 56432, 60959, 65722, 70727, 75980, 81487, 87254

Offset: 0

Bruno Berselli, Mar 11 2016, at the suggestion of Giuseppe Amoruso in BASE Cinque forum

For n>1, many consecutive terms of the sequence are generated by floor(sqrt(n^2 + 2)^3) + n^2 + 2.
It appears that this is a subsequence of A000037 (the nonsquares).
The primes in the sequence belong to A045326.
Inverse binomial transform is 2, 5, 8, 6, 0, 0, 0, ... (0 continued).

Bruno Berselli, Table of n, a(n) for n = 0..1000
MathsSmart, Number pattern and Puzzle - 7, 20, 47, 94, 167.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Subsequence of A001651, A047212.
Cf. A000037, A045326.
Cf. A027444: numbers of the form n^3+n*(n+1); A085490: numbers of the form n^3+(n-1)*n.
Cf. A008865: numbers of the form n+(n+1)*(n+2); A130883: numbers of the form n^2+(n+1)*(n+2).

Magma
```
[n^3+(n+1)*(n+2): n in [0..50]];
```

Table[n^3 + (n + 1) (n + 2), {n, 0, 50}]

Maxima
```
makelist(n^3+(n+1)*(n+2), n, 0, 50);
```
PARI
```
vector(50, n, n--; n^3+(n+1)*(n+2))
```
Sage
```
[n^3+(n+1)*(n+2) for n in (0..50)]
```

O.g.f.: (2 - x + 4*x^2 + x^3)/(1 - x)^4.
E.g.f.: (2 + x)*(1 + x)^2*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.
a(n+h) - a(n) + a(n-h) = n^3 + n^2 + (6*h^2+3)*n + (2*h^2+2) for any h. This identity becomes a(n) = n^3 + n^2 + 3*n + 2 if h=0.
a(h*a(n) + n) = (h*a(n))^3 + (3*n+1)*(h*a(n))^2 + (3*n^2+2*n+3)*(h*a(n)) + a(n) for any h, therefore a(h*a(n) + n) is always a multiple of a(n).
a(n) + a(-n) = 2*A059100(n) = A255843(n).
a(n) - a(-n) = 4*A229183(n).

A143053 A143051(A143051(n)).

Original entry on oeis.org

0, 1, 3, 4, 2, 8, 9, 5, 6, 7, 15, 16, 10, 11, 12, 13, 14, 24, 25, 17, 18, 19, 20, 21, 22, 23, 35, 36, 26, 27, 28, 29, 30, 31, 32, 33, 34, 48, 49, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 63, 64, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 80, 81, 65, 66, 67, 68, 69, 70

Offset: 0

Reinhard Zumkeller, Jul 20 2008

nonn

Permutation of the natural numbers, inverse: A143054;
a(A000290(n)) = A008865(n) for n>1;
a(A002522(n)) = A000290(n+1) for n>0.

Index entries for sequences that are permutations of the natural numbers

A156140 Accumulation of Stern's diatomic series: a(0)=-1, a(1)=0, and a(n+1) = (2e(n)+1)*a(n) - a(n-1) for n > 1, where e(n) is the highest power of 2 dividing n.

Original entry on oeis.org

-1, 0, 1, 3, 2, 7, 5, 8, 3, 13, 10, 17, 7, 18, 11, 15, 4, 21, 17, 30, 13, 35, 22, 31, 9, 32, 23, 37, 14, 33, 19, 24, 5, 31, 26, 47, 21, 58, 37, 53, 16, 59, 43, 70, 27, 65, 38, 49, 11, 50, 39, 67, 28, 73, 45, 62, 17, 57, 40, 63, 23, 52, 29, 35, 6, 43, 37, 68, 31, 87, 56, 81, 25, 94, 69

