A036404 -id:A036404 - cp's OEIS frontend

A036409 a(n) = ceiling(n^2/11).

0, 1, 1, 1, 2, 3, 4, 5, 6, 8, 10, 11, 14, 16, 18, 21, 24, 27, 30, 33, 37, 41, 44, 49, 53, 57, 62, 67, 72, 77, 82, 88, 94, 99, 106, 112, 118, 125, 132, 139, 146, 153, 161, 169, 176, 185, 193, 201, 210, 219, 228, 237, 246, 256, 266, 275, 286, 296

Offset: 0

N. J. A. Sloane

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,0,0,1,-2,1).

Cf. A036404, A036405, A036406, A036407, A036408.

seq(ceil(n^2/11),n=0..100); # Robert Israel, Apr 06 2016

Table[Ceiling[n^2/11], {n, 0, 57}] (* Michael De Vlieger, Apr 06 2016 *)
LinearRecurrence[{2,-1,0,0,0,0,0,0,0,0,1,-2,1},{0,1,1,1,2,3,4,5,6,8,10,11,14},60] (* Harvey P. Dale, Aug 29 2021 *)

concat(0, Vec(x*(1+x)*(1-x+x^2)*(1-x+x^2-x^3+x^4)*(1-x^2+x^4) / ((1-x)^3*(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10)) + O(x^50))) \\ Colin Barker, Apr 06 2016

a(n) = +2 a(n-1) -a(n-2) +a(n-11) -2 a(n-12) +a(n-13). - R. J. Mathar, Mar 11 2012
G.f.: x*(1+x)*(1-x+x^2)*(1-x+x^2-x^3+x^4)*(1-x^2+x^4) / ((1-x)^3*(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^10)). - Colin Barker, Apr 06 2016
a(m + 11 k) = a(m) + 11 k^2 + 2 m k. - Robert Israel, Apr 06 2016

A298950 Numbers k such that 5*k - 4 is a square.

Original entry on oeis.org

1, 4, 8, 17, 25, 40, 52, 73, 89, 116, 136, 169, 193, 232, 260, 305, 337, 388, 424, 481, 521, 584, 628, 697, 745, 820, 872, 953, 1009, 1096, 1156, 1249, 1313, 1412, 1480, 1585, 1657, 1768, 1844, 1961, 2041, 2164, 2248, 2377, 2465, 2600, 2692, 2833, 2929, 3076, 3176, 3329, 3433

Offset: 1

Bruno Berselli, Jan 30 2018

a(n) is a member of A140612. Proof: a(n) = n^2 + (n/2-1)^2 for even n, otherwise a(n) = (n-1)^2 + ((n+1)/2)^2; also, a(n) + 1 = (n-1)^2 + (n/2+1)^2 for even n, otherwise a(n) + 1 = n^2 + ((n-3)/2)^2. Therefore, both a(n) and a(n) + 1 belong to A001481.
Primes in sequence are listed in A245042.
Squares in sequence are listed in A081068.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A195162: numbers k such that 5*k + 4 is a square.
Subsequence of A001481, A020668, A036404, A140612.
Cf. A036666, A081068, A106833 (first differences), A245042.

List([1..60], n -> (10*n*(n-1)+(2*n-1)*(-1)^n+9)/8);

[(10*n*(n-1)+(2*n-1)*(-1)^n+9)/8: n in [1..60]];

Table[(10 n (n - 1) + (2 n - 1) (-1)^n + 9)/8, {n, 1, 60}]
LinearRecurrence[{1,2,-2,-1,1},{1,4,8,17,25},60] (* Harvey P. Dale, Sep 16 2022 *)

makelist((10*n*(n-1)+(2*n-1)*(-1)^n+9)/8, n, 1, 60);

Vec((1+x^2)*(1+3*x+x^2)/((1-x)^3*(1+x)^2)+O(x^60))

vector(60, n, nn; (10*n*(n-1)+(2*n-1)*(-1)^n+9)/8)

[(10*n*(n-1)+(2*n-1)*(-1)**n+9)/8 for n in range(1, 60)]

[(10*n*(n-1)+(2*n-1)*(-1)^n+9)/8 for n in (1..60)]

G.f.: x*(1 + x^2)*(1 + 3*x + x^2)/((1 - x)^3*(1 + x)^2).
a(n) = a(1-n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = (10*n*(n-1) + (2*n-1)*(-1)^n + 9)/8.
a(n) = A036666(n) + 1.

cp's OEIS Frontend

A036409 a(n) = ceiling(n^2/11).

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A298950 Numbers k such that 5*k - 4 is a square.

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