A201279 -id:A201279 - cp's OEIS frontend

A048395 Sum of consecutive nonsquares.

0, 5, 26, 75, 164, 305, 510, 791, 1160, 1629, 2210, 2915, 3756, 4745, 5894, 7215, 8720, 10421, 12330, 14459, 16820, 19425, 22286, 25415, 28824, 32525, 36530, 40851, 45500, 50489, 55830, 61535, 67616, 74085, 80954, 88235, 95940, 104081

Offset: 0

Patrick De Geest, Mar 15 1999

Relationship with natural numbers: a(4) = (first term + last term)*n = (10+15)*3 = (25)*3 = 75; a(5) = (17+24)*4 = (41)*4 = 164; ...
Also (X*Y*Z)/(X+Y+Z) of primitive Pythagorean triples (X,Y,Z=Y+1) as described in A046092 and A001844. - Lambert Herrgesell (zero815(AT)googlemail.com), Dec 13 2005
First differences are in A201279. - J. M. Bergot, Jun 22 2013 [Corrected by Omar E. Pol, Dec 26 2021]

			Between 3^2 and 4^2 we have 10+11+12+13+14+15 which is 75 or a(4).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for sequences related to sums of squares

Cf. A048396, A048397, A046092, A001844.

Cf. A001844, A046092, A086849, A199771, A201279.

a048395 0 = 0
a048395 n = a199771 (2 * n)  -- Reinhard Zumkeller, Oct 26 2015

Table[n(1+2*n(1+n)),{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,5,26,75},40] (* Harvey P. Dale, Nov 01 2013 *)

v0=[1,0,1]; M=[1, 2, 2; -2, -1, -2; 2, 2, 3];
g(v)=v[1]*v[2]*v[3]/(v[1]+v[2]+v[3]);
a(n)=g(v0*M^n);
for(i=0,50,print1(a(i),", ")) \\ Lambert Herrgesell (zero815(AT)googlemail.com), Dec 13 2005

a(n) = 2*n^3 + 2*n^2 + n.
a(n) = Sum_{j=0..n} ((n+j+2)^2 - j^2 + 1). - Zerinvary Lajos, Sep 13 2006
O.g.f.: x(x+5)(1+x)/(1-x)^4. - R. J. Mathar, Jun 12 2008
a(n) = A199771(2*n) for n > 0. - Reinhard Zumkeller, Nov 23 2011
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=5, a(2)=26, a(3)=75. - Harvey P. Dale, Nov 01 2013
E.g.f.: exp(x)*x*(5 + 8*x + 2*x^2). - Stefano Spezia, Jun 25 2022

A136392 a(n) = 6n^2 - 10n + 5.

Original entry on oeis.org

1, 9, 29, 61, 105, 161, 229, 309, 401, 505, 621, 749, 889, 1041, 1205, 1381, 1569, 1769, 1981, 2205, 2441, 2689, 2949, 3221, 3505, 3801, 4109, 4429, 4761, 5105, 5461, 5829, 6209, 6601, 7005, 7421, 7849, 8289, 8741, 9205, 9681, 10169, 10669, 11181, 11705, 12241

Offset: 1

Gary W. Adamson, Dec 28 2007

Binomial transform of [1, 8, 12, 0, 0, 0, ...].
Numbers k such that 6*k - 5 is the square of a number of the form 6*k - 5, contained in A199859. - Eleonora Echeverri-Toro, Nov 29 2011
Central terms of the triangle A033292. - Reinhard Zumkeller, Feb 06 2012
Sequence found by reading the line from 1, in the direction 1, 9, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Jul 18 2012

Kelvin Voskuijl, Table of n, a(n) for n = 1..10000
John Elias, Illustration: Centered nesting cube frames.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A001318, A016777, A033292, A033580, A059722, A199859, A201279, A204675.

a136392 n = 2 * n * (3*n - 5) + 5 -- Reinhard Zumkeller, Feb 06 2012

Table[6n^2-10n+5,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,9,29},50] (* Harvey P. Dale, Mar 05 2023 *)

a(n)=6*n^2-10*n+5 \\ Charles R Greathouse IV, Nov 29 2011

a(n) = n*(3*n - 2) + (n-1)*(3*n - 5), n > 1.
a(n) = n*A016777(n-1) + (n-1)*A016777(n-2).
a(n) = a(n-1) + 12*n - 16 (with a(1)=1). - Vincenzo Librandi, Nov 24 2010
G.f.: x*(1+x)*(1+5*x)/(1-x)^3. - Colin Barker, Jan 09 2012
a(n) = 1 + A033580(n-1). - Omar E. Pol, Jul 18 2012
a(n) = A059722(n) - A059722(n-1). - J. M. Bergot, Nov 02 2012
a(n) = A000567(n-1) + A000567(n). - Charlie Marion, May 29 2024
From Elmo R. Oliveira, Oct 31 2024: (Start)
E.g.f.: exp(x)*(2*x*(3*x - 2) + 5) - 5.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A319384 a(n) = a(n-1) + 2a(n-2) - 2a(n-3) - a(n-4) + a(n-5), a(0)=1, a(1)=5, a(2)=9, a(3)=21, a(4)=29.