Offset: 0

Arie Werksma (Werksma(AT)Tiscali.nl), Feb 04 2009

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Cf. A002487, A007814.
From Yosu Yurramendi, Mar 09 2018: (Start)
a(2^m +  0) = A000027(m),   m >= 0.
a(2^m +  1) = A002061(m+2), m >= 1.
a(2^m +  2) = A002522(m),   m >= 2.
a(2^m +  3) = A033816(m-1), m >= 2.
a(2^m +  4) = A002061(m),   m >= 2.
a(2^m +  5) = A141631(m),   m >= 3.
a(2^m +  6) = A084849(m-1), m >= 3.
a(2^m +  7) = A056108(m-1), m >= 3.
a(2^m +  8) = A000290(m-1), m >= 3.
a(2^m +  9) = A185950(m-1), m >= 4.
a(2^m + 10) = A144390(m-1), m >= 4.
a(2^m + 12) = A014106(m-2), m >= 4.
a(2^m + 16) = A028387(m-3), m >= 4.
a(2^m + 18) = A250657(m-4), m >= 5.
a(2^m + 20) = A140677(m-3), m >= 5.
a(2^m + 32) = A028872(m-2), m >= 5.
a(2^m -  1) = A005563(m-1), m >= 0.
a(2^m -  2) = A028387(m-2), m >= 2.
a(2^m -  3) = A033537(m-2), m >= 2.
a(2^m -  4) = A008865(m-1), m >= 3.
a(2^m -  7) = A140678(m-3), m >= 3.
a(2^m -  8) = A014209(m-3), m >= 4.
a(2^m - 16) = A028875(m-2), m >= 5.
a(2^m - 32) = A108195(m-5), m >= 6.
(End)

A156140 := proc(n)
    option remember ;
    if n <= 1 then
        n-1 ;
    else
        (2*A007814(n-1)+1)*procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A156140(n),n=0..80) ; # R. J. Mathar, Mar 14 2009

Fold[Append[#1, (2 IntegerExponent[#2, 2] + 1) #1[[-1]] - #1[[-2]] ] &, {-1, 0}, Range[73]] (* Michael De Vlieger, Mar 09 2018 *)

first(n)=my(v=vector(n+1)); v[1]=-1; v[2]=0; for(k=1,n-1,v[k+2]=(2*valuation(k,2)+1)*v[k+1] - v[k]); v \\ Charles R Greathouse IV, Apr 05 2016

fusc(n)=my(a=1, b=0); while(n>0, if(bitand(n, 1), b+=a, a+=b); n>>=1); b
a(n)=my(m=1,s,t); if(n==0, return(-1)); while(n%2==0, s+=fusc(n>>=1)); while(n>1, t=logint(n,2); n-=2^t; s+=m*fusc(n)*(t^2+t+1); m*=-t); m*(n-1) + s \\ Charles R Greathouse IV, Dec 13 2016

a <- c(0,1)
maxlevel <- 6 # by choice
for(m in 1:maxlevel) {
  a[2^(m+1)] <- m + 1
  for(k in 1:(2^m-1)) {
    r <- m - floor(log2(k)) - 1
    a[2^r*(2*k+1)] <- a[2^r*(2*k)] + a[2^r*(2*k+2)]
}}
a
# Yosu Yurramendi, May 08 2018

Let b(n) = A002487(n), Stern's diatomic series.
a(n+1)*b(n) - a(n)*b(n+1) = 1 for n >= 0.
a(2n+1) = a(n) + a(n+1) + b(n) + b(n+1) for n >= 0.
a(2n) = a(n) + b(n) for n >= 0.
a(2^n + k) = -n*a(k) + (n^2 + n + 1)*b(k) for 0 <= k <= 2^n.
b(2^n + k) = -a(k) + (n + 1)*b(k) for 0 <= k <= 2^n.
a(2^m + k) = b(2^m+k)*m + b(k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Mar 09 2018
a(2^(m+1)+2^m+1) = 2*m+1, m >= 0. - Yosu Yurramendi, Mar 09 2018
From Yosu Yurramendi, May 08 2018: (Start)
a(2^m) = m, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r = m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m.
(End)

A241199 Numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2.

Original entry on oeis.org

14, 19, 31, 38, 54, 63, 83, 94, 118, 131, 159, 174, 206, 223, 259, 278, 318, 339, 383, 406, 454, 479, 531, 558, 614, 643, 703, 734, 798, 831, 899, 934, 1006, 1043, 1119, 1158, 1238, 1279, 1363, 1406, 1494, 1539, 1631, 1678, 1774, 1823, 1923, 1974, 2078, 2131

Offset: 1

T. D. Noe, Apr 17 2014

From Robert Israel, Apr 28 2015: (Start)
Numbers n >= 14 such that 3*n + 7 is a square.
This is because
C(n,i+3) - 3*C(n,i+2) + 3*C(n,i+1) - C(n,i) = n!/((n-i)!*(i+3)!) * g(n,i)
where g(n,i) = (n-3-2*i) * ((n-3-2*i)^2 - 3*n - 7). (End)

			Binomial(14,k) = (1, 14, 91, 364, 1001, 2002, 3003, 3432) for k = 0..7. The 4 terms beginning with 91 equal 182 - 273*x + 182*x^2 for x = 1..4.

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Sequence A241200 gives the position of the first of the 4 terms. Sequence A008865 gives the terms greater than 2 for which 3 consecutive terms satisfy a linear relation.
A014206 is a related sequence. - Avi Friedlich, Apr 28 2015
Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).

map(k -> (3*k^2+8*k+3,3*k^2+10*k+6),[$1..100]); # Robert Israel, Apr 28 2015

Select[Range[2500], MemberQ[Differences[Binomial[#, Range[0, #/2]], 3], 0] &]
LinearRecurrence[{1,2,-2,-1,1},{14,19,31,38,54},50] (* Harvey P. Dale, Oct 29 2017 *)

Vec(-x*(6*x^4-3*x^3-16*x^2+5*x+14)/((x-1)^3*(x+1)^2) + O(x^100)) \\ Colin Barker, Apr 29 2015

a(n)=(6*n^2+42*n+55-(-1)^n*(2*n+7))/8 \\ Charles R Greathouse IV, Apr 15 2016

a(n) = (55-7*(-1)^n-2*(-21+(-1)^n)*n+6*n^2)/8. G.f.: -x*(6*x^4-3*x^3-16*x^2+5*x+14) / ((x-1)^3*(x+1)^2). - Colin Barker, Apr 18 2014 and Apr 29 2015
The terms appear to satisfy a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), with initial terms 14, 19, 31, 38, 54. - T. D. Noe, Apr 18 2014
Numbers are of the form A200182(3n+1) and A200182(3n-1). - Avi Friedlich, Apr 25 2015
a(2*k-1) = 3*k^2 + 8*k + 3, a(2*k) = 3*k^2 + 10*k + 6. - Robert Israel, Apr 28 2015

A163257 An interspersion: the order array of the even-numbered columns (after swapping the first two rows) of the double interspersion at A161179.

Original entry on oeis.org

1, 5, 2, 11, 6, 3, 19, 12, 8, 4, 29, 20, 15, 10, 7, 41, 30, 24, 18, 14, 9, 55, 42, 35, 28, 23, 17, 13, 71, 56, 48, 40, 34, 27, 22, 16, 89, 72, 63, 54, 47, 39, 33, 26, 21, 109, 90, 80, 70, 62, 53, 46, 38, 32, 25, 131, 110, 99, 88, 79, 69, 61, 52, 45, 37, 31, 155, 132, 120, 108

Offset: 1

Clark Kimberling, Jul 24 2009

A permutation of the natural numbers.
Beginning at row 6, the columns obey a 3rd-order recurrence:
c(n)=c(n-1)+c(n-2)-c(n-3)+1.
Except for initial terms, the first seven rows are A028387, A002378, A005563, A028552, A008865, A014209, A028873, and the first column, A004652.

			Corner:
1....5...11...19
2....6...12...20
3....8...15...24
4...10...18...28
The double interspersion A161179 begins thus:
1....4....7...12...17...24
2....3....8...11...18...23
5....6...13...16...25...30
9...10...19...22...33...38
Expel the odd-numbered columns and then swap rows 1 and 2, leaving
3....11...23...39
4....12...24...40
6....16...30...48
10...22...38...58
Then replace each of those numbers by its rank when all the numbers are jointly ranked.

Clark Kimberling, Doubly interspersed sequences, double interspersions and fractal sequences, The Fibonacci Quarterly 48 (2010) 13-20.

Cf. A163253, A163254, A163255, A163256, A163258.

Let S(n,k) denote the k-th term in the n-th row. Four cases:
S(1,k)=k^2+k-1
S(2,k)=k^2+k
if n>1 is odd, then S(n,k)=k^2+(n-1)k+(n-1)(n-3)/4
if n>2 is even, then S(n,k)= k^2+(n-1)k+n(n-4)/4.

A164314 Largest prime factor of n^2 - 2.