Original entry on oeis.org

1, 5, 9, 21, 29, 49, 61, 89, 105, 141, 161, 205, 229, 281, 309, 369, 401, 469, 505, 581, 621, 705, 749, 841, 889, 989, 1041, 1149, 1205, 1321, 1381, 1505, 1569, 1701, 1769, 1909, 1981, 2129, 2205, 2361, 2441, 2605, 2689, 2861, 2949, 3129, 3221, 3409, 3505, 3701, 3801, 4005, 4109, 4321, 4429, 4649, 4761, 4989, 5105, 5341, 5461

Offset: 0

Paul Curtz, Sep 18 2018

The two bisections A136392(n+1)=1,9,29,61, ... and A201279(n)=5,21,49, ... are in the hexagonal spiral based on 2*n+1:
.
                    67--65--63--61
                    /             \
                  69  33--31--29  59
                  /   /         \   \
                71  35  11---9  27  57
                /   /   /     \   \   \
              73  37  13   1   7  25  55
                  /   /   /   /   /   /
                39  15   3---5  23  53
                  \   \         /   /
                  41  17--19--21  51
                    \             /
                    43--45--47--49
.
A201279(n) - A136892(n) = 20*n.

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

Cf. A001844, A032766, A104585.
In the spiral: A003154(n+1), A080859, A126587, A136392, A201279, A227776.
Partial sums of A382154.

[(6*n^2 + 6*n + 5 - (2*n + 1)*(-1)^n)/4 : n in [0..50]]; // Wesley Ivan Hurt, Jan 19 2021

Table[(6 n^2 + 6 n + 5 - (2 n + 1)*(-1)^n)/4, {n, 0, 80}] (* Wesley Ivan Hurt, Jan 07 2021 *)

Vec((1 + x^2)*(1 + 4*x + x^2) / ((1 - x)^3*(1 + x)^2) + O(x^50)) \\ Colin Barker, Jun 05 2019

def A319384(n): return (n*(3*n+4)+3 if n&1 else n*(3*n+2)+2)>>1 # Chai Wah Wu, Mar 25 2025

a(2*n) = A136392(n+1), a(2*n+1) = A201279(n).
a(-n) = a(n).
a(2*n) + a(2*n+1) = 6*A001844(n).
a(n) = (6*n^2 + 6*n + 5 - (2*n + 1)*(-1)^n)/4. - Wesley Ivan Hurt, Oct 04 2018
G.f.: (1 + x^2)*(1 + 4*x + x^2) / ((1 - x)^3*(1 + x)^2). - Colin Barker, Jun 05 2019
a(n) = A104585(n) + A032766(n+1). - Alex W. Nowak, Jan 08 2021

More terms from N. J. A. Sloane, Mar 23 2025

A342138 Array T(n,k) = (n+k)(3n+3k-5)/2 + (3k+1), read by ascending antidiagonals.

Original entry on oeis.org

1, 0, 3, 2, 5, 8, 7, 10, 13, 16, 15, 18, 21, 24, 27, 26, 29, 32, 35, 38, 41, 40, 43, 46, 49, 52, 55, 58, 57, 60, 63, 66, 69, 72, 75, 78, 77, 80, 83, 86, 89, 92, 95, 98, 101, 100, 103, 106, 109, 112, 115, 118, 121, 124, 127, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156

Offset: 0

Michel Marcus, Mar 01 2021

This is an instance of a storing function on N^2 (injective) with density 1/3.

			Array begins:
   1  3   8  16  27 ...
   0  5  13  24  38 ...
   2 10  21  35  52 ...
   7 18  32  49  69 ...
  15 29  46  66  89 ...
  ...

John S. Lew and Arnold L. Rosenberg, Polynomial indexing of integer lattice-points I. General concepts and quadratic polynomials, Journal of Number Theory, 10(2):215-243, 1978. See Corollary 5.2 p. 211.

Cf. A005449 (first column), A104249 (first row), A140090 (second row), A201279 (diagonal).

T(n,k) = (n+k)*(3*n+3*k-5)/2 + (3*k+1);
matrix(8, 8, n, k, T(n-1, k-1))

cp's OEIS Frontend

A048395 Sum of consecutive nonsquares.

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A136392 a(n) = 6*n^2 - 10*n + 5.

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A319384 a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), a(0)=1, a(1)=5, a(2)=9, a(3)=21, a(4)=29.

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A342138 Array T(n,k) = (n+k)*(3*n+3*k-5)/2 + (3*k+1), read by ascending antidiagonals.

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PARI

A136392 a(n) = 6n^2 - 10n + 5.

A319384 a(n) = a(n-1) + 2a(n-2) - 2a(n-3) - a(n-4) + a(n-5), a(0)=1, a(1)=5, a(2)=9, a(3)=21, a(4)=29.

A342138 Array T(n,k) = (n+k)(3n+3k-5)/2 + (3k+1), read by ascending antidiagonals.