Original entry on oeis.org

2, 7, 7, 23, 17, 47, 31, 79, 7, 17, 71, 167, 97, 223, 127, 41, 23, 359, 199, 439, 241, 31, 41, 89, 337, 727, 23, 839, 449, 137, 73, 1087, 577, 1223, 647, 1367, 103, 31, 47, 73, 881, 1847, 967, 17, 151, 2207, 1151, 2399, 1249, 113, 193, 401, 47, 3023, 1567, 191

Offset: 2

Vladimir Joseph Stephan Orlovsky, Aug 12 2009

nonn

Alois P. Heinz, Table of n, a(n) for n = 2..10000

Cf. A002583, A014442, A069902.

a:= n-> max(numtheory[factorset](n^2-2)):
seq(a(n), n=2..60);  # Alois P. Heinz, Jul 22 2017

Table[FactorInteger[n^2 - 2][[-1, 1]], {n, 2, 57}] (* Michael De Vlieger, Jul 22 2017 *)

a(n) = vecmax(factor(n^2-2)[,1]); \\ Michel Marcus, Jul 22 2017

a(n) = A006530(A008865(n)).

Offset corrected by R. J. Mathar, Aug 21 2009

A220508 T(n,k) = n^2 + k if k <= n, otherwise T(n,k) = k*(k + 2) - n; square array T(n,k) read by ascending antidiagonals (n >= 0, k >= 0).

Original entry on oeis.org

0, 1, 3, 4, 2, 8, 9, 5, 7, 15, 16, 10, 6, 14, 24, 25, 17, 11, 13, 23, 35, 36, 26, 18, 12, 22, 34, 48, 49, 37, 27, 19, 21, 33, 47, 63, 64, 50, 38, 28, 20, 32, 46, 62, 80, 81, 65, 51, 39, 29, 31, 45, 61, 79, 99, 100, 82, 66, 52, 40, 30, 44, 60, 78, 98, 120

Offset: 0

Omar E. Pol, Feb 09 2013

This sequence consists of 0 together with a permutation of the natural numbers. The nonnegative integers (A001477) are arranged in the successive layers from T(0,0) = 0. The n-th layer start with T(n,1) = n^2. The n-th layer is formed by the first n+1 elements of row n and the first n elements in increasing order of the column n.
The first antidiagonal is formed by odd numbers: 1, 3. The second antidiagonal is formed by even numbers: 4, 2, 8. The third antidiagonal is formed by odd numbers: 9, 5, 7, 15. And so on.
It appears that in the n-th layer there is at least a prime number <= g and also there is at least a prime number > g, where g is the number on the main diagonal, the n-th oblong number A002378(n), if n >= 1.

			The second layer is [4, 5, 6, 7, 8] which looks like this:
  .  .  8
  .  .  7,
  4, 5, 6,
Square array T(0,0)..T(10,10) begins:
    0,   3,   8,  15,  24,  35,  48,  63,  80,  99, 120,...
    1,   2,   7,  14,  23,  34,  47,  62,  79,  98, 119,...
    4,   5,   6,  13,  22,  33,  46,  61,  78,  97, 118,...
    9,  10,  11,  12,  21,  32,  45,  60,  77,  96, 117,...
   16,  17,  18,  19,  20,  31,  44,  59,  76,  95, 118,...
   25,  26,  27,  28,  29,  30,  43,  58,  75,  94, 117,...
   36,  37,  38,  39,  40,  41,  42,  57,  74,  93, 114,...
   49,  50,  51,  52,  53,  54,  55,  56,  73,  92, 113,...
   64,  65,  66,  67,  68,  69,  70,  71,  72,  91, 112,...
   81,  82,  83,  84,  85,  86,  87,  88,  89,  90, 111,...
  100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110,...
  ...

Index entries for sequences that are permutations of the natural numbers

Column 1 is A000290. Main diagonal is A002378. Column 2 is essentially A002522. Row 1 is A005563. Row 2 gives the absolute terms of A008865.

Cf. A000005, A000040, A002061, A002620, A014206, A014209, A027688, A028387, A028552, A060736, A220516.

From Petros Hadjicostas, Mar 10 2021: (Start)
T(n,k) = (A342354(n,k) - 1)/2.
O.g.f.: (x^4*y^3 + 3*x^3*y^4 + x^4*y^2 - 10*x^3*y^3 - x^2*y^4 + 3*x^3*y^2 + x^2*y^3 - 4*x^3*y + 8*x^2*y^2 + 3*x^2*y + x*y^2 + x^2 - 10*x*y - y^2 + x + 3*y)/((1 - x)^3*(1 - y)^3*(1 - x*y)^2). (End)

Name edited by Petros Hadjicostas, Mar 10 2021

A230584 Either two less than a square or two more than a square.

Original entry on oeis.org

2, 3, 6, 7, 11, 14, 18, 23, 27, 34, 38, 47, 51, 62, 66, 79, 83, 98, 102, 119, 123, 142, 146, 167, 171, 194, 198, 223, 227, 254, 258, 287, 291, 322, 326, 359, 363, 398, 402, 439, 443, 482, 486, 527, 531, 574, 578, 623, 627, 674, 678, 727, 731, 782, 786, 839, 843, 898, 902, 959, 963

Offset: 1

Ralf Stephan, Oct 24 2013

Numbers n such that the polynomial x^4 - n*x^2 + 1 is reducible.
The corresponding factorizations are (x^2 + k*x - 1)*(x^2 - k*x - 1) == x^4 - (k^2 + 2)*x^2 + 1 and (x^2 + k*x + 1)*(x^2 - k*x + 1) == x^4 - (k^2 - 2)*x^2 + 1. - Joerg Arndt, Feb 07 2015
Union of A008865 and A059100.
For k > 1: a(2*k+1) - a(2*k) = 4 and a(2*k) - a(2*k-1) = k - 1; for n > 4: a(n) - a(n-2) = 2*floor(n/2) + 1 = A109613(n). - Reinhard Zumkeller, Feb 10 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008865, A059100, A000290, A109613.

import Data.List (transpose)
a230584 n = a230584_list !! (n-1)
a230584_list = 2 : 3 : concat
               (transpose [drop 2 a059100_list, drop 2 a008865_list])
-- Reinhard Zumkeller, Feb 10 2015

PARI
```
is(n)=issquare(n-2)||issquare(n+2)
```

A230584_vec(N)=Vec((2+x-x^2-x^3+2*x^5-x^6)/((1-x)^3*(1+x)^2)+O(x^N)) \\ M. F. Hasler, Oct 26 2013

From Colin Barker, Oct 24 2013: (Start)
  a(n) = (5-13*(-1)^n+2*(3+(-1)^n)*n+2*n^2)/8 for n>2.
  a(n) = (n^2+4*n-4)/4 for n>2 and even.
  a(n) = (n^2+2*n+9)/4 for n>2 and odd.
  a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5) for n>7.
  G.f.: x*(x^6-2*x^5+x^3+x^2-x-2) / ((x-1)^3*(x+1)^2). (End)
After the first two terms 0^2+2 = 2^2-2, 1^2+2, the squares are sufficiently spaced to ensure that the sequence continues 2^2+2, 3^2-2, 3^2+2, 4^2-2, 4^2+2,..., i.e., a(2n-1) = n^2+2, a(2n)=(n+1)^2-2. - M. F. Hasler, Oct 26 2013

A241200 For the n in A241199, the index of the first of 4 terms in binomial(n,k) that satisfy a quadratic relation.

Original entry on oeis.org

2, 4, 9, 12, 19, 23, 32, 37, 48, 54, 67, 74, 89, 97, 114, 123, 142, 152, 173, 184, 207, 219, 244, 257, 284, 298, 327, 342, 373, 389, 422, 439, 474, 492, 529, 548, 587, 607, 648, 669, 712, 734, 779, 802, 849, 873, 922, 947, 998, 1024, 1077, 1104, 1159, 1187

Offset: 1

T. D. Noe, Apr 17 2014

This value of k appears to approach n/2 as n grows larger.

			Binomial(14,k) = (1, 14, 91, 364, 1001, 2002, 3003, 3432) for k = 0..7. The 4 quadratic terms begin at k = 2.

Colin Barker, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008865 (binomial(n,k) has 3 consecutive terms in a linear relation).
Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).
Cf. A241199 (the values of n).

t = {}; Do[b = Binomial[n, Range[0, n/2]]; d = Differences[b, 3]; If[MemberQ[d, 0], AppendTo[t, Position[d, 0, 1, 1][[1, 1]] - 1]], {n, 3000}]; t
LinearRecurrence[{1,2,-2,-1,1},{2,4,9,12,19},60] (* Harvey P. Dale, Dec 18 2022 *)

Vec(x*(x^2-2)*(x^2+x+1)/((x-1)^3*(x+1)^2) + O(x^100)) \\ Colin Barker, Apr 29 2015

a(n) = (-11-5*(-1)^n-2*(-15+(-1)^n)*n+6*n^2)/16. G.f.: x*(x^2-2)*(x^2+x+1) / ((x-1)^3*(x+1)^2). - Colin Barker, Apr 18 2014 and Apr 29 2015

The terms appear to satisfy a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), with initial terms 2, 4, 9, 12, 19. - T. D. Noe, Apr 18 2014

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